Let m and n be natural numbers. What is the <strong>MINIMUM</strong> value of (m+n) such that 33m+22n is divisible by 121?

Explanation

33m+22n = 11(3m+2n). To be divisible by 121, 3m+2n must be a multiple of 11. Testing small natural numbers: for m=3, n=1, 3(3)+2(1) = 11. Thus, the minimum sum m+n = 3+1 = 4.

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