Remainder Theorem

In such a question, always remember that :

100/7 leaves remainder 2 so, (100^2)/7 will leave remainder 2*2 = 4. This is the remainder theorem.

Proceeding in this way, let us take the first part i.e., 2222^5555 / 7.

Now, 2222/ 7 =3. Therefore, 2222^5555 / 7 = 3^5555/ 7.
Now 3^5555 = 3^(5*1111) = 243^1111 (as 3^5= 243).
Therefore, 3^5555/ 7= 243^1111/ 7. Again, 243 / 7 = 5 and 5^1111 / 7 will still give 5 as remainder.
Therefore remainder of the first part is 5.

Taking the second part, 5555^2222/ 7, 5555/7= 4. So, 5555^2222/ 7 = 4^2222/7 = 4^(2*1111) / 7
Similarly as above, 4^ (2*1111) = 16^1111 and 16^1111 / 7 = 2^1111 / 7 ( As 16/7 will give remainder 2).
And remainder of the second part will be 2^1111/7 which will be 2.

So, sum of the remainders of both sides is 5+ 2= 7. This divided by 7 will give remainder 0 which is the answer.

If you have any queries regarding the method, feel free to ask.

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Thanks a lot.....

I have 1 more problem....Please let me know:

Find remainder when 63^2403 is divided by 29.

Please suggest me simple ways to find the answers to questions of this type.
This question reminds me a particular type of problem, but exactly have no clear idea at all.

Birottam Dutta wrote:In such a question, always remember that :

100/7 leaves remainder 2 so, (100^2)/7 will leave remainder 2*2 = 4. This is the remainder theorem.

Proceeding in this way, let us take the first part i.e., 2222^5555 / 7.

Now, 2222/ 7 =3. Therefore, 2222^5555 / 7 = 3^5555/ 7.
Now 3^5555 = 3^(5*1111) = 243^1111 (as 3^5= 243).
Therefore, 3^5555/ 7= 243^1111/ 7. Again, 243 / 7 = 5 and 5^1111 / 7 will still give 5 as remainder.
Therefore remainder of the first part is 5.

Taking the second part, 5555^2222/ 7, 5555/7= 4. So, 5555^2222/ 7 = 4^2222/7 = 4^(2*1111) / 7
Similarly as above, 4^ (2*1111) = 16^1111 and 16^1111 / 7 = 2^1111 / 7 ( As 16/7 will give remainder 2).
And remainder of the second part will be 2^1111/7 which will be 2.

So, sum of the remainders of both sides is 5+ 2= 7. This divided by 7 will give remainder 0 which is the answer.

If you have any queries regarding the method, feel free to ask.

------------------

If you find one of my posts helpful, please take a moment to click on the "Thank" icon.

R(2222/7) = 3 and R(5555/7) = 4

R = R(3^5555+4^2222)/7 = R(3*3^5554+2^4444)/7 = R(3*9^2777+2*8^1481)/7 = R(3*2^2777+2)/7
= R(4*3*8^925+2)/7 = R(12+2)/7 = 0