Question

The greatest number which can divide 1356, 1868 and 2764 leaving the same remainder 12 in each case, is the perfect square of:

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

The correct option is C 8

Let the greatest number be x.
Then, (1356 - 12), (1868 -12) and (2764 - 12) will be completely divisible by x.
Required number = H.C.F. of (1356 - 12), (1868 -12) and (2764 - 12)
$1344=2^6\times 3\times 7$
$1856=2^6\times 29$
$2764=2^6\times 43$
$\Rightarrow$ Common factors are $2^6=64$
$\therefore$ H.C.F is 64.
$64=8^2$
Therefore, the required number is 8.