Elitmus
Exam
Numerical Ability
Algebra
x lies from 121 < X >1331, x^2+1 when divided by 11 what will not be the remainder .
A.1 b.4 c.6 d.8
Read Solution (Total 6)
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- answer is d.
i have such used hit and trial method.
part 1:
see,x is greater than 121 means (11)^2 and less than 1331 i.e less than than (37)^2 [as 37^ is 1369,so take value till (36)^2=1296]
=>x could be from 12^2 to 36^2
part 2:
(x^2+1)/11
(i)put x=12 in the equation
( 12^2+1)/11-->R=2(so not beneficial as it not in the option)
go on checking....X=13,14,15,16....
(ii)x=15
R=6
(iii)x=16
R=4
(iv)think when would be remainder 0 ,it has to be 11 ka square or multiple of 11 ka square,as 11 ka square cannot be taken,next multiple of 11 would be 22,put x=(22^2+1)/11-->R=1
(u will have to think a bit)
now the only left option is d i.e 8
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if u know theoretically not by hit and trial plz share ur knowledge - 7 years agoHelpfull: Yes(8) No(2)
- x lies between 11^2 and 11^3
SUPPOSE
x=11^2
then x^2+1 become 1332
1332/11 gives rem 1
BUT X CANT BE 11^2
SO WRONG REM IS 1 - 7 years agoHelpfull: Yes(1) No(5)
- i have made a program for this question and then only 8 is not comming in the reminder
so the answer is d)8 - 7 years agoHelpfull: Yes(1) No(0)
- For any range x^2+1 when divided by 11 forms a cyclic pattern which is
1 2 5 10 6 6 10 5 2 1 1 2 5 10 6 6 10 5 2 1 and so on.
So we can see that 8 never occurs - 7 years agoHelpfull: Yes(1) No(3)
- 11*11
- 7 years agoHelpfull: Yes(0) No(2)
- #inlcude
int main()
{
int x,r;
x=122;
while(x - 6 years agoHelpfull: Yes(0) No(1)
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