SSC
Government Jobs Exams
Numerical Ability
Number System
The remainder ,when 17^37+29^37 is divided by 23 is
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- it is in the form of 2(a^x+b^x)/(a+b)..
when x=even there will be some remainder. and when x= odd.
remainder will be zero. - 11 years agoHelpfull: Yes(5) No(1)
- ZERO(0) is the remainder when 17^37+29^37 is divided by 23.
- 11 years agoHelpfull: Yes(3) No(0)
- @ GLORY ....wat i meant was..
consider a=17,b=29.
so (a^x+b^x)/(a+b)/2....where (a+b)/2 is nothing but 23.
not put x=any odd no.say 1 then u wil get some no remainder..
hope u understood.. - 11 years agoHelpfull: Yes(1) No(2)
- @ Naagar please explain it in detail, i did not get it.
- 11 years agoHelpfull: Yes(0) No(1)
- 17^((4*9)+1)+29^((4*9)+1)= (0+17)+(0+29)=46/23=2 remainder 0
- 11 years agoHelpfull: Yes(0) No(2)
- 17^37+29^37
17^1+29^1=46 here 17 cycle is 4 then 17^37=17^1,29 cycle is 2 then 29^37=29^1
46 divided by 23 we get remainder is 0 - 11 years agoHelpfull: Yes(0) No(3)
- Everybody sucks here :(
- 11 years agoHelpfull: Yes(0) No(0)
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