Question

What is the remainder when 50^51 is divided by 53?

Answer 1

the reminder will 47.17^50

Answer 2

Hey friend .
The correct answer is 35. I’ve cross checked this from wolfram alpha.

Following is the method :

In the first step we should note that 5050 and 5353are co-prime and that we can use the theorem given under.

If aa and bb are co-prime and ee is the Euler’s number for b, then remainder when akeb=1akeb=1, where kk is a natural number.

The Euler’s number for a prime is one less than the prime itself.

In the given case, e=52e=52 as 5353 is a prime. So, we can conclude that

rem (505253)=1(505253)=1

or, rem (50×505153)=1(50×505153)=1

Let the rem (505153)=R(505153)=R

then, rem (50×R53)=1(50×R53)=1

⇒⇒ 50R=53m+150R=53m+1

⇒⇒ R=53m+150R=53m+150

Now, we need to find a suitable mm, such that the RHS is a natural.

Clearly, we have to look for multiples ending in 33, as only then the last digit will be 99, which is required as we have to add 11 to it. After adding 11 the number must end with 00as the numerator has to be divisible by 5050. So, we can start the trials from 33, 1313, 2323 and so on.

It’s found that 53×33=174953×33=1749

⇒⇒ R=53×33+150R=53×33+150

⇒⇒ R=1749+150R=1749+150

⇒⇒ R=175050R=175050

⇒⇒ R=35
I hope it helps