11 ms·

Solving the cube root of 19,683 mentally

134 points by nigamanth 4y ago

kqr 4y ago
Slightly related musing: I can quickly approximate 1/3 of any number. I suspect this is because that's an operation I've had to use often. If I had needed cube roots equally often, would I have been just as good at mentally intuition roughly what number it has to be? Or are humans somehow wired to deal with linearities more intuitively than powers? Or is it our mathematical notation (Arabic numbers are linear, as opposed to some log scale) that creates that impression?
- 8192kjshad09- 4y ago
  There's some research indicating the opposite, that people intuitively think in logarithms. We perceive sound, light, and possibly even the passage of time logarithmicly[1] I don't think there has been an explanation proposed for why it's easier to do linear math in our heads, I have to imagine this is just the effect of school. [1]: https://www.nature.com/articles/s41598-020-75191-6 https://www.nature.com/articles/s41598-020-75191-6
  - anon___ 4y ago
    Everything you listed, and the linked article is about perception, not thinking. Paradoxically, our thinking is linear, and it seems to be useful to change that: https://hbr.org/2017/05/linear-thinking-in-a-nonlinear-world https://hbr.org/2017/05/linear-thinking-in-a-nonlinear-world.
- hgsgm 4y ago
  Approximate how precisely? What is 17463/3 ? To 2 sig figs, it's trivial because you only have to divide a digit by 3 and then interpolate the fractional part.
  - kqr 4y ago
    Exactly. So how come it's not trivial with the cube root? On your description it sounds like the numerals we use are the problem.
meling 4y ago
There is a typo in the second example: 2. Find the cube root of 653,503 It should be 658,503…
- scottlu2 4y ago
  Thank you. I was wondering that.
adastra22 4y ago
Only works if you know the number is a cube root though. I doesn’t help you find the “nearest” cube root.
- xracy 4y ago
  Yeah, I was hoping it was more of the latter trick, than the former.
bumbledraven 4y ago
We want to find 19683¹ᐟ³ mentally. First, note that, since the digit sum of 19683 is a multiple of 9, 19683 itself is a multiple of 9. Second, since 19683 is odd, its cube root will be odd as well. Therefore, the cube root of 19683 will be an odd multiple of 3. To find 19683¹ᐟ³, will take the base 10 log of 19683, divide by 3, and raise the result to the 10th power, approximating where convenient. To this end, it will be helpful to have approximate values for lg 2 and lg 5. Since 10³ ≈ 1024 = 2¹⁰ = 10^(lg[2]·10), we have 3 ≈ lg[2]·10. Solving, we find that lg 2 ≈ 0.3. Next, since 10 = 2·5, we have lg[10] = 1 = lg[2·5] = lg 2 + lg 5, so lg 5 = 1 − lg 2 ≈ 0.7. Using those approximations, we calculate: lg[19683¹ᐟ³] ≈ lg[20000¹ᐟ³] = lg[2·10000]/3 = (lg 2 + lg 10000)/3 ≈ (0.3 + 4)/3 ≈ 1.43. Since 25 = 5·5, we have lg 25 = lg 5 + lg 5 ≈ 1.4. This means we're looking for an odd multiple of 3 near 25. The closest number that fits the bill is 27. Cubing, we find that 27·27·27 = 19683, so we are done.
dmurray 4y ago
How to do this without relying on the fact that the number is a perfect cube: Learn a handful of logs base 10 (not much more work than the cube digit laws). Maybe you know 2,3,5 and e. 19,683 is equal to 20,000 log 20,000 is log 2 + log 10 + log 10 + log 10 + log 10 That's 0.30 + 1 + 1 + 1 + 1 which is 4.30 Divide by 3 (equivalent to taking cube root). That's 1.43 Recall that log e is 0.43. So 1.43 is log (e * 10). Hopefully you remember e = 2.71, so the number you're looking for is 27.1. If you remember that you rounded up to start, you can round back down and guess that it's more like 27.0. You can simplify this a little further by dividing out thousands and multiplying by 10s at the end, so you only need to find the cube root of 20 using the same process.
- colinbeveridge 4y ago
  I did it similarly: log(19,683) is about 4.3, so ln(19,683) is about 4.3 * 2.3, which is about 9.9. That jumps out as close to 9ln(3). So the natural log of the cube root is (presumably) 3 ln(3), making the cube root itself 27. Alternatively, the digit sum of 19,683 is 27, so the cube is a multiple of 9. Given that it's a perfect cube, 19,683 must be a multiple of 3^3 = 27. Dividing by that gives 729, or 9^3, so the cube root is 3 * 9=27. Lastly, it's possible to dredge up that e^3 is about 20, so (27.1828...)^3 is about 20,000, and it's plausible that 27 is the cube root we're after.
  - cperciva 4y ago
    My approach: ln(19683) ~ ln(20000) = ln(20) + 3 ln(10) ~ 3 + 3*2.3 = 9.9. 3.3 = 1.0 + 2.3 ~ ln(e) + ln(10). My mental landmarks are ln(2), ln(e), and ln(10), roughly 0.7, 1.0, and 2.3 respectively.
    - colinbeveridge 4y ago
      Nice. I like ln(3) and ln(5) (1.1 and 1.6) as well.
      - cperciva 4y ago
knaik94 4y ago
I know this more of a party trick than a practical skill, but I found it's tricks like this that gets students I tutor a little more interested in math. Knowing these tricks doesn't make you better at math, but understanding the underlying principle of focusing on significant/meaningful digits absolutely will.
- nigamanth 4y ago
  For students in higher grades this may very well pay off. Especially in MCQ
  - knaik94 4y ago
    If accuracy is important, students would be allowed to use a calculator. There is no longer practical value of being able to do this without one. So there's no point in testing it in school. Any teacher asking this kind of arithmetic question is wasting the time and energy of their students. I have never come across it in my last 10 years of tutoring high school or college math. If the abstract understanding of root properties is being asked, the question will be presented using variables. The cube root function is effectively equivalent to the square root function execpt the range includes negative values, when looking at it graphically.
    - dahart 4y ago
      Are we talking about the practical value of actively learning math, or the practical value for an average high school student who barely tolerates math? Sure, yes, maybe most people can’t be bothered in practice, but isn’t there a lot of practical value in learning number theoretic ideas and seeing relationships between digits, for example, for the person bound for a job or graduate degree in a STEM field? Seems practical for people interested in math history as well, no? I’m a fan of some of the math problems on Project Euler. This isn’t high school math, of course, but many problems are carefully designed such that blindly using a computer to solve them just doesn’t work. This forces you to think about the problem more deeply and use that to come up with tricks that reduce the search space. This is very much along the same lines as this blog post, but generalized to bigger harder problems. And FWIW, stuff like this - a deep interest in math and how there are often many different ways to solve a problem, and the study of digits and numbers and polynomials and roots and efficient ways to manipulate them - is quite core to my job and all the people I work with. I do mostly software at a company that makes hardware and software, and number-theoretic thinking and fun & “impractical” tricks put into practice are literally everywhere. I guess I’m saying I think that there actually is a lot of practical value in this kind of play, for the engaged audience, and once someone goes beyond basic high school & college math, in addition to the benefits of increased motivation that you mentioned. I wouldn’t be so quick to write it off as pointless, and only teach to the test. Maybe if we had more of this in high school, we’d have a better math education system. I’d agree we need to teach all kids how to use calculators, but I also think we need to teach kids how to think deeply about numbers, and have the skills to build tricks and do mental estimation, and send more of them into science and tech jobs with exploration skills emphasized and prioritized over calculator skills.
