What is the remainder of dividing 1995 × 1996 × 1997 × 1998 × 1999 by 13?

What is the remainder of dividing 1995 × 1996 × 1997 × 1998 × 1999 by 13?

1995 = 153 * 13 + 61996 = 153 * 13 + 71997 = 153 * 13 + 81998 = 153 * 13 + 91999 = 153 * 13 + 10, the multiple of 13 must be divisible by 13, only consider the result of multiplication of non 13 multiples, that is, 6 * 7 * 8 * 9 * 10 = 30240 = 13 * 2326 + 2, so the remainder is 2

Who can help to solve this problem "1991 × 1993 × 1995 × 1997 × 1999 divided by 13?"
This is a Mathematical Olympiad question in grade five of primary school. You are all right,

The remainder of 1991 × 1993 × 1995 × 1997 × 1999 divided by 13 is that 1991 divided by 13 is more than 2. Similarly, 1993 is more than 41995 more than 61997 more than 81999 more than 10. According to the law of congruence (see related contents of elementary number theory), the remainder of 1991 × 1993 × 1995 × 1997 × 1999 divided by 13 is 2 × 4 × 6 × 8 × 10 divided by

The remainder of 3 ^ 2013 divided by 7

3^2013 = 27^671
= (28-1)^671
= 28^671 - C(1,671)*28^670 + C(2,671)*28^669 -…… + C(670,671)*28 - 1
Except for the last "- 1", all of the above items can be divided by 7, so 3 ^ 2013 is divided by 7 and there are more than 6