2024 SSMO Team Round Problems/Problem 2
Problem
Find the sum of the three smallest positive integers 
where the last two digits of 
are 
.
Solution
From 
we have 
and 
From the first congruence, we have 
Now, let 
for integers 
Then, ![\[(5a+b)^4 = 5^4a^4+4\cdot5^3a^3b+6\cdot5^2a^2b^2+4\cdot5ab^3+b^4 \equiv 20ab^3+b^4\pmod{25}\]](http://latex.artofproblemsolving.com/8/4/5/845b20bba028a3e17761cf9963f859f231819b91.png)
so ![\[20ab^3+b^4 \equiv 1 \pmod{25}.\]](http://latex.artofproblemsolving.com/6/4/f/64f42a2f57cc1ae22680a2f4f128208ce4b21756.png)
Now, we will proceed using casework. Clearly, 
is not a multiple of 5. Now, note that
\begin{align*}
b\equiv1\pmod{5}&\implies a\equiv0\pmod{5},\\
b\equiv2\pmod{5}&\implies a\equiv1\pmod{5},\\
b\equiv3\pmod{5}&\implies a\equiv3\pmod{5},\text{ and }\\
b\equiv4\pmod{5}&\implies a\equiv4\pmod{5}.
\end{align*}
So, we have 
Combining this with 
we find the following solutions as residues mod 50: 
and 
So, the answer is 
~SMO_Team
