# Difficult Number System Solved QuestionAptitude Discussion

 Q. The largest number amongst the following that will perfectly divide $101^{100} – 1$ is:
 ✖ A. $100$ ✔ B. $10,000$ ✖ C. $100^{100}$ ✖ D. $100,000$

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.

## (5) Comment(s)

Deepak
()

we have a formula (1+x)^n = 1+nx. So 101^100=(1+100)^100-1=1+(100*100)-1=10^4 which is 10000

Abhinav
()

i understand

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

So you can safely conclude that $(101^10 - 1)$ to $(101^100 - 1)$ will be divisible by 1000.

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

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.