Question

Find the last two digits: $(201\times 202\times 203\times 204\times 246\times 247\times 248\times 249{)}^2$

• A. 36
• B. 56
• C. 76
• D. 16

Answer 1

The correct option is C 76

For this question we need to find the remainder of :
$(201\times 202\times 203\times 204\times 246\times 247\times 248\times 249)\times (201\times 202\times 203\times 204\times 246\times 247\times 248\times 249)$ divided by 100.
$\frac{(201\times 202\times 203\times 204\times 246\times 247\times 248\times 249)\times (201\times 202\times 203\times 204\times 246\times 247\times 248\times 249)}{100}=\frac{(201\times 202\times 203\times 204\times 246\times 247\times 248\times 249)(201\times 101\times 203\times 102\times 246\times 247\times 248\times 249)}{25}$
$\to \frac{(1\times 1\times 3\times 2\times -4\times -3\times -2\times -1)\times (1\times 2\times 3\times 4\times -4\times -3\times -2\times -1)}{25}=\frac{144\times 576}{25}$
$\to \frac{(19\times 1)}{25}$=remainder 19.
$19\times 4=76$ is the actual remainder (since we divided by 4 during the process of finding the remainder.