detrites 4y ago
I guessed it first try from the post title, based on: - it ends in 3, so the cube root must be a multiple of 3 - 30*3 is intuitively going to be 27,000, and 20*3 will be 8,000 - that leaves only 21, 24 and 27 to choose from, and given the exponential shape for 8k - 27k, I went with the largest Not a "solution", as I couldn't be certain without confirming, but interesting how close a logical deduction could get.
- ykonstant 4y ago
  One correction: there are many numbers that end in 3 but are not multiples of 3. It is the fact that its digit sum is divisible by 3 that lets us conclude this number itself is so. Perhaps that is what you meant?
  - deleted 4y ago
    [deleted]
  - detrites 4y ago
    Yes, as stated: it ends in 3, so the cube root must be a multiple of 3. The "cube root" part contains the implicit information that we have a product ending in 3, of three identical factors. I guess the missing assumption is that given the context of solving the cube root in ones head, it would more likely be a trick that only works with integers, and as such the correct answer might be expected to be one.
    - dmurray 4y ago
      > Yes, as stated: it ends in 3, so the cube root must be a multiple of 3. I don't get it. 7 cubed is 343. The article explains better what numbers might have cubes ending in 3.
      - detrites 4y ago
        Yes, the article is great (now I've read it). 37*3 also ends in 3 but is not divisible by 3. Thus, were the range not capped at 30 it may have been illegitimately discarded by the criteria. I think the 3 combined with the 5 digits and 20-30 range, made my guess a bit lucky. It's also contrived for that cube, so not useful for others, other than perhaps as a direction to go in (and likely arrive at the article).
Simon_O_Rourke 4y ago
Super neat!
deleted 4y ago
[deleted]
kazinator 4y ago
It should be close to cbrt(20000) = cbrt(20) cbrt(1000) = cbrt(20) * 10 = 2.x * 10 = 2x. The job is to find x. Assuming 19683 has an integer cube root, I would guess the last digit x to be 7, because that gives us the 3: 7 * 7 = 49, and 9 * 7 = 63. Thus, the answer is probably 27. Cubing 27 mentally to validate the answer is the harder thing for me. :)
- hgsgm 4y ago
  This is a summary of the article.
thrdbndndn 4y ago
https://math.stackexchange.com/questions/1142767/why-does-this-cube-root-trick-work https://math.stackexchange.com/questions/1142767/why-does-th... Also, I have an additional question. Obviously this trick relies on the fact that "the last digit of the cube of all numbers from 0 to 9 is distinct". Is this fact just coincidence, or is it bound to happen?
- rk06 4y ago
  0^3 = 0 1^3 = 1 2^3 = 8 3^3 = 27 4^3 = 64 5^3 = 125 6^3 = 216 7^3 = 343 8^3 = 512 9^3 = 729 As you can see, for decimal system, last digit of cubes is unique for all digits. As last digit of cube depends on last digit of cube root only (same for squares and so on), it is bound to be true for all numbers
  - thrdbndndn 4y ago
    You misunderstand my question. I was asking if this (cubes of number 0-9, the one you list, all have unique last digit in 0-9) a coincidence or not. (I know a number's cube's last digit is only affected by its own last digit.)
    - archi42 4y ago
      The question is in which case(s) `f(x) = x^n mod m` is a bijection for `0 <= x < m`. My intuition is to guess that `gcd(n,m) = 1` is at least a sufficient condition (correct math term? In german it's "hinreichende Bedingung"). Maybe it's the sole necessary condition. But I did only think about this for 5 to 10s and did not spend too much time on looking for a counter example, let alone a proof strategy (my number theory is a bit rusty -- ah, but on second thought, if the theory is correct, then I bet the Chinese remainder theorem is involved).
      - schoen 4y ago
        I asked the same question in this thread and got a couple of detailed answers: https://news.ycombinator.com/item?id=34652909 https://news.ycombinator.com/item?id=34652909
schoen 4y ago
I was surprised by the chart showing that there is a bijection between last digit of a number and last digit of the number's cube. In base 10, this works for every odd power but for no even power. (if you're given that a square ends in anything but 0 or 5, there are always two different possibilities for the last digit of the square root) Here are the powers (less than 100) for which this works in different bases (up to 25). Now I'm quite confused: what is the pattern here and what is the number-theoretic explanation for it? (Edit: It looks like the bases for which no power, except the trivial case of 1, has this property are exactly those bases that are divisible by a perfect square -- but I wouldn't have expected that and still don't know how to account for it.) 2 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99] 3 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 4 [1] 5 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 6 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 7 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 8 [1] 9 [1] 10 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 11 [1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99] 12 [1] 13 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 14 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 15 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 16 [1] 17 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 18 [1] 19 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 20 [1] 21 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 22 [1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99] 23 [1, 3, 5, 7, 9, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97] 24 [1] 25 [1]
Zacharias030 4y ago
Followup question, how to estimate/calculate compound interest in your head quickly? (Any direction, solving for the rate, or for the time, or the principal needed)
- nigamanth 4y ago
  Applying the formula isn't that hard, but this one could be hard to write about because it depends upon the amount of time the sum is compounding or the exponent in the equation. ^2 and ^3 are quite easy but when it comes to ^7/^9 there's no easy way to calculate mentally.
- bberrry 4y ago
  There is a method called the rule of 72. You can look it up but it's essentially: Years to Double = 72 / Annual Rate of Return
  - nigamanth 4y ago
    Still cannot be done mentally.
deleted 4y ago
