Problem 20 · 2013 Math Kangaroo
Hard
Number Theory
divisibilitycasework
Given a six-digit number whose digit sum is even and whose digit product is odd. Which of the following statements is true for this number?
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Answer: E — None of the statements (A)–(D) are correct.
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Hint 1 of 2
A product of digits is odd only when every single digit is odd.
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Hint 2 of 2
With all six digits odd, check each statement — and remember there are only five odd digits to choose from.
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Approach: deduce all digits are odd, then test each claim
- An odd digit product forces all six digits to be odd (any even digit would make the product even).
- Six odd digits sum to an even number, matching the condition — so such numbers exist (e.g. 111111).
- Now: (A) zero even digits, false; (C) six odd digits is an even count, false; (D) only five distinct odd digits exist, so six different is impossible, false.
- Every listed statement fails, so the answer is E.
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