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# Brilliant question

Let A be a number with 2001 digits such that A is a multiple of 10! Let B be the digit sum of A, C be the digit sum of B, and D be the digit sum of C. What is the unit’s digit of D?

Note by Alpha Beta
3 years, 7 months ago

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Let $$S(n)$$ denote the sum of the digits of $$n$$. Since $$A$$ has $$2001$$ digits, $$B = S(A) \le 9 \cdot 2001 = 18009$$.

Since $$B$$ has at most $$5$$ digits, $$C = S(B) \le 9 \cdot 5 = 45$$. By inspection, $$D = S(C) \le 3 + 9 = 12$$.

Since $$S(n) \equiv n \pmod{9}$$ for any integer $$n$$, we have $$D \equiv C \equiv B \equiv A \equiv 10! \cdot m \equiv 0 \pmod{9}$$.

Trivially, $$D \ge 1$$. Since $$1 \le D \le 12$$ and $$D \equiv 0 \pmod{9}$$, we have $$D = 9$$, and thus, the unit digit is $$9$$. · 3 years, 7 months ago

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Awesome buddy.THNX · 3 years, 7 months ago

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