12. Find the last two digits of $7 \times 19 \times 31 \times \cdots \times 1999$. (Here $7,19,31, \ldots, 1999$ form an arithmetic sequence of common difference 12.)
(1 mark)
12. Find the last two digits of $7 \times 19 \times 31 \times \cdots \times 1999$. (Here $7,19,31, \ldots, 1999$ form an arithmetic sequence of common difference 12.)
12. 75
12. Note that $7, 9, 31, \ldots, 1999$ form an arithmetic sequence with 167 terms and common difference 12. Since 12 and 25 are relatively prime, at least one of the terms is divisible by 25.
It remains to consider the product $(\bmod 4)$. Note that each term is congruent to $-1(\bmod 4)$. As there are 167 terms, the product is also congruent to $-1(\bmod 4)$. Since it is divisible by 25, the answer must be 75.