Problem 30 · 2016 Math Kangaroo
Stretch
Number Theory
factorizationfactor-pairs
The positive whole number N has exactly six different (positive) factors, including 1 and N. The product of five of these factors is 648. Which of these numbers is the sixth factor of N?
Show answer
Answer: C — 9
Show hints
Hint 1 of 3
Divisors pair up to multiply to N, so the product of all of them is a power of N.
Still stuck? Show hint 2 →
Hint 2 of 3
With six divisors, all six multiply to \(N^3\).
Still stuck? Show hint 3 →
Hint 3 of 3
Then the sixth factor is \(N^3\) divided by the given product 648.
Show solution
Approach: product of all divisors = N^(d/2)
- For \(N\) with 6 divisors, the product of all six is \(N^3\) (divisors pair up to give \(N\)).
- So the sixth factor is \(N^3 / 648\), meaning \(648\) must equal \(N^3\) divided by one divisor.
- Trying \(N = 18\): its factors \(1,2,3,6,9,18\) multiply to \(5832 = 18^3\), and \(5832 / 648 = 9\).
- So the missing sixth factor is 9 (C).
Mark:
· log in to save