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Why is the last digit of n^5 equal to the last digit of n? [video]

17 points by iamzlatan 9y ago

panda88888 9y ago
At work so I haven’t watched the video yet, but my quick rationale is, Let m = abs(n) It follows n = sign(n) * m We can work with m since sign(n) doesn’t affect the ones digit of n^5. More formally, n^5 = sign(n)^5 * abs(n)^5. Let m = 10 * b + a, where b = floor(m/10) a = m % 10 Essentially we extract the ones digit as a. m^5 = (10 * b + a)^5 If we expand the terms, only a^5 affects the ones digit. The remaining has at least one 10 as factor. Quick python script will verify a^5 has a as the ones digit for 0...9. for a in range(10): print(pow(a, 5) == a) Edit: fixed formatting
- iamzlatan 9y ago
  It's not exactly the same thing as he does in the video but it's also an interesting approach. Also in the video he assumes n is natural so no need to use sign(n).
- contravariant 9y ago
  There are some slight problems with your proof, firstly your proof doesn't really use the fact that you're raising a number to the 5th power, except for the last part and secondly the last part (as it is currently written) will print "false" for all numbers other than 1. Also I don't think you need to bother with the sign of the number. Python is one of the few languages that took the sensible approach and made statements like (a % 10 = (a + 10) % 10) true for all a.
  - panda88888 9y ago
    Ah thanks! I was replying on my phone and the mod operator was lost during transcription.
blt 9y ago
good example of an inductive proof, would be helpful for people learning about inductive proofs.
contravariant 9y ago
A somewhat more general proof involves the use of the Chinese remainder theorem and the fact that 4 is divisible by the order of the multiplicative group of both integers mod 2 and mod 5. Therefore x^5 = x mod 2 and x^5 = x mod 5 for all x, and as a consequence x^5 = x mod 10 for all x. This construction seems to fail if the base isn't square free. In particular in base 4 the number 2 has as powers 2, 10, 20, 100 etc.
- knappa 9y ago
  That's not true for x=2 : 2^4 = 6 mod 10 Also, 4 does not divide the order of the multiplicative group of Z/2Z. i.e. 4 does not divide 1. (Also, the claim was that x^5 = x mod 10 not x^5=1, but I assume that's a typo.)
  - contravariant 9y ago
    Hmm yes I suppose I was a bit sloppy, although I got the gist of it right somehow. I guess the problem here is that x^4 = 1 mod 2 is false if x = 0 mod 2, but x^5 = x mod 2 does hold for all numbers. I've fixed the problems you mentioned, thanks.

17 points by iamzlatan 9y ago

At work so I haven’t watched the video yet, but my quick rationale is, Let m = abs(n) It follows n = sign(n) * m We can work with m since sign(n) doesn’t affect the ones digit of n^5. More formally, n^5 = sign(n)^5 * abs(n)^5. Let m = 10 * b + a, where b = floor(m/10) a = m % 10 Essentially we extract the ones digit as a. m^5 = (10 * b + a)^5 If we expand the terms, only a^5 affects the ones digit. The remaining has at least one 10 as factor. Quick python script will verify a^5 has a as the ones digit for 0...9. for a in range(10): print(pow(a, 5) == a) Edit: fixed formatting

iamzlatan 9y ago
It's not exactly the same thing as he does in the video but it's also an interesting approach. Also in the video he assumes n is natural so no need to use sign(n).
contravariant 9y ago
There are some slight problems with your proof, firstly your proof doesn't really use the fact that you're raising a number to the 5th power, except for the last part and secondly the last part (as it is currently written) will print "false" for all numbers other than 1. Also I don't think you need to bother with the sign of the number. Python is one of the few languages that took the sensible approach and made statements like (a % 10 = (a + 10) % 10) true for all a.
- panda88888 9y ago
  Ah thanks! I was replying on my phone and the mod operator was lost during transcription.

blt 9y ago

good example of an inductive proof, would be helpful for people learning about inductive proofs.

contravariant 9y ago

A somewhat more general proof involves the use of the Chinese remainder theorem and the fact that 4 is divisible by the order of the multiplicative group of both integers mod 2 and mod 5. Therefore x^5 = x mod 2 and x^5 = x mod 5 for all x, and as a consequence x^5 = x mod 10 for all x. This construction seems to fail if the base isn't square free. In particular in base 4 the number 2 has as powers 2, 10, 20, 100 etc.

knappa 9y ago
That's not true for x=2 : 2^4 = 6 mod 10 Also, 4 does not divide the order of the multiplicative group of Z/2Z. i.e. 4 does not divide 1. (Also, the claim was that x^5 = x mod 10 not x^5=1, but I assume that's a typo.)
- contravariant 9y ago
  Hmm yes I suppose I was a bit sloppy, although I got the gist of it right somehow. I guess the problem here is that x^4 = 1 mod 2 is false if x = 0 mod 2, but x^5 = x mod 2 does hold for all numbers. I've fixed the problems you mentioned, thanks.