Question

Prove that for every integer n greater than 1 that \(7^n\) and \(2^n+5\) always have the same last digit.

Answer 1

7^mod10=2^n+5 mod 10?

Answer 2

7^n mod 10

Answer 3

Yeah, that's another way to phrase the question. =)

Answer 4

\[2^n\mod10\] cycles \(2,4,8,6,\dots\) and \(7^n\mod10\) cycles \(7,9,3,1,\dots\) so \(2+5=7,4+5=9,8+5=13,6+5=11\) and it checks out

Answer 5

Ahhh that's not wrong, but can anyone show it without using an exhaustive method? =P

Answer 6

\(7^n-2^n\equiv5\) but \(7^n-2^n=(7-2)(7^{n-1}+2^{n-1}+7^{n-2}\cdot2+7\cdot2^{n-2}+\dots)\)

Answer 7

hmm so now to show \(\sum\limits_{k=0}^{n-1} 7^k\cdot2^{n-1-k}\equiv1\pmod{10}\)

Answer 8

I didn't do it this way, but I think it might be easier to consider leaving the 5 in the expression since 5*2 makes for a nice simplification.

Answer 9

Here's my proof:

\(7^n \equiv (2+5)^n \mod 10\)

If we expand this binomial out, we'll have terms with 2*5 in them, which in mod 10 are all 0, so we can throw out all those terms that are combined of 2 and 5.

\((2+5)^n \equiv 2^n+5^n \mod 10\)

Now we know \(5^2 \equiv 5 \mod 10\) so we can replace all powers of 5 with just 5 in mod 10, so here we go:

\(7^n \equiv 5+2^n \mod 10\)