Find the last two digits of 7 × 19 × 31 ×......× 1999.

Let $\displaystyle P = 7 \times 19 \times 31 \times \cdots \times 1999 = \prod_{n=0}^{166} (12n+7)$ . We need to find $P \bmod 100$ . Let us use Chinese remainder theorem by finding $P \bmod 4$ and $P \bmod 25$ .

• Since $12n + 7 \equiv 2n + 2 \equiv 0 \text{ (mod 5)}$ or $12n + 7$ is divisible by $5$ when $n=4, 9, 14, ...$ , $\implies P \equiv 0 \text{ (mod 25)}$ .
• And we have $\displaystyle P \equiv \prod_{n=0}^{166} (12n+7) \equiv \prod_{n=0}^{166} 3 \equiv 3^{167} \equiv (4-1)^{167} \equiv -1 \equiv 3 \text{ (mod 4)}$ .
• $\implies 4k+3 \equiv 0 \text{ (mod 25)}$ , where $k$ is an integer. $\implies k = 18$ and $P \equiv 4(18) + 3 \equiv \boxed{75} \text{ (mod 100)}$ .

All the numbers in the list are odd, and more than one is a multiple of $5$ , so the product is an odd multiple of $25$ and ends in either $25$ or $75$ .

There is an odd number of entries in the list, and each one is of the form $4k+3$ , so their product is also. This eliminates $25$ as a possibility, and means the answer must be $\boxed {75}$ .