Number System
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Q.

The largest number amongst the following that will perfectly divide $101^{100} – 1$ is:

 A.

$100$

 B.

$10,000$

 C.

$100^{100}$

 D.

$100,000$

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Solution:
Option(B) is correct

The easiest way to solve such problems for objective exam purposes is trial and error or by back substituting answers in the choices given.

$101^2 = 10,201$. 

$101^2 - 1 = 10,200$. This is divisible by 100. 

Similarly try for $101^3 - 1 = 1,030,301 - 1 = 1,030,300$.

So you can safely conclude that $(101^1 - 1)$ to $(101^9 - 1)$ will be divisible by 100.

$(101^{10} - 1)$ to $(101^{99} - 1)$ will be divisible by 1000. 

Therefore, $(101^{100} - 1)$ will be divisible by 10,000.

Edit: For an alternative solution using binomial expansion, check comment by Anuj Mittal.


(3) Comment(s)


Soumik
 ()

I think it will be 100. As $101^{100} -1 = (100+1)^{100} -1$

$=(100^{100}+^{100}C_1 100^{99} + ...+1-1$

$= 100$ (.....)


Prateek
 ()

Soumik you are right, but question says LARGEST number. One of the option choices clearly mentions $10,000$ as a choice. Which if true will indicate that is has to be divisible by $100$ as well.


Anuj Mittal
 ()

we can also do this question by binomial theorem.

consider $101^100=(100+1)^100$ and apply binomial expansion the last two no.s in the expansion we get are 10000 and 1 and all the other no.s in the expansion are greater than 10000 and are also multiple of 10000 , so the last 5 digits we get are 10001 and subtracting 1 from it gives 10000.