Difficulty 47

let x=47k+11 x^2=47k^2+121+2k 47 11 since 121 is not fully divisible by 47,so the reminder will be 121-2*47=27

The simple number to be assumed as x is 11 itself. because 11 | 47 = 11. so, 121-94 = 27. <b>

you dont need formula for these things.. its a kinda problem to be solved without pen and paper.

Let the number be $x$ .

Now ATQ $\large x\equiv 11\quad mod\quad 47$ .

Therefore, ${ x }^{ 2 }\equiv { 11 }^{ 2 }\quad mod\quad 47\quad =\quad 121mod47$ . And $121\equiv 27\quad mod\quad 47$ Hence answer is 27

Its so simple. Remember the euclids division lemma. Now, when we square both the sides we get 47 47q + 2 11 47q + 11 11 = x^2 Since 121 is the remainder and it should be less than 47 we divide it by 47 which gives us the remainder 27.

Let the number be 58 i.e. 47+11(remainder). Thus 58 58 = 3364. 3364 = 71 47+27. Where 27 is the remainder.

When they say a number, You believe them, assume a number which when divided by 47 gives remainder 11 and do what they say, square it and find remainder when divided by 47.. As simple as that!!. As Bharanidharan said, you should not use pen/paper for it :)

x=47k+11

${ x }^{ 2 }$ = ${ 47k }^{ 2 }$ + $2\times 47k\times 11$ + $11^{2}$

Since the last term is the only term not divisible by 47

121 mod 47 = 27

if x%47=11; => x=47k +11; (k is some positive integer)

=>x^{2} = (47k +11)^{2}

=> x^{2}= (47k)^{2} +121 + 2x47xk

=> x^{2} = (47k)^{2} +47x2+27 + 2x47xk

=> x^{2}= 47(47k^{2} + 2 + 2k) + 27

=> x^{2}%47=27.

hence, 27 is the remainder left when x^2 is divided by 47