A) 37
B) 36
C) 39
D) 30
Correct Answer: C
Solution :
LCM of 9, 10 and 15 = 90. The multiple of 90 are also divisible by 9, 10 or 15. \[\therefore \] \[21\times 90=1890\] will be divisible by them. \[\therefore \] Now, 1897 will be the number that will give remainder 7. \[1936-1897\] Required number = \[1936-1897\] = 39You need to login to perform this action.
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