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2012 Math Kangaroo

Problem 6

Problem 6 · 2012 Math Kangaroo Medium
Algebra & Patterns substitution

In a list of five numbers the first number is 2 and the last one is 12. The product of the first three numbers is 30, of the middle three 90 and of the last three 360. What is the middle number in that list?

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Answer: C — 5
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Hint 1 of 2
The first three multiply to 30 and the first number is 2, so you know the product of numbers 2 and 3.
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Hint 2 of 2
Use the last-three product and the known last number to pin down number 4, then chain back.
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Approach: peel products from both ends
  1. Label the list \(a,b,c,d,e\) with \(a=2\) and \(e=12\). Since \(abc=30\), we get \(bc = 15\).
  2. From \(bcd=90\): \(d = \frac{90}{bc} = \frac{90}{15} = 6\); and \(cde=360\) gives \(cd = \frac{360}{12} = 30\).
  3. Then \(c = \frac{cd}{d} = \frac{30}{6} = 5\), so the middle number is \(5\), choice C.
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