Discussion
Let \(x\) be the least number which when divided by 11, 21, 24 and 28, leaves a remainder of 10 in each case. If \(x\) is divisible by 37, what will be the sum of the digits of \(x\)?
Answer: C
The sum of the digits of \(x\) is 16. The least number divisible by 11, 21, 24 and 28 is their LCM: \[ \text{LCM}(21,24,28) = 168,\quad 168 \times 11 = 1848 \] So \(x = 1848k + 10\). Since \(1848 = 37\times 49 + 35\), we have \(1848 \equiv -2 \pmod{37}\), and \(x\) divisible by 37 needs \(10 - 2k \equiv 0\), so \(k = 5\) (the least positive value). \[ x = 1848\times 5 + 10 = 9250 = 37\times 250 \] \[ 9 + 2 + 5 + 0 = 16 \] The options 13, 15 and 17 do not match the digit sum of 9250.
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