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Last two digits of Cubes
Posted: Posted October 14th, 2017 by The Fly

Show that no cube \(N^3\) of a positive whole number \(N = 1, 2, 3, \dots \) can end with 10. (It also cannot end with 26 or 55, among others.)

Have Fun,
The Fly


There are 3 Replies

PS, by 'end' I mean like the number 82910 ends with 10 (its last two digits).

Posted October 14th, 2017 by The Fly

My attempt:

If \(N^3\) ends in 10, that means it is divisible by 10. That means that both 2 and 5 divide \(N^3\). If we decompose \(N\) into its prime factors \(N = {p_1}^{e_1}{p_2}^{e_2} \dots {p_k}^{e_k}\), then \(N^3 = {p_1}^{3e_1}{p_2}^{3e_2} \dots {p_k}^{3e_k}\). We know that 2 and 5 have to divide \(N\), which implies that there are at least two other factors of 2 and two other factors of 5 that divide \(N^3\). That means that \(N^3\) is at least divisible by 1000, which would make it end in three zeros, not one. Thus we reach a contradiction, so \(N^3\) can't end in 10.

Edited October 15th, 2017 by EN

Yes, you got it, although you don't need the prime decomposition but only the fact that 2 and 5 are primes that would have to divide N, hence 10 divides N, so that N cubed would end with 2 zeros and not 10. The case with N cubed ending with 26 and 55 can't be done the same way it seems to me.

Posted October 17th, 2017 by The Fly
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