Question

How to find the remainder when 51^203 is divided by 7? (using remainder theorem)

Answer 1

Remainder theorem?

Answer 2

Yes, got it from http://www.pagalguy.com/forum/quantitative-questions-and-answers/78107-official-quant-thread-cat-2012-a-820.html

Answer 3

For this, you would want to use Fermat's Little Theorem. This states that if \(p\) is prime, and \(p\nmid a\) then \[a^{p-1}\equiv 1 \mod p\]

Answer 4

could be please provide some insight on how go abt this

Answer 5

First, though, we need to reduce 51 modulo 7. \[51\equiv 2 \mod 7\]Next, we have that \[2^6 \equiv 1 \mod 7\]Now, notice that \(203 = 33\cdot 6+5\) so \(203\equiv 5 \mod 6\). Therefore, \[\large 51^{203}\equiv 2^{203}\equiv (2^6)^{33}\cdot 2^5 \mod 7\]Now we can use Fermat's little theorem to say that this is the same as \[1^{33}\cdot 2^5 \equiv 2^5 \mod 7\]Hence, the solution of the original problem is \[32\equiv 4 \mod 7\]so the remainder is 4

Answer 6

Did that explanation make sense?

Answer 7

just chking it..

Answer 8

could not get this: "now, notice that 203=33⋅6+5 so 203≡5mod6"

Answer 9

We're clear that \(203=33\cdot 6 +5\) correct?

Answer 10

Yes

Answer 11

This means that if you were to divide 6 into 203, you would get 33 with a remainder of 5.