Problem 24 · 2020 Math Kangaroo
Stretch
Number Theory
divisibilityfactorization
A positive integer N is divisible by every integer from 2 to 11 except for two of them. Among the pairs \((6,7)\), \((7,8)\), \((8,9)\), \((9,10)\) and \((10,11)\), how many could be that pair of exceptions?
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Answer: B — 2
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Hint 1 of 2
If N misses two numbers, the LCM of the remaining nine must fail to divide each missing one.
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Hint 2 of 2
Test each candidate pair by building the LCM of the others.
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Approach: check each pair against the LCM of the kept numbers
- For a pair to be the exception, N can be the LCM of the other nine numbers and must NOT be divisible by either removed number.
- Testing each pair, only (7,8) and (8,9) work; the others leave an LCM still divisible by one of the removed numbers.
- So the count of possible exception pairs is 2.
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