Question

Find the last two digits of: $15\times 37\times 63\times 51\times 97\times 17.$

• A. 35
• B. 45
• C. 55
• D. 85

Answer 1

The correct option is A 35

$15\times 37\times 63\times 51\times 97\times 17$ on division by 100 would give us a remainder that would be equal to its' last 2 digits. First we can divide the numerator and the denominator by 5 to get the expression:
$\frac{3\times 37\times 63\times 51\times 97\times 17}{20}$ using remainder theorem $\to \frac{3\times 17\times 3\times 11\times 17\times 17}{20}=\frac{51\times 33\times 289}{20}$ using remainder theorem $\to \frac{11\times 13\times 9}{20}=\frac{143\times 9}{20}\to \frac{3\times 9}{20}=\frac{27}{20}\to$ Remainder =7. Hence, the required remainder $=7\times 5=35,$ which would also be the last two digits of the given number.