Problem 11 · 2009 AMC 8
Medium
Number Theory
gcd
The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43. Some of the 30 sixth graders each bought a pencil, and they paid a total of $1.95. How many more sixth graders than seventh graders bought a pencil?
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Answer: D — 4 more.
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Hint 1 of 2
The number of buyers in each grade must be a WHOLE number = (total paid) ÷ (one price). So the single pencil price has to divide BOTH totals evenly. Work in cents: 143 and 195.
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Hint 2 of 2
A number dividing both 143 and 195 is a common divisor — factor each and look for what they share.
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Approach: the price is a common divisor of both totals
- Switch to cents so everything's whole: $1.43 = 143¢, $1.95 = 195¢. The price p must divide both (buyers = total ÷ p has to come out whole).
- Factor: 143 = 11 × 13 and 195 = 3 × 5 × 13. Their only shared factors are 1 and 13.
- Rule out p = 1¢: at 1¢ each, 195 sixth graders would have bought — but there are only 30. So p = 13¢.
- Seventh graders: 143 ÷ 13 = 11. Sixth graders: 195 ÷ 13 = 15. Difference = 15 − 11 = 4.
- Why this transfers: "equal items, total cost, unknown unit price" means the price divides every total — reach for common divisors (gcd), then use side conditions (like "at most 30 kids") to pick the right one.
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