Problem 18 · 2010 Math Kangaroo
Stretch
Number Theory
primesgrouping
Each star in the expression 1 ∗ 2 ∗ 3 ∗ 4 ∗ 5 ∗ 6 ∗ 7 ∗ 8 ∗ 9 ∗ 10 is either replaced by a “+” or a “×”. Let N be the biggest number possible that can be obtained this way. What is the smallest prime factor of N?
Show answer
Answer: E — Another number
Show hints
Hint 1 of 3
Multiplying by 1 is wasteful — adding the 1 instead makes the result bigger.
Still stuck? Show hint 2 →
Hint 2 of 3
So N = 1 + (2×3×…×10); the product is even, so the +1 makes N odd.
Still stuck? Show hint 3 →
Hint 3 of 3
An odd N can't have factor 2; check whether 3, 5 or 7 divide it before settling on the answer.
Show solution
Approach: maximise, then factor the result
- Each factor from 3 to 10 should be multiplied, but for the 1 note that 1+P > 1×P, so the 1 is added.
- Thus N = 1 + 2×3×4×…×10 = 1 + 3&,628&,800 = 3&,628&,801.
- Since the product is even, N is odd, so 2 is out; and 3&,628&,800 is a multiple of 3, 5 and 7, so N = product + 1 is divisible by none of them either.
- Factoring, 3&,628&,801 = 11 × 329&,891, so its smallest prime factor is 11 — another number.
Mark:
· log in to save