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by gmatmachoman » Mon Feb 15, 2010 10:40 am
The largest number amongst the following that will perfectly divide {(101)^100} - 1 is


A.100
B.10,000
C.100100
D.100,000
E.1000
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by shashank.ism » Mon Feb 15, 2010 10:54 am
gmatmachoman wrote:The largest number amongst the following that will perfectly divide {(101)^100} - 1 is


A.100
B.10,000
C.100100
D.100,000
E.1000
This question has been discussed in the post :
https://www.beatthegmat.com/largest-numb ... 15519.html
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by vijay_venky » Mon Feb 15, 2010 11:01 pm
Actually this method could be solved by using the binomial expansion as well.

(a+b)^n=nc0*a^n.b^0+nc1*a^n-1*b^1+.......+ncn-1*a^1*b^n-1+ncn*a^0*b^n.

Going by this expansion,the 101^100 expansion looks like

100c0*100^100+100c1*100^99+......+100c99*100+100c100.

101^100-1 means the last digit of 101^100 that is 100c100 has to be deleted.
Let us look at the least number,(the number with least number of zeros). Because all other numbers are of higher powers pf 100, 100c99.100 should be the least number. This is 10000. So this expansion contains 4 zeros at the end and so divisible by 10000.

Typical CAT problem.
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by komal » Tue Feb 16, 2010 8:46 am
gmatmachoman wrote:The largest number amongst the following that will perfectly divide {(101)^100} - 1 is


A.100
B.10,000
C.100100
D.100,000
E.1000
101^2 = 10201. 101^2 - 1 = 10200. This is divisible by 100. Similarly try for 101^3 - 1 = 1030301 - 1 = 1030300.

So you can safely conclude that (101^1 - 1) to (101^9 - 1) will be divisible by 100 & then (101^10 - 1) to (101^99 - 1) will be divisible by 1000 & therefore (101^100 - 1) will be divisible by 10,000.

(B) is the answer
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