Problem 30 · 2018 Math Kangaroo
Stretch
Number Theory
divisibilitydigit-sum
Archimedes has calculated \(15!\) and the result is on the board, but two of the digits—the second and the tenth—cannot be read: \(1\,?\,0\,7\,6\,7\,4\,3\,6\,?\,0\,0\,0\). What are the two missing digits? (Remark: \(15! = 15 \cdot 14 \cdot 13 \cdots 2 \cdot 1\).)
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Answer: E — 3 and 8
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Hint 1 of 2
15! is divisible by 9 and by 11, which constrain the hidden digits through digit tests.
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Hint 2 of 2
Use the rule for 9 (digit sum) and the alternating rule for 11.
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Approach: apply divisibility tests for 9 and 11 to recover the hidden digits
- Write 15! as 1 a 0 7 6 7 4 3 6 b 0 0 0 with hidden 2nd digit a and 10th digit b.
- Divisibility by 9: the digit sum 34+a+b must be a multiple of 9, forcing a+b = 11.
- Divisibility by 11: the alternating digit sum forces b−a = 5.
- Solving gives a = 3, b = 8 — the missing digits are 3 and 8.
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