Problem 30 · 2015 Math Kangaroo
Stretch
Number Theory
divisibilityplace-value
Independently from each other, Bill and Bob substitute the letters in the word KANGAROO with numbers, so that the resulting numbers are multiples of 11. They both substitute different letters with different digits and same letters with the same digits (K ≠ 0). Bill obtains the biggest possible number and Bob the smallest possible. In both cases one letter is substituted with the same digit. Which digit is that?
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Answer: D — 5
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Hint 1 of 2
For divisibility by 11 use the alternating digit sum; in KANGAROO the repeated A and O sit in opposite positions and cancel.
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Hint 2 of 2
Build the largest number greedily from the front and the smallest greedily from the front, then see which letter lands on the same digit both times.
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Approach: alternating-digit-sum test for divisibility by 11, then optimise
- Labelling the positions of K A N G A R O O, the alternating sum reduces to \(K+N-G-R\) (the two A's and two O's cancel), so divisibility by 11 means \(K+N-G-R\equiv 0 \pmod{11}\).
- Pushing the largest digits to the front gives Bill's maximum 98758066 (K=9, A=8, N=7, G=5, R=0, O=6), and pushing the smallest gives Bob's minimum 10250933 (K=1, A=0, N=2, G=5, R=9, O=3).
- Comparing the two, the letter G takes the digit 5 in both, so the shared digit is 5 (D).
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