Let x be the least number which when divided by 11, 21, 24 and 28 leaves a…
2023
Let x be the least number which when divided by 11, 21, 24 and 28 leaves a remainder of 10 each time. If x is divisible by 37, what is the sum of the digits of x?
- A.
13
- B.
15
- C.
16
- D.
17
Attempted by 53 students.
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Correct answer: C
We need the smallest x such that x gives remainder 10 when divided by 11, 21, 24 and 28, and x is divisible by 37.
Key idea: x − 10 must be a common multiple of 11, 21, 24 and 28.
Compute the least common multiple: 11 is prime; 21 = 3·7; 24 = 2^3·3; 28 = 2^2·7. Taking the highest powers gives LCM = 2^3·3·7·11 = 1848.
So x = 1848k + 10 for some integer k ≥ 0.
Require x divisible by 37: 1848k + 10 ≡ 0 (mod 37). Reduce 1848 mod 37: 1848 ≡ 35 ≡ −2 (mod 37).
Thus −2k + 10 ≡ 0 ⇒ 2k ≡ 10 (mod 37) ⇒ k ≡ 5 (mod 37). The smallest nonnegative k that works is k = 5.
Then x = 1848·5 + 10 = 9250. Sum of digits = 9 + 2 + 5 + 0 = 16.
Answer: 16