# 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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