question_answer

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

A)  189     

B)  182

C)  175                             

D)  91

Correct Answer: B

Solution :

LCM of 5, 10, 12, 15

\[\therefore \]      LCM \[=2\times 3\times 5\times 2=60\]

\[\therefore \]      Number \[=60\text{ }k+2\]

Now, the required number should be divisible by 7.

Now,

\[60\,\,k+2=7\times 8k+4k+2\]

If we put k = 3, \[(4k+2)\]is equal to 14 which is exactly divisible by 7.

\[\therefore \] Required number age\[=60\times 3+2=182\]