7 ms·

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 2

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).
    - nigamanth 4y ago
      No, he meant that the cube root must be a multiple of 3. 27 is a multiple of 3.
      - jstanley 4y ago
        But 7 isn't a multiple of 3.
    - mtlmtlmtlmtl 4y ago
      The difference is that 343 is not a multiple of 3, which excludes its cube root from being a multiple of 3.
      - less_less 4y ago
        Right. The correct argument is that 19683 is a *multiple* of 3, because its digit sum 27 is a multiple of 3. So if it's a perfect cube, then the cube root must be a multiple of 3. https://news.ycombinator.com/item?id=34652053 https://news.ycombinator.com/item?id=34652053 Also, if it's a perfect cube, since it ends in 3, the cube root must end in 7 (cubing integers is 1:1 on the units digit, and 7->3). So it must be 27, 57, 87, ... Since it's too small for the cube root to be 57, 27 is the answer.
madaxe_again 4y ago
Your method is good, but in my view overthinks it - I just looked at it, and let what seemed like the right solution just pop into my head - and it was. It felt like 27 would be the answer. I know that seems woolly as all hell, but that’s the nature of running stuff at a level that can’t be consciously interacted with or directly queried. A quick conscious mental multiplication verified it. Intuition is incredibly underrated - I have learned to trust the first damned fool thing that pops into my head, before any conscious thought has had time to get moving and therefore interfere. It’s usually correct, regardless of context - and when it goes wrong, it’s because I accidentally allowed myself to think and reason. It’s in the same category as “walking away from a problem”, in my view - you can’t solve it, you go grab a coffee, and while you’re looking for the sugar the answer is just suddenly there in your head, because you stopped running the CPU-intensive conscious process, and let the background workers catch up. Conscious thought is great for producing outputs for other humans to then work with. Unconscious through this great for… everything else. We do incredibly complex things all the time - catching a ball, walking upright, looking in the direction of that sudden sound - but typically run all of the gnarly logic in an abstraction layer within a virtual machine, rather than on the hardware that was designed to run it natively. No end of trouble at school where I wouldn’t just not show my working, but couldn’t - but the real world fortunately doesn’t particularly care about the means, just the ends.
- sarabande 4y ago
  I think you can profit from this intuition if you've already consciously sharpened your arithmetic skills over the years. I doubt the average person can rely on their untrained intuition for large square roots.
- timwaagh 4y ago
  That's pretty amazing do you have a lot of experience mentally calculating cubes?
  - madaxe_again 4y ago
    Not particularly since school, no - but I do have a lot of experience with quieting my conscious mind and extracting answers from the silent mass. Maths, facts, connections and correspondences, engineering, debugging, directions, the time, you name it - there’s a whole suite of stuff in there that you just have to learn how to interact with, and nobody is taught. When it’s something important, it can be verified with good old fashioned cogitation. One of my favourite tricks is just knowing, to +- 1 minute, what time it is. I don’t set an alarm, I just plan to wake at whatever time. If someone asks the time I give it without checking. It actually works better when I’m asleep, as I will wake +- 10 seconds from the intended time, which shows just how much consciousness interferes. Honestly, it’s laziness - thinking takes far too much effort and is much too slow, and 80% right in 1% of the execution time is a “good enough” optimisation.
deleted 4y ago
[deleted]
Fricken 4y ago
I solved it deductively pretty much the way you did, but I wouldn't have approached it that way at all if I couldn't assume that the cube root of 19,683 would be a whole number.
- hgsgm 4y ago
  The method is completely wrong and is gets the correct answer by sheer luck.
  - deleted 4y ago
    [deleted]
  - Fricken 4y ago
    If we've guessed that the number will be in the mid to high 20s, then 27 is the only number one could logically arrive at. 24, 25, 26, and 28 are easy to eliminate because 19,863 is neither an even number nor a multiple of 5. That's all there is to it.
hgsgm 4y ago
How can you confidently believe this > - it ends in 3, so the cube root must be a multiple of 3 And also believe this > that leaves only 21, 24 and 27 to choose from, without noticing that 21^3 ends in 1 and 24^3 is even? It's like you're doing math in a while different axiom system. But not notice that this >
- detrites 4y ago
  Those are good observations that would immediately arrive at the correct answer quicker given the establishing of bounds as described. I was doing math in an axiom system made for the problem. Possibly not a good idea in math, but in programming it's kind of what we do. (I don't guarantee it's bug-free or works with different input...)
  - nigamanth 4y ago
    Don't mix up programming with math. They use a different approach altogether, to solve any word problem through a program chances are you'll use variables and convert it into a linear/quadratic/cubic system of equations, when in Mathematics you can solve it in an easier manner.

