2017 AMC 12A Problem 5

Attempt Problem 5 of the 2017 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12A solutions, or check the answer key.

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

5.

At a gathering of 3030 people, there are 2020 people who all know each other and 1010 people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
240240
245245
290290
480480
490490
Answer: B
Concepts:graph theorybasic counting
Small hint:
The 2020 acquaintances only shake hands with the 1010 strangers
Big hint:
Add up each person’s handshakes and divide by 22
Solution:
Each of the 2020 people who know each other shakes hands with only the 1010 strangers. Each of the 1010 strangers shakes hands with all 2929 other people.
Summing handshake counts and dividing by 22 (each handshake involves two people) gives 12(20⋅10+10⋅29)=12(200+290)=245. \begin{aligned} &\dfrac{1}{2}(20\cdot10+10\cdot29) \\ &=\dfrac{1}{2}(200+290)=245. \end{aligned}
Thus, the correct answer is B.
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