2025 AMC 8 Problem 16

Attempt Problem 16 of the 2025 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

Five distinct integers from 11 to 1010 are chosen, and five distinct integers from 1111 to 2020 are chosen. No two numbers differ by exactly 10.10. What is the sum of the ten chosen numbers?
9595
100100
105105
110110
115115
Answer: C
Concepts:bijectionpairing and grouping
Small hint:
Pair each number from 11 to 1010 with the number 1010 larger
Big hint:
The chosen upper numbers correspond to the unchosen lower numbers
Video solution:
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Written solution:
Call the integers from 11 to 1010 inclusive the lower range, and call the integers from 1111 to 2020 inclusive the higher range.
Each of the 55 distinct numbers chosen from the lower range blocks out the number in the higher range that is exactly 1010 more than itself. There are only 1010 numbers in the higher range, so there are only 10−5=510 - 5 = 5 numbers not yet blocked.
We need to choose 55 distinct numbers from the higher range, so the numbers chosen from the higher range are precisely those which are not yet blocked. They are each exactly 1010 more than a not-chosen number in the lower range.
So, the sum of the 55 distinct numbers chosen from the higher range is exactly 5×10=505 \times 10 = 50 more than the sum of the 55 not-chosen numbers in the lower range.
The sum of all 1010 chosen numbers is therefore equal to 5050 plus the sum of all chosen and not-chosen numbers in the lower range 1+2+⋯+10.1 + 2 + \dots + 10.
The sum of the numbers from 11 to 1010 is 10(10+1)2=55,\frac{10 (10+1)}{2} = 55, so the answer is 50+55=105,50 + 55 = 105, or choice C.
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