How many numbers exist from 1 to 100 such that on division by both 3 and 4, the remainder is 2 in each case?
How many numbers exist from 1 to 100 such that on division by both 3 and 4, the remainder is 2 in each case?
Option 1 -
8
Option 2 -
9
Option 3 -
10
Option 4 -
12
Asked by Shiksha User
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1 Answer
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Correct Option - 2
Detailed Solution:In this problem, we are looking for the LCM of the divisors, that is, 3 and 4 or any of their multiple to which the desired remainder 2 can be added. The smallest such number is 2. The others will be 14, 26, 38, 50, 62, 74, 86 and 98, that is there are 9 such numbers
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