That's one big number!

Let \(N\) be the number that consists of 61 consecutive 3's, so \(N = \underbrace{333\ldots333}_{61 \, 3's}\). Let \(M\) be the number that consists of 62 consecutive 6's, so \(M=\underbrace{6666\ldots666}_{62 \, 6's}\). What is the digit sum of \(N\times M\)?

Details and assumptions

The digit sum of a number is the sum of all its digits. For example the digit sum of 1123 is \(1 + 1 + 2 + 3 = 7\).

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