Elitmus
Exam
Numerical Ability
Number System
1.what is the maximum possible remainder when a number 104^n is divided by 51 ? (Assume n be any positive integer)
(a)29 (b) 32
(c) 96 (d)96
Read Solution (Total 9)
-
- here is Sol URL---> http://postimg.org/image/3xvnp5b7d/
- 9 years agoHelpfull: Yes(35) No(0)
- Ans (b)=32
let n=1 then 104/51= 2(rem)
for getting a max rem put n=5, because rem(2)*5= 32 which is less than 51
- 9 years agoHelpfull: Yes(27) No(0)
- since 51×2=102
104-102=2
=>2^n will be the remainder
since n is a positive integer and viewing the options we can say that 32 is the only possible answer 😊 - 9 years agoHelpfull: Yes(8) No(0)
- Ans---b----32
- 9 years agoHelpfull: Yes(5) No(0)
- (b)
unit place of (104)^n is either 4 or 6.
then when is divide by 51 then reminder having unit place will be either 2 or 4.
so here 32 will be answer....
for exam, n = 1, 104/51 reminder is 2.
but for n =2, 104*104/51 = 10816/51 reminder will be 4 and so on.....
so for proper answer, unit place of reminder should be either 2 or 4... - 9 years agoHelpfull: Yes(2) No(0)
- ans is 171 naveen umar u forget to add these no. 110,220,330,440,550,660,770,880,990
162 + 9=171 - 9 years agoHelpfull: Yes(1) No(5)
- ans= (b)=32
- 9 years agoHelpfull: Yes(0) No(0)
- ans is (b)
- 9 years agoHelpfull: Yes(0) No(0)
- naveen bhai thank u for upload image
- 8 years agoHelpfull: Yes(0) No(0)
Elitmus Other Question