[deleted]
red_Seashell_32 4y ago
offtopic, but your website is hard to read on mobile - width of text makes up less than 33% of screen, and whitespace on sides is too wide :/ (though unsure how many of your traffic primarly reads article on mobile, so it might not even be an impactful issue)
xchip 4y ago
Note that where the article says 653,503 it should be 658,503
bohadi 4y ago
My solution was similar, represent. First we observe x is between 20^3=8e3 and 30^3=27e3. Then note that the least significant (singles) digit of x^3 is solely determined by the singles digit of x [only holds in the integers]. Luckily we had cached that the pattern for cubes is 123456789 -> 187456329 Finally x^3 = 19,683 ends in 3 therefore x = 27 we are done. ------ It turns out there is a unique mapping for the first digit of x^n, following one of four patterns: n mod 4 = 1 123456789 -> 123456789 n mod 4 = 2 -> 149656941 n mod 4 = 3 -> 187456329 n mod 4 = 0 -> 161656161 So this trick would not always be as useful, where the mapping is not surjective for squares, fourth powers -- all the even powers. I imagine there is a nice visual geometric explanation why those mappings take their symmetric pattern. You may ask yourself, is there also a rule for the larger place digits of x^3, x^n? Yes, we naturally observe that the 10s digit of x^n is similarly wholly determined by the 10s and 1s digits of x. So now we want n maps on [0,9]x[0,9] to [0,9]. The python snippet may help to explore this: where the row 3, 25th digit shows that when x ends in 25, x cubed has 10s digit 2 >>> print(*zip(range(50), [''.join([str(x**i).zfill(2)[-2:-1] for x in range(100)]) for i in range(50)]), sep='\n') (0, '0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000') (1, '0000000000111111111122222222223333333333444444444455555555556666666666777777777788888888889999999999') (2, '0000123468024692582604827272840628529642086432100000001234680246925826048272728406285296420864321000') (3, '0002621412032947913506462278580963075571028082329405076719170824429630014127735304680250760785873799') (4, '0018529096043612327208547274580272321634069092581000185290960436123272085472745802723216340690925810') (5, '0034227064053927756900342270640539277569003422706405392775690034227064053927756900342270640539277569') (6, '0062925444068032162802087278020826123086044452926000629254440680321628020872780208261230860444529260') (7, '0028823456070107573304842270100167479397084062167405238739510206025238098927751506624298920345671179') ... (19, '0086429478094787353708082276960769671755062002581405814799730442823032030327719102646212500125075319') (20, '0070727070007072707000707270700070727070007072707000707270700070727070007072707000707270700070727070') (21, '0050025000011167116102722272220333873383049442944405550755050616621666077727772708388238880999479949') (22, '0000123468024692582604827272840628529642086432100000001234680246925826048272728406285296420864321000') (23, '0002621412032947913506462278580963075571028082329405076719170824429630014127735304680250760785873799') (24, '0018529096043612327208547274580272321634069092581000185290960436123272085472745802723216340690925810') ... (35, '0060227434056527793900602274340565277939006022743405652779390060227434056527793900602274340565277939') (36, '0032925054061432187802967276920878123416045052923000329250540614321878029672769208781234160450529230') (37, '0076823046079307576304102274800137479107085462182405718735410298025268091527798506324296020359671329') (38, '0048321482080252902406667276660420925208028412384000483214820802529024066672766604209252080284123840') (39, '0086429478094787353708082276960769671755062002581405814799730442823032030327719102646212500125075319') (40, '0070727070007072707000707270700070727070007072707000707270700070727070007072707000707270700070727070') (41, '0050025000011167116102722272220333873383049442944405550755050616621666077727772708388238880999479949') (42, '0000123468024692582604827272840628529642086432100000001234680246925826048272728406285296420864321000') (43, '0002621412032947913506462278580963075571028082329405076719170824429630014127735304680250760785873799') ... ----- Apologies for the wall of text but the output is included here to show the periodicity is 20 in the 10s digit for powers of n. What do you think the periodicity in the 100s digit is? Okay let's use our cache to just lookup the least two digits of say, 41679296 ^ 35 (without doing any real compute): 1s digit 35 mod 4 = 3 187456329 index 6 -> 6 10s digit (35, '0060227434056527793900602274340565277939006022743405652779390060227434056527793900602274340565277939') index ------96 -> 7 The last two digits of 41679296 ^ 35 = 497981393104388234083687472206165994017494365065741791800324926291638211001469403054062791593737655590988883756006534609416313532750908441977019034732014717649617778735038244085515193249673259731575280032440237649262981884959117005449913087392615145373869733708824576 are indeed 76, cool
_Microft 4y ago
My method is certainly not as sophisticated as others here but since it was different from these others, I will post it as well. The solution should obviously be somewhere between 20 and 30. Since I knew 25^2 = 625 already, I multiplied by 25 to arrive at 25^3 = 15625. From there I used the approximation (1 + x)^n = 1 + nx which works if x << 1. 19683 = 15625*(1 + x) = 25^3*(1 + x) with x ~= 4000/15625 ~= 0.25. This stretches the condition ("x << 1") for the approximation a lot but so be it. (1 + 1/4)^1/3 ~= 1 + 1/12. 25*(1 + 1/12) ~= 27 Good enough ;)
atemerev 4y ago
I learned this method at the age of 7, and was playing a math savant for a few weeks.
timwaagh 4y ago
Interesting to try 140.something is what I'm guessing. Edit: Ah wait cube root. That is the square root I just tried. 26.something is what my mental calculations gave me. I thought it was going to be 27. Turns out it is. I made an error somewhere which is why you gotta do this on paper. As for method I kind of brie brute forced it. started with 100 cubed which is hundred thousand then 20 cubed which is 8000 then 30 cubed which is 3 cubed times 10 cubed which is 27k. 25 cubed which I don't remember but was less then, then I knew it could not be 26 because the number is not even. I'd deal with fractions later. so I tried 27 cubed. 27 cubed is 3^3^3=2733^5=8133^4=24333^3= 72927=72920+7297 72920=72902=14400+180=14580 7297=4900+140+63=5103 Add them and 19683 comes out Now try that in your head and it's a good recipe for frying one's brain.
- sp332 4y ago
  Much too large. 100 cubed is 1,000,000.
- colinbeveridge 4y ago
  For the square root, I get 140.3. My heuristic is to find a nearby square (here, 140^2 = 19600) and divide the discrepancy (83) by double the root of the nearby square (280). This is the number to adjust by. [^0] [^0] The Maclaurin series expansion of (A^2 + x)^(1/2) starts A + x/(2A), which is where this comes from. 83/280 is close to 84/280, which is 12/40 or 0.3. If I was trying to show off [^1], I might adjust that -- 1/28 is a quarter of a seventh, about 0.0357, so I can take 0.004 off of my adjustment to get 0.296 or 0.297. [^1] I mean, show off more. (The actual answer is 140.29611 -- the "0.3" adjustment is right to one part in 36,000; "0.296" is right to one part in 1.2 million.)
