Multiple choice

Let x be the least 4-digit number which when divided by 2, 3, 4, 5, 6 and 7 leaves a remainder of 1 in each case. If x lies between 2800 and 3000, then what is the sum of the digits of x?

  1. 15

  2. 16

  3. 12

  4. 13
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B Correct answer
Explanation

LCM of 2, 3, 4, 5, 6, 7 is 420. Multiples of 420: 420, 840, 1260, 1680, 2100, 2520, 2940, 3360. Number x = 420k + 1. For 2800 < x < 3000, 2940 + 1 = 2941. Sum of digits = 2+9+4+1 = 16.

AI explanation

The number x must be 1 more than a common multiple of 2, 3, 4, 5, 6, and 7. The least common multiple of these divisors is 420, so the number takes the form 420k + 1. Testing multipliers to find a value between 2800 and 3000 yields k = 7, giving x = 2941. The sum of the digits is 2 + 9 + 4 + 1 = 16.

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