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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 4 years, 11 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 \(9\).

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Awesome buddy.THNX

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TopNewestLet \(S(n)\) 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 \(9\).

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Awesome buddy.THNX

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