2 will be ans

take 6 for ex 6^100 will end up in 6 add one and see XXX7 divide by 6-1 = 5 get remainder 2...its an dirty question never to be tested in gmat, trick it contains is vision to see the divisibility with 5

in general the remainder of 1/n will be 1 for n>, the remainder of 1/(n-1) will be 1 the remainder of n/n-1 will be 1

so breaking the rational number in parts we get:

1/(n-1) + n^100/(n-1)

the first part will have a remainder of 1 and remainder of second part is 1 so sum of the remainders 2 - as long as the divisor is >2 then we will always have remainder =2

would go for C

did a long division
n^100+1/n-1=n^99+n^98..........1, leaves a remainder 2
(there emerges a pattern after the first 2 steps)

Since n leaves a remainder 1 when divided by n-1, the remainder of n^100 when divided by n-1 is also 1.

Hence the remainder of 1 + n^100 = 2

dtweah wrote:If n is a positive integer, n > 4, then what is the remainder if 1+ n^100 is divided by n – 1?
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4

Lots of interesting perspectives to this question. Here is the official OA:

c) This may be observed from the identity
n^100 + 1 = (n – 1)(n99 + n98 + n97 + … + n + 1) + 2. (Note: n can be any positive integer except 1).

My Approach:

I love to turn stuff into binomial to figure out remainders. My goal is to make the first term of a binomial equal to the denominator and aim for 1 as the second term. Using this procedure I write

((n-1 +1)^100)/n-1 + 1/n-1

Once in this form I know all terms of binomial will be factors of n-1 exept last term which will be 1, leaving 2/n-1.

Take for example

(17^2000 +5)/12. What is the remainder. Using the logic above

(17^2000) /12 +5/12

(12 + 5)^2000 /12 + 5/12

Everything in the first term evens out except the last term which will be 5^2000. Reason: You will have 2000Cn where n=0,1,2....2000 and these will be multiplied by 12^2000, 12^1999....12^0 and 5^1......5^2000. So you know all terms are factors of 12 except the last term.

I do similar thing to 5^2000/12=25^1000 /12 =(24+1)^1000 /12 and again everything evens out except 1. The beauty of 1 is that 1^n =1.

So you will have some interger G + 1/12 + 5/12 = G+ 1/2, with remainder 1

With some practise you can get remainders of such fractions with exponents very quickly without writing anything.

maihuna wrote:2 will be ans

take 6 for ex 6^100 will end up in 6 add one and see XXX7 divide by 6-1 = 5 get remainder 2...its an dirty question never to be tested in gmat, trick it contains is vision to see the divisibility with 5

maihuna,

why did u take 6?

if i take 5, i get the remainder 1..

how do u know which is the smart number to choose?