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Weekly Problem 6 - 2008

Stage: 3 and 4 Challenge Level: Challenge Level:1

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

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

This problem is taken from the UKMT Mathematical Challenges.

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