2003 AMC 12B Problem 7

Below is the professionally curated solution for Problem 7 of the 2003 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

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Concepts:system of equationsDiophantine Equationbounding to limit cases

Difficulty rating: 1430

7.

Penniless Pete's piggy bank has no pennies in it, but it has 100100 coins, all nickels, dimes, and quarters, whose total value is $8.35.\$8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?

00

1313

3737

6464

8383

Solution:

Let n,n, d,d, qq be the numbers of nickels, dimes, quarters. Then n+d+q=100n + d + q = 100 and n+2d+5q=167n + 2d + 5q = 167 (dividing the value equation by 55).

Subtracting gives d+4q=67,d + 4q = 67, so d=674q.d = 67 - 4q.

The largest dd is at q=0,q = 0, giving d=67d = 67 (with n=33n = 33). The smallest occurs at q=16,q = 16, giving d=3d = 3 (with n=81n = 81). The difference is 673=64.67 - 3 = 64.

Thus, the correct answer is D.

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