Problem 19 · 2011 Math Kangaroo
Hard
Number Theory
cryptarithmfactorization
What is the smallest possible positive whole-number value of the expression \(\dfrac{K\cdot A\cdot N\cdot G\cdot A\cdot R\cdot O\cdot O}{G\cdot A\cdot M\cdot E}\) if different letters stand for different digits, none equal to 0, and equal letters stand for equal digits?
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Answer: B — 2
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Hint 1 of 2
Cancel the letters common to top and bottom, leaving K·A·N·R·O·O over M·E.
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Hint 2 of 2
Make the denominator's two digits divide the numerator while keeping the quotient as small as possible.
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Approach: cancel, then choose digits to minimise the integer quotient
- After cancelling one A and one G, the value is (K·A·N·R·O²)/(M·E) with eight distinct nonzero digits.
- Choosing O = 1 and letting M·E absorb the rest, e.g. K,A,N,R = 2,3,4,6 and M,E = 8,9, gives 2·3·4·6/(8·9) = 2.
- No assignment gives a positive value below 2, so the minimum is 2.
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