Question : Let $x$ be the least number which, when divided by 5, 6, 7, and 8, leaves a remainder of 3 in each case, but when divided by 9, leaves no remainder. The sum of the digits of $x$ is:
Option 1: 21
Option 2: 22
Option 3: 18
Option 4: 24
Correct Answer: 18
Solution :
Given: The least number which, when divided by 5, 6, 7, and 8, leaves a remainder of 3, and when divided by 9, leaves no remainder.
LCM of 5, 6, 7, 8 = (2 × 5 × 3 × 7 × 2 × 2) = 840
It leaves the remainder of 3 in each case ⇒ 840 + 3 = 843
Now, check if it is divisible by 9 ⇒ $\frac{843}{9}$
It is not divisible by 9.
So, the next number is 840 × 2 + 3 = 1683
Now, check if it is divisible by 9 ⇒ $\frac{1683}{9}$ = 187
So, the sum of the digits of the number 1683 is (1 + 6 + 8 + 3) = 18
Hence, the correct answer is 18.
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