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Numerical Ability
Number System
if 8^25 divided by 7 then what is reminder
a]25
b]1
c]0
d].....
Read Solution (Total 13)
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- 8^25/7
8 can be written as 7*1+1
(7*1+1)^25/7
(ax+1)^n/a always gives the remainder 1
here a=7,x=1,n=25
Hence remainder is 1
Ans : 1
- 10 years agoHelpfull: Yes(59) No(0)
- 8^1 = 8
8^2 = 4
8^3 = 2
8^4 = 6
this pattern contiues 8,4,2,6 8,4,2,6
8^25 = 8. 8 when divided by 7 gives remainder 1
answer is b - 10 years agoHelpfull: Yes(13) No(0)
- We can use cyclic remainder theorem
8^1 % 7 = 1
8^2 % 7 = 1
8^3 % 5 = 1
8^4 % 5 = 1
.
.
.
Again
8^n % 7 = 1
So the 1st cycle is upto the power of "4"...
So the remainder of 8^35 % 8 will be (4k + n)th value...
25 => 4*6+1 [k=6, n=1]
If n=1, rem=1
If n=2, rem=1
If n=3, rem=1
If n=4, rem=1
Hence the remainder is 1
Ans : 1
- 10 years agoHelpfull: Yes(7) No(0)
- if (a+1)^n/a then remainder is always 1
- 10 years agoHelpfull: Yes(4) No(0)
- 8^25 = 2^75
2,4,8,16,32,64,128,256,512,... are powers of 2. Starting with
power remainder
2^3 1
2^4 2
2^5 4
2^6 1
hence for power 3,6,9,12..... remainder is 1.
whenever 2^3n is divided by 7.. remainder is 1. - 10 years agoHelpfull: Yes(4) No(0)
- We can use cyclic remainder theorem
8^1 % 7 = 1
8^2 % 7 = 1
8^3 % 7 = 1
8^4 % 7 = 1
.
.
.
Again
8^n % 7 = 1
So the 1st cycle is upto the power of "4"...
So the remainder of 8^25 % 7 will be (4k + n)th value...
25 => 4*6+1 [k=6, n=1]
If n=1, rem=1
If n=2, rem=1
If n=3, rem=1
If n=4, rem=1
Hence the remainder is 1
Ans : 1
- 10 years agoHelpfull: Yes(3) No(4)
- we know that
(a+1)^n/a will always give 1 as remainder for all natural no of a and n.
so ans=1
(7+1)^25/7=1 - 10 years agoHelpfull: Yes(0) No(0)
- ans:
since 8/7=1(rem)
64/7=1(rem)
512/7=1(rem)
:
:
8^25=1(rem) - 10 years agoHelpfull: Yes(0) No(0)
- option (b) is correct
8^25/7=1^25/7=1/25
so if 1 divided by 7 then rem=1 - 10 years agoHelpfull: Yes(0) No(0)
- by formula 8^25/7=(7+1)^25/7=1
- 10 years agoHelpfull: Yes(0) No(0)
- ans=1
8^1/7=remainder=1
8^2/7=remainder=1
8^3/7=remainder=1
so 8^25/7=remainder=1
- 10 years agoHelpfull: Yes(0) No(0)
- 25/4 here 7 cylicity is 4 hence remainder is 1
- 8 years agoHelpfull: Yes(0) No(0)
- Here reminder is. 1
- 6 years agoHelpfull: Yes(0) No(0)
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