13another simple objective method... [6]
substitute n={ur lucky no.*} [3]............ (*make sure dat it is as small as possible) [4]
lets take a simple one..
what is the remainder when
(1+2+3+.... n)n is divided by n-1?
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17 Answers
13
62think of this one
what is the remainder when
(1+a)n = 1+a.{polynomial of a}
hence when divided by a, the remainder will be 1
more generally
(x+a)n = xn+a.{polynomial of a}
hence when divided by a, the remainder will be same as remiander of xn on division by a
13applying binomial expansion... every term of the expansion (1+Σ(n-1) )n will be divisible by (n-1) except the first term which is nC0.1n = 1... so the remainder will be the remainder obtained by dividing 1 with n-1...
62Honey..
what is the remainder when 3 is divided by 2...
go back a few classes.. when we did not study fractions..
when we used to do quotient and remainder....
I know why u are confused :)
Or if this is not ur doubt.. then please re ask..
13@honey...
now after concluding dat d remainder is 1/n-1 ... find d remainder when 1 is divided by n-1... since remainder must not have divisor in its denominator(basic condition) ... so 1=(n-1).0 + 1... hence remainder is one...
hope u got it now... [1]
1yeah mak u are correct ,,, i chceked that way only....
aur agar isme options rehta... toh solve bhi nahi karte.. :P put kar dete :P
par nishant bhaiya is obviously intelligent enuf to make these all subjective :P
1bt whn u divide........isn't it whole [1 + Σ(n-1)]/n-1
so hw cld u get remainder 1?????
in tht way remainder shd be 1/n-1
13[nn.(n+1)n / 2n] / (n-1)
[nn.(n-1+2)n / 2n] / (n-1)
expanding and dividing with n-1 implies remainder of d entire term = reminder of nn/n-1...
(n-1+1)n / n-1 .... simplifying and dividing gives remainder of given term = remainder of 1/n-1 which is 1...
let me know if i'm wrong...
62honey .. maq is right :)
If you dont understand this one.. do try to understand it .. it is not very complex.. but very useful indeed !
13yeah got it... 1+2+3+...+n = 1 + Σ(n-1)...
since Σ(n-1) is divisible by n-1... the remainder is 1... [1]
62hmm.. this is good.. but did u make a simpler solution complex?
I mean there is a still more simpler solution :)