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 of7 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, and of these only 625 is a positive integer power of 5.

So $5^{4} + 6^{0} + 7^{2} + 11^{3} = 2006$ . Therefore $j + k + l + m$ = 4 + 0 + 2 +3 = 9.