madcaptenor 4y ago
I just recognized that it’s 3^9, but that’s dumb luck on my part.
- canadianfella 4y ago
  [dead]
cdelsolar 4y ago
I got 27 in about 10 seconds by doing a quick binary search and noticing the number ended in 3 (so it’s gotta end in 7). It’s in between 20 (8000) and 30 (27000).
sbaiddn 4y ago
For those of us who have tried but failed to memorize our logs (obviously the best method to ballpark it) : For the n-the square root just group the digits by n starting from the ones. That will tell you how many digits you're dealing with. Simplish math will give you the first digit off of the last grouping (simple since you only have to ^n the numbers from 1-10) Simple counting and a bit of math and I have the order of mag and first digit. 2_ At this point I can either think of a decent trick to iterate the second digit (quick Taylor? [1]), or just guesstimate a linear fit and call it a 5. 25 is within 10%. [1] instead of Taylor I would do Pascal's triangle on the cube with the next term to calculate (30-x)^3 (I use 30 - x and not 20+x since I want x as small as possible and I know the answer is closer to 30 than 20. Pascal's triangle: 1(30)^3 + 3(30)^2x + 330x^2 + 1x^3 Drop the higher two terms. 3^3 + 33^2x = 19 19-27 = 27x X = 8/27 = 1/3 (- 1/27) 30-3.3 = 26.66 I use the sign of the next term of pascal's triangle to see if Ive under or over estimated 33*(-0.3)^2 -> positive So my x is too large So my answer is an underestimate. (Cube root of 19000) greater than but close to 26.6
graderjs 4y ago
Guessing: 1000 is 10*10*10 this is close to 20K, so we have 10, now we need cube of 20. Between 2 and 3, take 2.5 so 25? No... it's 27.
_steady 4y ago
Loads of chat about remembering logs etc, I was expecting to see more number theory here as it’s a really lovely and applicable part of maths. (As far as doing the calc, I thought it has to end in 7 becaus3 it’s the only one with a cub3 residue of 3 mod 10, which is what they say in post. Then 20 and 30 cubed are 8000 and 27000 respectively and assuming we’re restricting to Z 27 is the only possible answer)
less_less 4y ago
If you know that the number is a perfect cube, you could also try Hensel's lemma or CRT, or Euler, all of which are aaaaalmost doable in your head for 2 digits, but maybe not quite. Suppose you are trying to find cbrt(x), and you know the answer is a 2-digit integer. =============== Hensel: first you have to remove any powers of 5 and 2. You know the cube root is even if the number is even, and a multiple of 5 if the number is a multiple of 5 (ends in a 0 or 5). So factor those out first. It now ends in 1,3,7 or 9. Find the cube root w of x mod 10, and t = 1/3x mod 10, by trial and error. Eg, x=19683 ends in 3, so w ends in 7 (since 7^3 ends in 3); t = 1/3x ends in 9 since 39x ends in 1. Calculate the tens digit of the error term e = x - w^3, which is always a multiple of 10. In this case it's ...683 - 343 === 40, and apply the adjustment w ==> w + wte. You only need to calculate the 10s digit of wte. In this case, it's the 10's digit of 7*9*40 = 63*40 is 20. That's all, the answer is 27. =============== CRT: Find the answer mod 10 and mod 11. As before, the answer here mod 10 is 7. For mod 11, you can reduce it to a 2-digit problem by applying %99 first, i.e. shifting each digit over by 2 until it's a 10s or 1s digit, and adding. We have 19683 mod 11 = (1 + 96 + 83) mod 11 = 180 mod 11 = 81 mod 11 = 4. Find the cube root mod 11 by exhaustion, or by memorization, or by taking the 7th power of the input (Fermat's theorem). In this case, it's 5, because 5^3 mod 11 = 125 mod 11 =(1+25) mod 11 = 4. Solve using the CRT: if a number is y mod 10 and z mod 11, then it's y + 10(y-z mod 11). In this case, it's 7 + 10(7-5 mod 11) = 27. ================ Euler: the answer is always the same as the last 2 digits of x^7. Calculate eg x^2, x^3, x^6, x^7 with four multiplications. You only need the last two digits. Probably this is the easiest approach by hand. ================ If you want to solve on a computer, then Hensel mod 2^n (continuing the pattern appropriately for more bits) is probably the fastest bet.
- nigamanth 4y ago
  > all of which are aaaaalmost doable in your head for 2 digits, but maybe not quite. You might be good at Mathematics, this is like a standard trick the average student can do. There are lots of other approaches but the easiest one to do mentally is this.
croes 4y ago
>Notice how, the cubes of 1, 4, 5, 6, and 9 always end with the number itself. The cube of 3 ends with 7 and cube of 7 end with 3 The cube of 2 ends with 8 and cube of 8 end with 2 Cube-buddies
deleted 4y ago
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jameshart 4y ago
Another approach is to recognize that if you know a number is a perfect cube, and you can recognize any prime factors it contains, you can pull out the cube of those from the number. Like, by inspection (sum of digits) this number has 3 as a factor. That means 3 has to also be a prime factor of its cube root which means 27 has to be a factor of the cube. So we can divide out 27, leaving us with the smaller problem of finding 3(cbrt(729)), which you might recognize as being 3*9, or spot that you can repeat the trick to give you 9(cbrt(27))… Or you might notice that if 19683 has to contain 27 as a factor, that doesn’t leave a lot of room for any other cubed prime factors to exist in the number, and do a trial multiplication to confirm that 27 just is the cube root. Basically comes down to, if you have a comfortable relationship with how numbers relate to their factors, you can develop a number of instincts for how to recognize shapes within them.
chasing 4y ago
20 cubed is obviously too low. 30 cubed too high. 7 is the only number that, when cubed, ends in 3. 27 it is.
- camgunz 4y ago
  Yeah this is what I do; what number ^3 gives you a 3: it's 7. Bracket it in the 10s place, try it out.
- yebyen 4y ago
  I went with "28 sounds about right" then saw it was clearly just a bit too high.
- rich_sasha 4y ago
  30 cubed is 27,000 or about 35% too much. Thus we solution is about 10% smaller than 30. 27 gets you there.
Gravityloss 4y ago
My mental process: Well it's less than 100 since 100^2 = 10,000 so that goes over. It's over 10 since 10^3=1000. So this narrows it to more than 10 and less than 100. log-wise 30 is about midway between 10 and 100. 900 x 30 is is 27000 so it's less than 30. 20? 400 x 20 = 8000 so bigger than that. Then it gets annoying to compute the cubes so close enough to say more than 20 and less than 30.
dsr_ 4y ago
In my head: oh, look, a multiple of 7. What's 30 cubed? 27 and three zeros, so not far away. What's 27 cubed? Oh, that's it. One of the things that you should learn is that the problems that humans set for each other usually involve integers or well known numbers.
_a_a_a_ 4y ago
27, but I only got that because it was explicitly given that 19,683 was a cube.
omnster 4y ago
Reminded me of the story about Feynman being challenged by an abacist to compute cubic root https://www.ecb.torontomu.ca/~elf/abacus/feynman.html https://www.ecb.torontomu.ca/~elf/abacus/feynman.html

