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2023 AMC 10B Problem 11

Problem 11 of 25IntermediateNumber TheoryCombinatorics

Suzanne went to the bank and withdrew $800.\$800. The teller gave her this amount using $20\$20 bills, $50\$50 bills, and $100\$100 bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?

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Solution

Let a,b,c≥1a, b, c \ge 1 count the $20,$50,$100\$20, \$50, \$100 bills. Then 20a+50b+100c=800,20a + 50b + 100c = 800, which divides down to 2a+5b+10c=80.2a + 5b + 10c = 80. Both 2a2a and 10c10c are even, so 5b5b is too, forcing b=2t.b = 2t. Now a=40−5t−5c≥1a = 40 - 5t - 5c \ge 1 means t+c≤7.t + c \le 7. With t,c≥1,t, c \ge 1, the pairs number 1+2+⋯+6=21.1 + 2 + \cdots + 6 = 21. Thus, B is the correct answer.
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Tagged: Diophantine Equation · parity · basic counting

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