# Sum of digits

$\large N = 13 \times 17 \times 41 \times 829 \times 56659712633$ It is know that $$N$$ is an 18-digit number, with nine of the ten digits from 0 to 9 each appearing twice. Find the sum of the digits of $$N$$.

Honor Bonus: Do it without calculating the product.

##### This problem is taken from the 7th Hong Kong Pui Ching Invitational Mathematics Competition, year 2008.
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