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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Question : Let $x$ be the least number of 4 digits that when divided by 2, 3, 4, 5, 6 and 7 leaves a remainder of 1 in each case. If $x$ lies between 2000 and 2500, then what is the sum of the digits of $x$?
Question : What is the sum of the digits of the least number which when divided by 15,18, and 36 leaves the same remainder 9 in each case and is divisible by 11?
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Question : What is the least number which when divided by 15, 18 and 36 leaves the same remainder 9 in each case and is divisible by 11?
Question : The least number, which when divided by 5, 6, 7 and 8 leaves a remainder of 3, but when divided by 9, leaves no remainder, is:
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