Fermat little theorem says, ap−1p remainder is 1.
ie., 3010 or 810when divided by 11 remainder is 1.
The unit digit of 7287 is 8 (using cyclicity of unit digits) Click here
So 7287 = 10K + 8
30(10K+8)11=(3010)K.30811=1k.30811
8811=22411=(25)4.2411=1611=5

307287mod11307287mod11
=(22+8)7287mod11=(22+8)7287mod11
=87287+22kmod11=87287+22kmod11 where k is some natural number
=87287mod11=87287mod11 as 22 is divisible by 11
using Euler's theorem, we know that
810mod11=1810mod11=1
So now we need to find
7287mod107287mod10 to proceed
using a process similar to steps 2, 3, and 4, we arrive at
287mod10287mod10
Now take a look at 2's powers mod 10:
21mod10=222mod10=423mod10=824mod10=625mod10=226mod10=421mod10=222mod10=423mod10=824mod10=625mod10=226mod10=4
As you can see, the answer repeats every 4 powers. So,
287mod10=23mod10=8287mod10=23mod10=8
back to the original question,
87287mod11=88mod1187287mod11=88mod11
=644mod11=94mod11=812mod11=644mod11=94mod11=812mod11
=42mod11=16mod11=5

This should take some thought. Let's start with 30 and work our way up. When you divide 30 by 11, you get a quotient of 2 and a remainder of 8.

How about 302302 ? It's 30 times 30. Each 30 divided by 11 gives a remainder of 8. So

30⋅30=(2⋅11+8)(2⋅11+8)=∗⋅11+8230⋅30=(2⋅11+8)(2⋅11+8)=∗⋅11+82

where * is some number we don't care about. And 82=6482=64 gives a remainder of 9 when divided by 11. Aha! The remainder for 302302 is the same as the remainder for 82.82.

Does that work for all powers? Is the remainder for 30n30n the same as the remainder for 8n8n ? After a few minutes, you'll see it's true. That means we can reduce the original question to

How do you determine what remainder 8728787287 gives when divided by 11?

Let's make a table of the remainders of various powers of 8.

8882983684485108638728858978101811888283848586878889810811896410325718

Aha! It repeats every 10.

That means if we want to know what 8728787287 is, we only have to look at the exponent 72877287 and ignore everything except the last digit, that is, the remainder after dividing it by 10. (Funny, isn't it? We started dividing by 11, but now we're going to divide the exponent by 10.)

We know that dividing 72 by 10 gives 2, and we can use the same principle we found about to reduce the problem of finding the remainder of 72877287 to that of finding the remainder of 287.287.

We can make a table again, this time for the remainders of various powers of 2 when divided by 10.

2222423824625226427828622223242526272824862486

This time it repeats every 4. So to find the remainder of 287287 after dividing by 10, we only need the remainder of 87 after dividing by 4. That's 3. Since 2323 gives a remainder of 8 after dividing by 10, so 287287 also gives a remainder of 8 after dividing by 10. And so does 7287.7287.

Now we can go back to the earlier question. 8728787287 when divided by 11 gives the same as 8888 when divided by 11, and according to our first table, that remainder was 5.

Therefore, we have the answer to the original question. The remainder of 307287307287 when divided by 11 is 5.

Fermat little theorem says, ap−1pap−1p remainder is 1.
ie., 30103010 or 810810when divided by 11 remainder is 1.
The unit digit of 72877287 is 8 (using cyclicity of unit digits) Click here
So 72877287 = 10K + 8
30(10K+8)11=(3010)K.30811=1k.3081130(10K+8)11=(3010)K.30811=1k.30811
8811=22411=(25)4.2411=1611=5