Elitmus
Exam
Category
find maximum remainder when (104)^n divided by 51.
a.32 b.26 c.2 d.16
Read Solution (Total 10)
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- a.32
(104)^n / 51
= (51*2 +2)^n / 51
=> rem = 2^n
2^n must be less than 51
so, for maxm value of rem, n=5
max rem = 2^5 = 32 - 10 years agoHelpfull: Yes(23) No(0)
- ans is c .
- 10 years agoHelpfull: Yes(3) No(5)
- Ans is. a)32
- 10 years agoHelpfull: Yes(3) No(1)
- (102 + 2) mod 51 gives remainder 2 and for the same (104)^n mod 51 give (2)^n remainder and till power of 5 its gives remainder in greater term..i.e (2)^5 = 32 bcoz as power exceeds the no will become divisible by 51 ....like 2^5*2 or 2^5*2*2
- 10 years agoHelpfull: Yes(2) No(2)
- factors of 51=3*17
104/3=34 quotient so 3*34=102...,2 as remainder
104/17=6 quotient so17*6=102 ...,2 as remainder - 10 years agoHelpfull: Yes(1) No(4)
- must be 32. As when we divide 104 from 51 getting 2 as remainder so max multilpe of 2 upto 51 is 32 thats why 32 is ans..
- 10 years agoHelpfull: Yes(1) No(0)
- 32 is ans...
- 10 years agoHelpfull: Yes(1) No(0)
- (104)^n %51= (51+53)^n%51= 53^n%51= (51+2)^n%51= 2^n%51.
n should be 5 so gives max rem. as 32... - 10 years agoHelpfull: Yes(0) No(0)
- a.32
(104)^n/51
=(2)^n/51
so max power of n=5 because 2^5=32,
so a is the right ans - 10 years agoHelpfull: Yes(0) No(0)
- ans-32
(102+2)^n
remainder=2^n
now for the largest remainder 2^n must be less than 51
power of 2 which is less than 51 is 2^5 i.e
32 - 10 years agoHelpfull: Yes(0) No(0)
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