2012 AMC 8 Problem 15

Attempt Problem 15 of the 2012 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 8 solutions, or check the answer key.

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15.

The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?

40 and 50 40\text{ and }50

51 and 55 51\text{ and }55

56 and 60 56\text{ and }60

61 and 65 61\text{ and }65

66 and 99 66\text{ and }99

Answer: D
Concepts:least common multiplemodular arithmetic
Difficulty rating: 1240
Video solution:
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Written solution:

Let the number be x.x. Since it leaves a remainder of 22 when divided by 3,4,5,6,3,4,5,6, we know x2x-2 is a multiple of 3,4,5,6.3,4,5,6. This means x2x-2 is a multiple of lcm(3,4,5,6)lcm(3,4,5,6) which is 60.60. Therefore, x2x-2 must be a multiple of 60.60. The next number such that this occurs is when x2=60    x=62.x-2 = 60 \implies x = 62 .

Thus, the answer is D .

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