Problem 14 · 2021 Math Kangaroo
Stretch
Number Theory
divisibility
2021 has a remainder of 5 when divided by 6, by 7, by 8, and by 9. How many positive integers, less than 2021, have this property?
Show answer
Answer: A — 4
Show hints
Hint 1 of 2
A number leaving remainder 5 under all of 6,7,8,9 is 5 more than a common multiple of them.
Still stuck? Show hint 2 →
Hint 2 of 2
The least common multiple of 6,7,8,9 is 504; count multiples of 504 (plus 5) below 2021.
Show solution
Approach: reduce to multiples of the lcm
- Such a number has the form 504k + 5, since lcm(6,7,8,9) = 504.
- For k = 0,1,2,3 this gives 5, 509, 1013, 1517 — all below 2021.
- k = 4 would give 2021 itself, which is not less than 2021.
- So there are 4 such integers, choice (A).
Mark:
· log in to save