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How many number of integers are there between 2008 and 20082008 including 2008 and 20082008 such that the sum of the digits in the square is 42?
Correct Answer: 0
A number and its digit sum leave the same remainder on division by 9. A perfect square leaves remainder 0, 1, 4 or 7 on division by 9, but $$42$$ leaves remainder 6. So no square has digit sum 42, and the number of such integers is 0.
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