What is the remainder when ((55)^15!)^188 is divided by 17

Why did you plug the remainder of 4 into the problem? I understand 55/17 gives remainder of 4 and I see where you went with the problem but I don't understand why you put the remainder back into the problem? Im lost!

When a number a is divided by another number b, the quotient would be some number q and we would get a remainder r. Since we need to find the remainder, we can ignore the quotient and feed in the remainder. On feeding in the remainder and raising it to the power, we can find the total remainder.

Hope this helps! Let me know if you need any further notification.

Thanks Kris, I appreciated it! I am just trying to apply the concept to something else but how would I go about this problem? What is the remainder when ((17)^10)^123) is divided by 3? 17/3 has a remainder of 2, feeding in 2 into the problem how would I then find the remainder? Thanks again!

We need to find the remainder when ((17)^10)^123) is divided by 3
As you suggested let us divide, 17/3 and feed in the remainder back into the question.

((2)^10)^123)= (1024)^123/3
You can divide 1024 by 3. The remainder is 1. Hence 1^123=1

This was possible because the remainder of 17/3 was a small number,2

Another way of doing this is

((2)^10)^123)=((3-1)^10)^123
When you divide the data inside the bracket by 3, 3 would get completely divided by 3 and hence can be ignored.

((-1)^10)^123
Any negative number raised to an even integer= positive number

1^123=1

Hope that helps!

[quote="Richard0715"]Thanks Kris, I appreciated it! I am just trying to apply the concept to something else but how would I go about this problem? What is the remainder when ((17)^10)^123) is divided by 3? 17/3 has a remainder of 2, feeding in 2 into the problem how would I then find the remainder? Thanks again!

I would always try to play with 1 wherever it is possible to reduce the time spent on calculations. In your question, 17 can be written as (18-1) where 18 is divisible by 3 and the rest about -1 is already explained by Kris01.

Hope this helps

Hello.. I understand till the point you plugged in 4 in the expression.
But post that aren't you only simplifying the expression (4^15!)^188 ? instead of calculating the remainder when divided by 17.
I hope I have expressed my doubt clearly. Basically I just lost the context after you plugged in 4 in the expression though I understood why it is done.
But after this, I feel you have merely simplified the expression whereas our main task was to find the remainder.
Kindly elaborate.

Remainder of 55/17=4

The problem reduces to ((4)^15!)^188 divided by 17

(4^(4*2*3*5*6*...15)^188)

We see the remainders of 4^4 by 17 is 1.

So the remainder for the whole expression is 1.

Great explanation!!!
I was able to solve till this step:4^{(15!*188)} and then I was struck!

Can we solve your example above like this:

(2^10)^123 = 2^1230 ... 2 has cycle of 2, 4 , 8 , 6 .. means 4..1230/4 = 370 + 2...2 remaining means last digit will be 4.. 4/3 leaves a remainder of 1.

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