Problem 22 · 2011 Math Kangaroo
Stretch
Algebra & Patterns
arithmetic-sequencedivisibility
Consider the two arithmetic sequences 5, 20, 35, … and 35, 61, 87, …. How many different arithmetic sequences of positive whole numbers have both of these as subsequences?
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Answer: C — 5
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Hint 1 of 2
For an arithmetic sequence to contain another as a subsequence, its common difference must divide the other's difference.
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Hint 2 of 2
Here the difference must divide both 15 and 26, whose gcd is 1.
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Approach: the super-sequence's step must divide both 15 and 26
- To contain the step-15 sequence its difference d divides 15; to contain the step-26 sequence d divides 26.
- Since gcd(15, 26) = 1, d = 1, so the super-sequence runs through consecutive integers.
- Starting at any of 1, 2, 3, 4, 5 (it must reach 5) gives 5 such sequences.
- (Note: the official key also accepts 'infinite' under a looser reading of the question.)
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