Multiple choice

The smallest possible integer that leaves remainders 2, 3, and 4 when divided by 3, 4, and 5, respectively, is

  1. 40

  2. 47

  3. 59

  4. 77
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C Correct answer
Explanation

The number n satisfies n = 3k-1, n = 4m-1, n = 5p-1. Thus n+1 is a multiple of LCM(3, 4, 5) = 60. The smallest positive integer is 60-1 = 59.

AI explanation

Notice that the difference between each divisor and its remainder is 1, since 3 minus 2 is 1, 4 minus 3 is 1, and 5 minus 4 is 1. The least common multiple of 3, 4, and 5 is 60. Since the number is one less than a common multiple of 3, 4, and 5, calculate 60 minus 1 to get 59.

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