134 points by nigamanth 4y ago

kqr 4y ago

Slightly related musing: I can quickly approximate 1/3 of any number. I suspect this is because that's an operation I've had to use often. If I had needed cube roots equally often, would I have been just as good at mentally intuition roughly what number it has to be? Or are humans somehow wired to deal with linearities more intuitively than powers? Or is it our mathematical notation (Arabic numbers are linear, as opposed to some log scale) that creates that impression?

8192kjshad09- 4y ago
There's some research indicating the opposite, that people intuitively think in logarithms. We perceive sound, light, and possibly even the passage of time logarithmicly[1] I don't think there has been an explanation proposed for why it's easier to do linear math in our heads, I have to imagine this is just the effect of school. [1]: https://www.nature.com/articles/s41598-020-75191-6 https://www.nature.com/articles/s41598-020-75191-6
- anon___ 4y ago
  Everything you listed, and the linked article is about perception, not thinking. Paradoxically, our thinking is linear, and it seems to be useful to change that: https://hbr.org/2017/05/linear-thinking-in-a-nonlinear-world https://hbr.org/2017/05/linear-thinking-in-a-nonlinear-world.
hgsgm 4y ago
Approximate how precisely? What is 17463/3 ? To 2 sig figs, it's trivial because you only have to divide a digit by 3 and then interpolate the fractional part.
- kqr 4y ago
  Exactly. So how come it's not trivial with the cube root? On your description it sounds like the numerals we use are the problem.

meling 4y ago

There is a typo in the second example: 2. Find the cube root of 653,503 It should be 658,503…

scottlu2 4y ago
Thank you. I was wondering that.

adastra22 4y ago

Only works if you know the number is a cube root though. I doesn’t help you find the “nearest” cube root.

xracy 4y ago
Yeah, I was hoping it was more of the latter trick, than the former.

bumbledraven 4y ago

We want to find 19683¹ᐟ³ mentally. First, note that, since the digit sum of 19683 is a multiple of 9, 19683 itself is a multiple of 9. Second, since 19683 is odd, its cube root will be odd as well. Therefore, the cube root of 19683 will be an odd multiple of 3. To find 19683¹ᐟ³, will take the base 10 log of 19683, divide by 3, and raise the result to the 10th power, approximating where convenient. To this end, it will be helpful to have approximate values for lg 2 and lg 5. Since 10³ ≈ 1024 = 2¹⁰ = 10^(lg[2]·10), we have 3 ≈ lg[2]·10. Solving, we find that lg 2 ≈ 0.3. Next, since 10 = 2·5, we have lg[10] = 1 = lg[2·5] = lg 2 + lg 5, so lg 5 = 1 − lg 2 ≈ 0.7. Using those approximations, we calculate: lg[19683¹ᐟ³] ≈ lg[20000¹ᐟ³] = lg[2·10000]/3 = (lg 2 + lg 10000)/3 ≈ (0.3 + 4)/3 ≈ 1.43. Since 25 = 5·5, we have lg 25 = lg 5 + lg 5 ≈ 1.4. This means we're looking for an odd multiple of 3 near 25. The closest number that fits the bill is 27. Cubing, we find that 27·27·27 = 19683, so we are done.

dmurray 4y ago

How to do this without relying on the fact that the number is a perfect cube: Learn a handful of logs base 10 (not much more work than the cube digit laws). Maybe you know 2,3,5 and e. 19,683 is equal to 20,000 log 20,000 is log 2 + log 10 + log 10 + log 10 + log 10 That's 0.30 + 1 + 1 + 1 + 1 which is 4.30 Divide by 3 (equivalent to taking cube root). That's 1.43 Recall that log e is 0.43. So 1.43 is log (e * 10). Hopefully you remember e = 2.71, so the number you're looking for is 27.1. If you remember that you rounded up to start, you can round back down and guess that it's more like 27.0. You can simplify this a little further by dividing out thousands and multiplying by 10s at the end, so you only need to find the cube root of 20 using the same process.

colinbeveridge 4y ago
I did it similarly: log(19,683) is about 4.3, so ln(19,683) is about 4.3 * 2.3, which is about 9.9. That jumps out as close to 9ln(3). So the natural log of the cube root is (presumably) 3 ln(3), making the cube root itself 27. Alternatively, the digit sum of 19,683 is 27, so the cube is a multiple of 9. Given that it's a perfect cube, 19,683 must be a multiple of 3^3 = 27. Dividing by that gives 729, or 9^3, so the cube root is 3 * 9=27. Lastly, it's possible to dredge up that e^3 is about 20, so (27.1828...)^3 is about 20,000, and it's plausible that 27 is the cube root we're after.
- cperciva 4y ago
  My approach: ln(19683) ~ ln(20000) = ln(20) + 3 ln(10) ~ 3 + 3*2.3 = 9.9. 3.3 = 1.0 + 2.3 ~ ln(e) + ln(10). My mental landmarks are ln(2), ln(e), and ln(10), roughly 0.7, 1.0, and 2.3 respectively.
  - colinbeveridge 4y ago
    Nice. I like ln(3) and ln(5) (1.1 and 1.6) as well.
    - cperciva 4y ago

knaik94 4y ago

I know this more of a party trick than a practical skill, but I found it's tricks like this that gets students I tutor a little more interested in math. Knowing these tricks doesn't make you better at math, but understanding the underlying principle of focusing on significant/meaningful digits absolutely will.

nigamanth 4y ago
For students in higher grades this may very well pay off. Especially in MCQ
- knaik94 4y ago
  If accuracy is important, students would be allowed to use a calculator. There is no longer practical value of being able to do this without one. So there's no point in testing it in school. Any teacher asking this kind of arithmetic question is wasting the time and energy of their students. I have never come across it in my last 10 years of tutoring high school or college math. If the abstract understanding of root properties is being asked, the question will be presented using variables. The cube root function is effectively equivalent to the square root function execpt the range includes negative values, when looking at it graphically.
  - dahart 4y ago
    Are we talking about the practical value of actively learning math, or the practical value for an average high school student who barely tolerates math? Sure, yes, maybe most people can’t be bothered in practice, but isn’t there a lot of practical value in learning number theoretic ideas and seeing relationships between digits, for example, for the person bound for a job or graduate degree in a STEM field? Seems practical for people interested in math history as well, no? I’m a fan of some of the math problems on Project Euler. This isn’t high school math, of course, but many problems are carefully designed such that blindly using a computer to solve them just doesn’t work. This forces you to think about the problem more deeply and use that to come up with tricks that reduce the search space. This is very much along the same lines as this blog post, but generalized to bigger harder problems. And FWIW, stuff like this - a deep interest in math and how there are often many different ways to solve a problem, and the study of digits and numbers and polynomials and roots and efficient ways to manipulate them - is quite core to my job and all the people I work with. I do mostly software at a company that makes hardware and software, and number-theoretic thinking and fun & “impractical” tricks put into practice are literally everywhere. I guess I’m saying I think that there actually is a lot of practical value in this kind of play, for the engaged audience, and once someone goes beyond basic high school & college math, in addition to the benefits of increased motivation that you mentioned. I wouldn’t be so quick to write it off as pointless, and only teach to the test. Maybe if we had more of this in high school, we’d have a better math education system. I’d agree we need to teach all kids how to use calculators, but I also think we need to teach kids how to think deeply about numbers, and have the skills to build tricks and do mental estimation, and send more of them into science and tech jobs with exploration skills emphasized and prioritized over calculator skills.