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).
  - nigamanth 4y ago
    No, he meant that the cube root must be a multiple of 3. 27 is a multiple of 3.
    - jstanley 4y ago
      But 7 isn't a multiple of 3.
  - mtlmtlmtlmtl 4y ago
    The difference is that 343 is not a multiple of 3, which excludes its cube root from being a multiple of 3.
    - less_less 4y ago
      Right. The correct argument is that 19683 is a *multiple* of 3, because its digit sum 27 is a multiple of 3. So if it's a perfect cube, then the cube root must be a multiple of 3. https://news.ycombinator.com/item?id=34652053 https://news.ycombinator.com/item?id=34652053 Also, if it's a perfect cube, since it ends in 3, the cube root must end in 7 (cubing integers is 1:1 on the units digit, and 7->3). So it must be 27, 57, 87, ... Since it's too small for the cube root to be 57, 27 is the answer.

madaxe_again 4y ago

Your method is good, but in my view overthinks it - I just looked at it, and let what seemed like the right solution just pop into my head - and it was. It felt like 27 would be the answer. I know that seems woolly as all hell, but that’s the nature of running stuff at a level that can’t be consciously interacted with or directly queried. A quick conscious mental multiplication verified it. Intuition is incredibly underrated - I have learned to trust the first damned fool thing that pops into my head, before any conscious thought has had time to get moving and therefore interfere. It’s usually correct, regardless of context - and when it goes wrong, it’s because I accidentally allowed myself to think and reason. It’s in the same category as “walking away from a problem”, in my view - you can’t solve it, you go grab a coffee, and while you’re looking for the sugar the answer is just suddenly there in your head, because you stopped running the CPU-intensive conscious process, and let the background workers catch up. Conscious thought is great for producing outputs for other humans to then work with. Unconscious through this great for… everything else. We do incredibly complex things all the time - catching a ball, walking upright, looking in the direction of that sudden sound - but typically run all of the gnarly logic in an abstraction layer within a virtual machine, rather than on the hardware that was designed to run it natively. No end of trouble at school where I wouldn’t just not show my working, but couldn’t - but the real world fortunately doesn’t particularly care about the means, just the ends.

sarabande 4y ago
I think you can profit from this intuition if you've already consciously sharpened your arithmetic skills over the years. I doubt the average person can rely on their untrained intuition for large square roots.
timwaagh 4y ago
That's pretty amazing do you have a lot of experience mentally calculating cubes?
- madaxe_again 4y ago
  Not particularly since school, no - but I do have a lot of experience with quieting my conscious mind and extracting answers from the silent mass. Maths, facts, connections and correspondences, engineering, debugging, directions, the time, you name it - there’s a whole suite of stuff in there that you just have to learn how to interact with, and nobody is taught. When it’s something important, it can be verified with good old fashioned cogitation. One of my favourite tricks is just knowing, to +- 1 minute, what time it is. I don’t set an alarm, I just plan to wake at whatever time. If someone asks the time I give it without checking. It actually works better when I’m asleep, as I will wake +- 10 seconds from the intended time, which shows just how much consciousness interferes. Honestly, it’s laziness - thinking takes far too much effort and is much too slow, and 80% right in 1% of the execution time is a “good enough” optimisation.

deleted 4y ago

[deleted]

Fricken 4y ago

I solved it deductively pretty much the way you did, but I wouldn't have approached it that way at all if I couldn't assume that the cube root of 19,683 would be a whole number.

hgsgm 4y ago
The method is completely wrong and is gets the correct answer by sheer luck.
- deleted 4y ago
  [deleted]
- Fricken 4y ago
  If we've guessed that the number will be in the mid to high 20s, then 27 is the only number one could logically arrive at. 24, 25, 26, and 28 are easy to eliminate because 19,863 is neither an even number nor a multiple of 5. That's all there is to it.

hgsgm 4y ago

How can you confidently believe this > - it ends in 3, so the cube root must be a multiple of 3 And also believe this > that leaves only 21, 24 and 27 to choose from, without noticing that 21^3 ends in 1 and 24^3 is even? It's like you're doing math in a while different axiom system. But not notice that this >

detrites 4y ago
Those are good observations that would immediately arrive at the correct answer quicker given the establishing of bounds as described. I was doing math in an axiom system made for the problem. Possibly not a good idea in math, but in programming it's kind of what we do. (I don't guarantee it's bug-free or works with different input...)
- nigamanth 4y ago
  Don't mix up programming with math. They use a different approach altogether, to solve any word problem through a program chances are you'll use variables and convert it into a linear/quadratic/cubic system of equations, when in Mathematics you can solve it in an easier manner.