Known: a = 19911991 Q: what is the remainder of a divided by 13? Clear process, easy to understand, thank you!

If 1991 is divided by 13 and the remainder is 2, then the remainder of 19911991 divided by 13 is equal to the remainder of 21991 divided by 13, which is 819911991. The remainder of 1991 divided by 13 is equal to the remainder of 81991 divided by 13, which is 0. Then, the next number, which is composed of four 1991 numbers, is the same as the remainder of 1991, which is 2

What is the remainder of 9 divided by 13 in 1997? Question 2: what is the remainder of 1991 divided by 74?
Let me add. It's 99999999... There are 1997 nines. It's not 1997 times 9
It's 19911991, 19911991... 1991. 1991. Not 1991 by 1991
It's that simple. I've already done it

999999÷13=76923
1997÷6=332…… five
99999÷13=7692…… three
199119911991÷37=5381619243
1991÷3=663…… two
19911991÷37=538161…… thirty-four
And 538161 (74 / 37) = 269080 one
1×37+34=71
The remainder is 3 and 71

What is the remainder of 1991 divided by 13? Please write why

Let 19911991.1991 (1991) = a, then
1991 / 13 remainder 2
21991 / 13 remainder 8
81991 / 13 remainder 0
That is to say, 199119911991 divided by 13 can be divisible
The remainder of 1991 / 3 is 2
So the remainder of a / 13 = 19911991, the remainder of a / 13 = 8

It is known that a = 19911991.1991 (a total of 1991), a divided by 13, what is the remainder?
It is known that a = 19911991.1991 (a total of 1991), a divided by 13, what is the remainder?

Find the law: first, except the first four 1991, we can see: 19911991, 19911991 / 13 = 153169163070153, more than 2
It is found that the rule is 153169163070 repeated, there are 12 numbers in total, and there are 1991 times 4 = 7964 numbers in a, using 7964 minus 1 to remove an ordinal bit, and then dividing by 12 to get 663 cycles, the remaining 7, that is, the last bit of the number is 1, so the remainder is 8

Known: a = 19911991 Q: what is the remainder of a divided by 13? Clear process, easy to understand, thank you!