2004 AMC 8 Problem 19

Below is the professionally curated solution for Problem 19 of the 2004 AMC 8, 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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Concepts:least common multiple modular arithmetic

Difficulty rating: 1220

19.

A whole number larger than $2$ leaves a remainder of $2$ when divided by each of the numbers $3, 4, 5,$ and $6.$ The smallest such number lies between which two numbers?

$40\ \text{and}\ 49$

$60 \text{ and}\ 79$

$100\ \text{and}\ 129$

$210\ \text{and}\ 249$

$320\ \text{and}\ 369$

Solution:

Let $x$ be the number. Then $x - 2$ is divisible by $3, 4, 5,$ and $6.$

The least common multiple of these numbers is $60,$ which makes $x$ $62.$

Thus, B is the correct answer.

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