detrites 4y ago

I guessed it first try from the post title, based on: - it ends in 3, so the cube root must be a multiple of 3 - 30*3 is intuitively going to be 27,000, and 20*3 will be 8,000 - that leaves only 21, 24 and 27 to choose from, and given the exponential shape for 8k - 27k, I went with the largest Not a "solution", as I couldn't be certain without confirming, but interesting how close a logical deduction could get.

ykonstant 4y ago
One correction: there are many numbers that end in 3 but are not multiples of 3. It is the fact that its digit sum is divisible by 3 that lets us conclude this number itself is so. Perhaps that is what you meant?
- deleted 4y ago
  [deleted]
- detrites 4y ago
  Yes, as stated: it ends in 3, so the cube root must be a multiple of 3. The "cube root" part contains the implicit information that we have a product ending in 3, of three identical factors. I guess the missing assumption is that given the context of solving the cube root in ones head, it would more likely be a trick that only works with integers, and as such the correct answer might be expected to be one.
  - dmurray 4y ago
    > Yes, as stated: it ends in 3, so the cube root must be a multiple of 3. I don't get it. 7 cubed is 343. The article explains better what numbers might have cubes ending in 3.
    - detrites 4y ago
      Yes, the article is great (now I've read it). 37*3 also ends in 3 but is not divisible by 3. Thus, were the range not capped at 30 it may have been illegitimately discarded by the criteria. I think the 3 combined with the 5 digits and 20-30 range, made my guess a bit lucky. It's also contrived for that cube, so not useful for others, other than perhaps as a direction to go in (and likely arrive at the article).

Simon_O_Rourke 4y ago

Super neat!

deleted 4y ago

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kazinator 4y ago

It should be close to cbrt(20000) = cbrt(20) cbrt(1000) = cbrt(20) * 10 = 2.x * 10 = 2x. The job is to find x. Assuming 19683 has an integer cube root, I would guess the last digit x to be 7, because that gives us the 3: 7 * 7 = 49, and 9 * 7 = 63. Thus, the answer is probably 27. Cubing 27 mentally to validate the answer is the harder thing for me. :)

hgsgm 4y ago
This is a summary of the article.

thrdbndndn 4y ago

https://math.stackexchange.com/questions/1142767/why-does-this-cube-root-trick-work https://math.stackexchange.com/questions/1142767/why-does-th... Also, I have an additional question. Obviously this trick relies on the fact that "the last digit of the cube of all numbers from 0 to 9 is distinct". Is this fact just coincidence, or is it bound to happen?

rk06 4y ago
0^3 = 0 1^3 = 1 2^3 = 8 3^3 = 27 4^3 = 64 5^3 = 125 6^3 = 216 7^3 = 343 8^3 = 512 9^3 = 729 As you can see, for decimal system, last digit of cubes is unique for all digits. As last digit of cube depends on last digit of cube root only (same for squares and so on), it is bound to be true for all numbers
- thrdbndndn 4y ago
  You misunderstand my question. I was asking if this (cubes of number 0-9, the one you list, all have unique last digit in 0-9) a coincidence or not. (I know a number's cube's last digit is only affected by its own last digit.)
  - archi42 4y ago
    The question is in which case(s) `f(x) = x^n mod m` is a bijection for `0 <= x < m`. My intuition is to guess that `gcd(n,m) = 1` is at least a sufficient condition (correct math term? In german it's "hinreichende Bedingung"). Maybe it's the sole necessary condition. But I did only think about this for 5 to 10s and did not spend too much time on looking for a counter example, let alone a proof strategy (my number theory is a bit rusty -- ah, but on second thought, if the theory is correct, then I bet the Chinese remainder theorem is involved).
    - schoen 4y ago
      I asked the same question in this thread and got a couple of detailed answers: https://news.ycombinator.com/item?id=34652909 https://news.ycombinator.com/item?id=34652909

schoen 4y ago

I was surprised by the chart showing that there is a bijection between last digit of a number and last digit of the number's cube. In base 10, this works for every odd power but for no even power. (if you're given that a square ends in anything but 0 or 5, there are always two different possibilities for the last digit of the square root) Here are the powers (less than 100) for which this works in different bases (up to 25). Now I'm quite confused: what is the pattern here and what is the number-theoretic explanation for it? (Edit: It looks like the bases for which no power, except the trivial case of 1, has this property are exactly those bases that are divisible by a perfect square -- but I wouldn't have expected that and still don't know how to account for it.) 2 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99] 3 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 4 [1] 5 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 6 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 7 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 8 [1] 9 [1] 10 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 11 [1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99] 12 [1] 13 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 14 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 15 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 16 [1] 17 [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99] 18 [1] 19 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 20 [1] 21 [1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97] 22 [1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99] 23 [1, 3, 5, 7, 9, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97] 24 [1] 25 [1]

Zacharias030 4y ago

Followup question, how to estimate/calculate compound interest in your head quickly? (Any direction, solving for the rate, or for the time, or the principal needed)

nigamanth 4y ago
Applying the formula isn't that hard, but this one could be hard to write about because it depends upon the amount of time the sum is compounding or the exponent in the equation. ^2 and ^3 are quite easy but when it comes to ^7/^9 there's no easy way to calculate mentally.
bberrry 4y ago
There is a method called the rule of 72. You can look it up but it's essentially: Years to Double = 72 / Annual Rate of Return
- nigamanth 4y ago
  Still cannot be done mentally.

