Find the digit of the units for the number $(5)(7^{29})+(8)(9^{72})$

Question

I was checking the following number theory exercise:

Find the digit of the units for the number

$(5)(7^{29})+(8)(9^{72})$

Here goes what I have:

$5 \equiv -5$ $(mod$ $10)$

$7^{29}\equiv $ $7(mod$ $10)$

$8 \equiv -2$ $(mod$ $10)$

$9^{72}\equiv $ $1(mod$ $10)$

Then $-5+7-2+1 = 1$. So $1$ should be the digit of the units, but the answer is three.

Any help will be really appreciated.

Answer 1

$9^n$ has a units digit of $9$ if $n$ is odd, and a units digit of $1$ if $n$ is even. This is pretty easy to prove/test. So, $8\cdot 9^{72}$ has a units digit of $8$.

We do a similar thing for $7$

$7^1$ = 7

$7^2$ = 49, so units digit is $9$

$7^3$ has units digit of $3$, since $7^2$ had a units digit of $9$ and $7*9 = 63$.

$7^4$ has a units digit of $1$, since $7*3 = 21$.

So, there is a pattern in the units digits of exponents of $7$. The pattern seems to be $1,7,9,3,1,7,...$. Then, since $29 \ \text{mod} 4 = 1$, we can say that $7^{29}$ has the same units digit as $7^1$, so it ends in $7$. Multiplied by $5$ gives us a units digit of $5$. We add the two units digits together to get $13$, so the number has a units digit of $3$.

Answer 2

HInt:

$7\equiv1\pmod2,7^n\equiv1^n\equiv1$

$\implies5\cdot7^n\equiv1\cdot5\pmod{2\cdot5}$

$9\equiv-1\pmod5\implies9^{2m}\equiv(-1)^{2m}$

$\implies8\cdot9^{2m}\equiv8\cdot1\pmod{8\cdot5}\equiv8\pmod{2\cdot5}$

Answer 3

All of the answers so far are correct, but they needlessly look at the value of the factor $7^{29}$. Any odd multiple of $5$ has a $5$ in the units place, and $7^{29}$ is plainly odd. No need to analyze any further. Add that $5$ to the $8$ found by the others, and the answer is $3$ a little more directly.