TITLE: Can the decimal digits of $5^a + 7^b$ end in $82$?
QUESTION [5 upvotes]: I had a question in my exam:

Let a and b be a pair of positive integers.  In computing $5^a + 7^b$, one gets a number with two least significant decimal digits equal to $82$.  Is it possible?  Justify your answer.

How can I solve it?
My understanding of this question is that $5^a+7^b=82 \pmod {100}$ then I solve for $a = 1$ to $5$ and $b = 1$ to $5$ separately which gives me this pattern:

$5^1=5 \pmod {100}$, $5^2=25 \pmod {100}$, $5^3= 25  \pmod {100}$ and so on.
$7^0=1 \pmod {100}$, $7^1= 7 \pmod {100}$, $7^2=49 \pmod {100}$, $7^3=43 \pmod {100}$, $7^4=1 \pmod {100}$, $7^5=7 \pmod {100}$ and so on.

So the sum of its numbers can never be $82$; maximum it goes to $49+25$.  I just want to confirm am I right?  Or is there any other way to solve this?
Thanks
SJ

REPLY [4 votes]: Yes, the proof is fine. Simpler: $\,{\rm mod}\ 20\!:\  5^{\large \Bbb N}\! + 7^{\large \Bbb N}\equiv\,\{5\} + \{1,7,9,3\} \not\ni\, 2$