deleted 4y ago

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red_Seashell_32 4y ago

offtopic, but your website is hard to read on mobile - width of text makes up less than 33% of screen, and whitespace on sides is too wide :/ (though unsure how many of your traffic primarly reads article on mobile, so it might not even be an impactful issue)

xchip 4y ago

Note that where the article says 653,503 it should be 658,503

bohadi 4y ago

My solution was similar, represent. First we observe x is between 20^3=8e3 and 30^3=27e3. Then note that the least significant (singles) digit of x^3 is solely determined by the singles digit of x [only holds in the integers]. Luckily we had cached that the pattern for cubes is 123456789 -> 187456329 Finally x^3 = 19,683 ends in 3 therefore x = 27 we are done. ------ It turns out there is a unique mapping for the first digit of x^n, following one of four patterns: n mod 4 = 1 123456789 -> 123456789 n mod 4 = 2 -> 149656941 n mod 4 = 3 -> 187456329 n mod 4 = 0 -> 161656161 So this trick would not always be as useful, where the mapping is not surjective for squares, fourth powers -- all the even powers. I imagine there is a nice visual geometric explanation why those mappings take their symmetric pattern. You may ask yourself, is there also a rule for the larger place digits of x^3, x^n? Yes, we naturally observe that the 10s digit of x^n is similarly wholly determined by the 10s and 1s digits of x. So now we want n maps on [0,9]x[0,9] to [0,9]. The python snippet may help to explore this: where the row 3, 25th digit shows that when x ends in 25, x cubed has 10s digit 2 >>> print(*zip(range(50), [''.join([str(x**i).zfill(2)[-2:-1] for x in range(100)]) for i in range(50)]), sep='\n') (0, '0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000') (1, '0000000000111111111122222222223333333333444444444455555555556666666666777777777788888888889999999999') (2, '0000123468024692582604827272840628529642086432100000001234680246925826048272728406285296420864321000') (3, '0002621412032947913506462278580963075571028082329405076719170824429630014127735304680250760785873799') (4, '0018529096043612327208547274580272321634069092581000185290960436123272085472745802723216340690925810') (5, '0034227064053927756900342270640539277569003422706405392775690034227064053927756900342270640539277569') (6, '0062925444068032162802087278020826123086044452926000629254440680321628020872780208261230860444529260') (7, '0028823456070107573304842270100167479397084062167405238739510206025238098927751506624298920345671179') ... (19, '0086429478094787353708082276960769671755062002581405814799730442823032030327719102646212500125075319') (20, '0070727070007072707000707270700070727070007072707000707270700070727070007072707000707270700070727070') (21, '0050025000011167116102722272220333873383049442944405550755050616621666077727772708388238880999479949') (22, '0000123468024692582604827272840628529642086432100000001234680246925826048272728406285296420864321000') (23, '0002621412032947913506462278580963075571028082329405076719170824429630014127735304680250760785873799') (24, '0018529096043612327208547274580272321634069092581000185290960436123272085472745802723216340690925810') ... (35, '0060227434056527793900602274340565277939006022743405652779390060227434056527793900602274340565277939') (36, '0032925054061432187802967276920878123416045052923000329250540614321878029672769208781234160450529230') (37, '0076823046079307576304102274800137479107085462182405718735410298025268091527798506324296020359671329') (38, '0048321482080252902406667276660420925208028412384000483214820802529024066672766604209252080284123840') (39, '0086429478094787353708082276960769671755062002581405814799730442823032030327719102646212500125075319') (40, '0070727070007072707000707270700070727070007072707000707270700070727070007072707000707270700070727070') (41, '0050025000011167116102722272220333873383049442944405550755050616621666077727772708388238880999479949') (42, '0000123468024692582604827272840628529642086432100000001234680246925826048272728406285296420864321000') (43, '0002621412032947913506462278580963075571028082329405076719170824429630014127735304680250760785873799') ... ----- Apologies for the wall of text but the output is included here to show the periodicity is 20 in the 10s digit for powers of n. What do you think the periodicity in the 100s digit is? Okay let's use our cache to just lookup the least two digits of say, 41679296 ^ 35 (without doing any real compute): 1s digit 35 mod 4 = 3 187456329 index 6 -> 6 10s digit (35, '0060227434056527793900602274340565277939006022743405652779390060227434056527793900602274340565277939') index ------96 -> 7 The last two digits of 41679296 ^ 35 = 497981393104388234083687472206165994017494365065741791800324926291638211001469403054062791593737655590988883756006534609416313532750908441977019034732014717649617778735038244085515193249673259731575280032440237649262981884959117005449913087392615145373869733708824576 are indeed 76, cool

_Microft 4y ago

My method is certainly not as sophisticated as others here but since it was different from these others, I will post it as well. The solution should obviously be somewhere between 20 and 30. Since I knew 25^2 = 625 already, I multiplied by 25 to arrive at 25^3 = 15625. From there I used the approximation (1 + x)^n = 1 + nx which works if x << 1. 19683 = 15625*(1 + x) = 25^3*(1 + x) with x ~= 4000/15625 ~= 0.25. This stretches the condition ("x << 1") for the approximation a lot but so be it. (1 + 1/4)^1/3 ~= 1 + 1/12. 25*(1 + 1/12) ~= 27 Good enough ;)

atemerev 4y ago

I learned this method at the age of 7, and was playing a math savant for a few weeks.

timwaagh 4y ago

Interesting to try 140.something is what I'm guessing. Edit: Ah wait cube root. That is the square root I just tried. 26.something is what my mental calculations gave me. I thought it was going to be 27. Turns out it is. I made an error somewhere which is why you gotta do this on paper. As for method I kind of brie brute forced it. started with 100 cubed which is hundred thousand then 20 cubed which is 8000 then 30 cubed which is 3 cubed times 10 cubed which is 27k. 25 cubed which I don't remember but was less then, then I knew it could not be 26 because the number is not even. I'd deal with fractions later. so I tried 27 cubed. 27 cubed is 3^3^3=2733^5=8133^4=24333^3= 72927=72920+7297 72920=72902=14400+180=14580 7297=4900+140+63=5103 Add them and 19683 comes out Now try that in your head and it's a good recipe for frying one's brain.

sp332 4y ago
Much too large. 100 cubed is 1,000,000.
colinbeveridge 4y ago
For the square root, I get 140.3. My heuristic is to find a nearby square (here, 140^2 = 19600) and divide the discrepancy (83) by double the root of the nearby square (280). This is the number to adjust by. [^0] [^0] The Maclaurin series expansion of (A^2 + x)^(1/2) starts A + x/(2A), which is where this comes from. 83/280 is close to 84/280, which is 12/40 or 0.3. If I was trying to show off [^1], I might adjust that -- 1/28 is a quarter of a seventh, about 0.0357, so I can take 0.004 off of my adjustment to get 0.296 or 0.297. [^1] I mean, show off more. (The actual answer is 140.29611 -- the "0.3" adjustment is right to one part in 36,000; "0.296" is right to one part in 1.2 million.)

madcaptenor 4y ago

I just recognized that it’s 3^9, but that’s dumb luck on my part.

canadianfella 4y ago
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cdelsolar 4y ago

I got 27 in about 10 seconds by doing a quick binary search and noticing the number ended in 3 (so it’s gotta end in 7). It’s in between 20 (8000) and 30 (27000).

