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.

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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
Difficulty rating: 1480
Video solution:
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Written solution:

Let a,b,ca,b,c 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 n56.n\ge56. Then b+4c86.b+4c\le86. Also a=1422b5c20, a=142-2b-5c\le20, so 2b+5c122.2b+5c\ge122. Combining these inequalities gives c16,c\le16, but then 2b+5c2(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,b=19,c=17:a=19, b=19, 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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