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Find the smallest number of two digits which on being divided by 2, 3, 4 and 5 leaves 0, 1, 2 and 3 as remainder respectively.
- A55
- B58
- C61
- D65
Correct answer: B
Discussion 1
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CCommunity
58
LCM of 2, 3, 4 and 5 is 60.
The smallest number of two digits is 10.
But we also see that the smallest number divisible by 2, 3, 4 and 5 is 60. Therefore, 10 has no significance here.
The difference between divisor and remainder is 2 in each case. So, we subtract 2 from 60.
60 - 2 = 58
LCM of 2, 3, 4 and 5 is 60.
The smallest number of two digits is 10.
But we also see that the smallest number divisible by 2, 3, 4 and 5 is 60. Therefore, 10 has no significance here.
The difference between divisor and remainder is 2 in each case. So, we subtract 2 from 60.
60 - 2 = 58