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?
Option 1: 1269
Option 2: 1071
Option 3: 1089
Option 4: 1080
Correct Answer: 1089
Solution : 15 = 3 × 5 18 = 2 × 3 × 3 36 = 2 × 2 × 3 × 3 LCM of 15, 18, 36 = 2 × 2 × 3 × 3 × 5 = 180 Required number = $180x + 9$ So, $180x + 9$ is a multiple of 11. ⇒ $176x + 4x + 9$ is a multiple of 11 ⇒ $4x+9$ is a multiple of 11 ⇒ $4×6+9 = 33$ is a multiple of 11 ⇒ $x = 6$ $\therefore$ The required number = 180 × 6 + 9 = 1089 Hence, the correct answer is 1089.
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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?
Option 1: 15
Option 2: 16
Option 3: 18
Option 4: 17
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
Question : Find the least number which when divided by 4, 9, 12, and 15, leaves the remainder 3 in each case.
Option 1: 360
Option 2: 183
Option 3: 193
Option 4: 180
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:
Option 1: 1677
Option 2: 1683
Option 3: 2523
Option 4: 3363
Question : A number, when divided by 15 and 18 every time, leaves 3 as a remainder, the least possible number is:
Option 1: 83
Option 2: 103
Option 3: 39
Option 4: 93
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