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

Posted October 14th, 2017 by The Fly
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
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
The Fly
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