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The smallest number, which when divided by 5, 10, 12 and 15. leaves remainder 2 in each case: but when divided by 7 leaves no remainder, is

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The smallest number, which when divided by 5, 10, 12 and 15. leaves remainder 2 in each case: but when divided by 7 leaves no remainder, is
1). 189
2). 182
3). 175
4). 91

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2 answers

3 vote
Answered by on | Votes 3 |
option 2 is the right answer

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2 vote
Answered by on | Votes 2 |
LCM of 5, 10, 12, 15 LCM = 2 × 3 × 5 × 2 = 60 \ Number = 60k + 2 Now, the required number should be divisible by 7. Now, 60k + 2 = 7 × 8k + 4k + 2 If we put k = 3, (4k + 2) is equal to 14 which is exactly divisible by 7. Required number = 60 × 3 + 2 = 182




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