2004 AMC 8 Problem 19

Attempt Problem 19 of the 2004 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 8 solutions, or check the answer key.

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19.

A whole number larger than 22 leaves a remainder of 22 when divided by each of the numbers 3,4,5,3, 4, 5, and 6.6. The smallest such number lies between which two numbers?

40 and 4940\ \text{and}\ 49

60 and 7960 \text{ and}\ 79

100 and 129100\ \text{and}\ 129

210 and 249210\ \text{and}\ 249

320 and 369320\ \text{and}\ 369

Answer: B
Concepts:least common multiplemodular arithmetic
Difficulty rating: 1220
Solution:

Let xx be the number. Then x2x - 2 is divisible by 3,4,5,3, 4, 5, and 6.6.

The least common multiple of these numbers is 60,60, which makes xx 62.62.

Thus, B is the correct answer.

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