Question : Let $x$ be the least number which when divided by 8, 9, 12, 14 and 36 leaves a remainder of 4 in each case, but $x$ is divisible by 11. The sum of the digits of $x$ is
Option 1: 5
Option 2: 6
Option 3: 9
Option 4: 4
Correct Answer: 4
Solution :
Given numbers = 8, 9, 12, 14, and 36
LCM(8, 9, 12, 14, 36) = 2 × 2 × 2 × 3 × 3 × 7 = 504
Let the required number = (504k + 4) which is divisible by 11,
⇒ (504k + 4) → $\frac{ (9k + 4)}{11}$
Putting k = 1, 2, 3, 4,……..
when k = 2
⇒ $\frac{(9 × 2 + 4)}{11}$ = $\frac{22}{11}$ = 2
Now,
504k + 4 is divisible by 11 when k = 2
Required number = 504 × 2 + 4
= 1008 + 4
= 1012
Sum of digits of 1012 = 1 + 0 + 1 + 2 = 4
Hence, the correct answer is 4.
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