Problem 3

The digits of 2024 have the property that $7^2 + 7^0 + 7^2 + 7^4 = 2500$ ends in two zeroes. Find how many sets of solutions there are for single digits $a$, $b$, $c$ and $d$ so that $7^a + 7^b + 7^c + 7^d$ ends in two zeroes. Note that $(a, b, c, d) = (2, 0, 2, 4)$ and $(a, b, c, d) = (0, 2, 2, 4)$ would be two different sets of solutions.