2023 AMC 8 Problem 14

Attempt Problem 14 of the 2023 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 2023 AMC 8 solutions, or check the answer key.

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

14.

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of 55-cent, 1010-cent, and 2525-cent stamps, with exactly 2020 of each type. What is the greatest number of stamps Nicolas can use to make exactly $7.10\$7.10 in postage?
(Note: The amount $7.10\$7.10 corresponds to 77 dollars and 1010 cents. One dollar is worth 100100 cents.)
4545
4646
5151
5454
5555
Answer: E
Concepts:optimizationmodular arithmetic
Small hint:
Use as many low-value stamps as possible
Big hint:
All 55- and 1010-cent stamps total 300300 cents, leaving a nonmultiple of 2525
Video solution:
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Written solution:
Let a,a, b,b, cc be the numbers of 55-, 1010-, and 2525-cent stamps, and let n=a+b+c.n=a+b+c. Dividing the value equation by 55 gives a+2b+5c=142, a+2b+5c=142, so n+b+4c=142.n+b+4c=142.
Suppose n≥56.n\ge56. Then b+4c≤86.b+4c\le86. Also a=142−2b−5c≤20, a=142-2b-5c\le20, so 2b+5c≥122.2b+5c\ge122. Combining these inequalities gives c≤16,c\le16, but then 2b+5c≤2(20)+5(16)=120,2b+5c\le2(20)+5(16)=120, a contradiction. Thus at most 5555 stamps can be used.
The bound is attainable with a=19,a=19, b=19,b=19, c=17:c=17: 19(5)+19(10)+17(25)=710. 19(5)+19(10)+17(25)=710. Hence the greatest possible number of stamps is 19+19+17=55.19+19+17=55.
Thus, E is the correct answer.
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