sbaiddn 4y ago

For those of us who have tried but failed to memorize our logs (obviously the best method to ballpark it) : For the n-the square root just group the digits by n starting from the ones. That will tell you how many digits you're dealing with. Simplish math will give you the first digit off of the last grouping (simple since you only have to ^n the numbers from 1-10) Simple counting and a bit of math and I have the order of mag and first digit. 2_ At this point I can either think of a decent trick to iterate the second digit (quick Taylor? [1]), or just guesstimate a linear fit and call it a 5. 25 is within 10%. [1] instead of Taylor I would do Pascal's triangle on the cube with the next term to calculate (30-x)^3 (I use 30 - x and not 20+x since I want x as small as possible and I know the answer is closer to 30 than 20. Pascal's triangle: 1(30)^3 + 3(30)^2x + 330x^2 + 1x^3 Drop the higher two terms. 3^3 + 33^2x = 19 19-27 = 27x X = 8/27 = 1/3 (- 1/27) 30-3.3 = 26.66 I use the sign of the next term of pascal's triangle to see if Ive under or over estimated 33*(-0.3)^2 -> positive So my x is too large So my answer is an underestimate. (Cube root of 19000) greater than but close to 26.6

graderjs 4y ago

Guessing: 1000 is 10*10*10 this is close to 20K, so we have 10, now we need cube of 20. Between 2 and 3, take 2.5 so 25? No... it's 27.

_steady 4y ago

Loads of chat about remembering logs etc, I was expecting to see more number theory here as it’s a really lovely and applicable part of maths. (As far as doing the calc, I thought it has to end in 7 becaus3 it’s the only one with a cub3 residue of 3 mod 10, which is what they say in post. Then 20 and 30 cubed are 8000 and 27000 respectively and assuming we’re restricting to Z 27 is the only possible answer)

less_less 4y ago

If you know that the number is a perfect cube, you could also try Hensel's lemma or CRT, or Euler, all of which are aaaaalmost doable in your head for 2 digits, but maybe not quite. Suppose you are trying to find cbrt(x), and you know the answer is a 2-digit integer. =============== Hensel: first you have to remove any powers of 5 and 2. You know the cube root is even if the number is even, and a multiple of 5 if the number is a multiple of 5 (ends in a 0 or 5). So factor those out first. It now ends in 1,3,7 or 9. Find the cube root w of x mod 10, and t = 1/3x mod 10, by trial and error. Eg, x=19683 ends in 3, so w ends in 7 (since 7^3 ends in 3); t = 1/3x ends in 9 since 39x ends in 1. Calculate the tens digit of the error term e = x - w^3, which is always a multiple of 10. In this case it's ...683 - 343 === 40, and apply the adjustment w ==> w + wte. You only need to calculate the 10s digit of wte. In this case, it's the 10's digit of 7*9*40 = 63*40 is 20. That's all, the answer is 27. =============== CRT: Find the answer mod 10 and mod 11. As before, the answer here mod 10 is 7. For mod 11, you can reduce it to a 2-digit problem by applying %99 first, i.e. shifting each digit over by 2 until it's a 10s or 1s digit, and adding. We have 19683 mod 11 = (1 + 96 + 83) mod 11 = 180 mod 11 = 81 mod 11 = 4. Find the cube root mod 11 by exhaustion, or by memorization, or by taking the 7th power of the input (Fermat's theorem). In this case, it's 5, because 5^3 mod 11 = 125 mod 11 =(1+25) mod 11 = 4. Solve using the CRT: if a number is y mod 10 and z mod 11, then it's y + 10(y-z mod 11). In this case, it's 7 + 10(7-5 mod 11) = 27. ================ Euler: the answer is always the same as the last 2 digits of x^7. Calculate eg x^2, x^3, x^6, x^7 with four multiplications. You only need the last two digits. Probably this is the easiest approach by hand. ================ If you want to solve on a computer, then Hensel mod 2^n (continuing the pattern appropriately for more bits) is probably the fastest bet.

nigamanth 4y ago
> all of which are aaaaalmost doable in your head for 2 digits, but maybe not quite. You might be good at Mathematics, this is like a standard trick the average student can do. There are lots of other approaches but the easiest one to do mentally is this.

croes 4y ago

>Notice how, the cubes of 1, 4, 5, 6, and 9 always end with the number itself. The cube of 3 ends with 7 and cube of 7 end with 3 The cube of 2 ends with 8 and cube of 8 end with 2 Cube-buddies

deleted 4y ago

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jameshart 4y ago

Another approach is to recognize that if you know a number is a perfect cube, and you can recognize any prime factors it contains, you can pull out the cube of those from the number. Like, by inspection (sum of digits) this number has 3 as a factor. That means 3 has to also be a prime factor of its cube root which means 27 has to be a factor of the cube. So we can divide out 27, leaving us with the smaller problem of finding 3(cbrt(729)), which you might recognize as being 3*9, or spot that you can repeat the trick to give you 9(cbrt(27))… Or you might notice that if 19683 has to contain 27 as a factor, that doesn’t leave a lot of room for any other cubed prime factors to exist in the number, and do a trial multiplication to confirm that 27 just is the cube root. Basically comes down to, if you have a comfortable relationship with how numbers relate to their factors, you can develop a number of instincts for how to recognize shapes within them.

chasing 4y ago

20 cubed is obviously too low. 30 cubed too high. 7 is the only number that, when cubed, ends in 3. 27 it is.

camgunz 4y ago
Yeah this is what I do; what number ^3 gives you a 3: it's 7. Bracket it in the 10s place, try it out.
yebyen 4y ago
I went with "28 sounds about right" then saw it was clearly just a bit too high.
rich_sasha 4y ago
30 cubed is 27,000 or about 35% too much. Thus we solution is about 10% smaller than 30. 27 gets you there.

Gravityloss 4y ago

My mental process: Well it's less than 100 since 100^2 = 10,000 so that goes over. It's over 10 since 10^3=1000. So this narrows it to more than 10 and less than 100. log-wise 30 is about midway between 10 and 100. 900 x 30 is is 27000 so it's less than 30. 20? 400 x 20 = 8000 so bigger than that. Then it gets annoying to compute the cubes so close enough to say more than 20 and less than 30.

dsr_ 4y ago

In my head: oh, look, a multiple of 7. What's 30 cubed? 27 and three zeros, so not far away. What's 27 cubed? Oh, that's it. One of the things that you should learn is that the problems that humans set for each other usually involve integers or well known numbers.

_a_a_a_ 4y ago

27, but I only got that because it was explicitly given that 19,683 was a cube.

omnster 4y ago

Reminded me of the story about Feynman being challenged by an abacist to compute cubic root https://www.ecb.torontomu.ca/~elf/abacus/feynman.html https://www.ecb.torontomu.ca/~elf/abacus/feynman.html