IMC 2015 Q10
10. What is the remainder when \(2^2 \times 3^3 \times 5^5 \times 7^7\) is divided by 8?
- A 2
- B 3
- C 4
- D 5
- E 7
Intermediate Mathematical Challenge 2015, Question 10. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 4
\(2^2 \times 3^3 \times 5^5 \times 7^7\) is of the form \(2^2 \times\) an odd number. It therefore has the form \(4(2n + 1) = 8n + 4\) where \(n\) is a positive integer and so leaves a remainder of 4 when divided by 8.
Solution © UK Mathematics Trust, from the IMC 2015 solutions. Longer versions with investigations: Solutions and investigations (PDF).