IMC 2006 Q25

First note that as \(j\), \(l\) and \(m\) are all non-negative, the values of \(5^j\), \(7^l\) and \(11^m\) are all odd. However, the sum \(5^j + 6^k + 7^l + 11^m\) is even, so we deduce that \(6^k\) cannot be even and hence \(k = 0\), that is \(6^k = 1\). Now, for all positive integer values of \(j\) and \(m\), the units digit of \(5^j + 6^0 + 11^m\) is 5 + 1 + 1, that is 7. So the units digit of \(7^l\) is 9 and we deduce that \(l = 2\) since 7, 49 and 343 are the only positive integer powers of 7 less than 2006. We now have \(5^j + 6^0 + 7^2 + 11^m = 2006\), that is \(5^j + 11^m = 1956\). The only positive integer powers of 11 less than 2006 are 11, 121 and 1331. These would require the value of \(5^j\) to be 1945, 1835 and 625 respectively, and of these only 625 is a positive integer power of 5.
So \(5^4 + 6^0 + 7^2 + 11^3 = 2006\).