2017 AMC 8

Complete problem set with solutions and individual problem pages

Problem 12 Medium

The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?

  • A.

    2\text{ and }19

  • B.

    20\text{ and }39

  • C.

    40\text{ and }59

  • D.

    60\text{ and }79

  • E.

    80\text{ and }124

Answer:D

Solution 1

Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder. The \text{LCM}(4,5,6) is 60. Since 60+1=61, that is in the range of \boxed{\textbf{(D)}\ \text{60 and 79}}.

 

Solution 2

Call the number we want to find n. We can say that

n \equiv 1 \mod 4

n \equiv 1 \mod 5

n \equiv 1 \mod 6.

We can also say that n-1 is divisible by 4,5, and 6. Therefore, n-1=\text{LCM}(4,5,6)=60, so n=60+1=61 which is in the range of \boxed{\textbf{(D)}\ \text{60 and 79}}.

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