Multiple choice

Let y be the least 4 digit-number between 2000 and 2500 that when divided by 2, 3, 4, 5, 6 and 7 leaves a remainder of 1 in each case. What is the sum of the digits of y?

  1. 9

  2. 15

  3. 10

  4. 4
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D Correct answer
Explanation

The number y must be 1 more than a multiple of LCM(2, 3, 4, 5, 6, 7) = 420. Multiples of 420 are 420, 840, 1260, 1680, 2100, 2520. The range 2000-2500 includes 2100. Thus, y = 2100 + 1 = 2101. The sum of the digits is 2 + 1 + 0 + 1 = 4.

AI explanation

The required number y must be 1 more than a common multiple of 2, 3, 4, 5, 6, and 7. The least common multiple of these divisors using prime factorization is 420, making the general form of such numbers 420k + 1. Testing values for k within the range of 2000 to 2500 gives k = 5, which results in y = 2101. The sum of its digits is 2 + 1 + 0 + 1 = 4.

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