When (67^67 + 67) is divided by 68 then what is the remainder?
When (67^67 + 67) is divided by 68 then what is the remainder?
When (6767 + 67) is divided by 68 then what is the remainder?
Last edited by Ronnie on Sun Jan 24, 2016 5:13 pm, edited 1 time in total.

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Re: When (67^67 + 67) is divided by 68 then what is the remainder?
There are mainly two methods to solve this problem.
1) By polynomial method
Let us see the polynomial expansions...
an + bn = (a + b)(b0an1  b1an2 + ...  bn2a1 + bn1a0) (For n>1 and odd)
an  bn = (a  b)(b0an1 + b1an2 + ... + bn2a1 + bn1a0)
So from above polynomial expansions we can say that
(xn + 1) is divisible by (x + 1) only when n is odd
So (6767 + 1) is divisible by (67 + 1)
i.e (6767 + 1) is divisible by 68
Now for the given problem we can write
(6767 + 67) = (6767 + 1) + 66
but (6767 + 1) is divisible by 68
So when (6767 + 1) + 66 is divided by 68 then it will give 66 as remainder.
2)By trial and error method
22 + 2 divided by 3
= 4 + 2
=6
so when 6 divided by 3 then remainder = 0
So by doing calculations like this we can get
22 + 2 divided by 3 then remainder = 0
33 + 3 divided by 4 then remainder = 2
44 + 4 divided by 5 then remainder = 0
55 + 5 divided by 6 then remainder = 4
So by observing above examples we can say
xx + x is divided by (x+1) then the remainder is (x1) where x is odd number
So now we can say when 6767 + 67 is divided by 68 then remainder is 66.
1) By polynomial method
Let us see the polynomial expansions...
an + bn = (a + b)(b0an1  b1an2 + ...  bn2a1 + bn1a0) (For n>1 and odd)
an  bn = (a  b)(b0an1 + b1an2 + ... + bn2a1 + bn1a0)
So from above polynomial expansions we can say that
(xn + 1) is divisible by (x + 1) only when n is odd
So (6767 + 1) is divisible by (67 + 1)
i.e (6767 + 1) is divisible by 68
Now for the given problem we can write
(6767 + 67) = (6767 + 1) + 66
but (6767 + 1) is divisible by 68
So when (6767 + 1) + 66 is divided by 68 then it will give 66 as remainder.
2)By trial and error method
22 + 2 divided by 3
= 4 + 2
=6
so when 6 divided by 3 then remainder = 0
So by doing calculations like this we can get
22 + 2 divided by 3 then remainder = 0
33 + 3 divided by 4 then remainder = 2
44 + 4 divided by 5 then remainder = 0
55 + 5 divided by 6 then remainder = 4
So by observing above examples we can say
xx + x is divided by (x+1) then the remainder is (x1) where x is odd number
So now we can say when 6767 + 67 is divided by 68 then remainder is 66.
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