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Question: The number \[{101^{100}} - 1\] is divisible by A. 100 B. 1000 C. 10000 D. 100000...

The number 101100−1{101^{100}} - 1 is divisible by
A. 100
B. 1000
C. 10000
D. 100000

Explanation

Solution

In this particular type of question first, expand (101)100{\left( {101} \right)^{100}} using binomial expansion to eliminate -1 from the given equation in the question. Then take the maximum common part outside to check the divisibility of the given number with the available options.

Complete answer:
We know that the binomial expansion of (a+b)n=an+(nC1)an−1b1+(nC2)an−2b2+........+(nCn−1)a1bn−1+bn{\left( {a + b} \right)^n} = {a^n} + \left( {{}^n{C_1}} \right){a^{n - 1}}{b^1} + \left( {{}^n{C_2}} \right){a^{n - 2}}{b^2} + ........ + \left( {{}^n{C_{n - 1}}} \right){a^1}{b^{n - 1}} + {b^n} , where a and b are the constant terms.
And according to the formula of combinations nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}
Therefore, we can write 101100−1{101^{100}} - 1 as (1+100)100−1{\left( {1 + 100} \right)^{100}} - 1
Now applying the binomial expansion formula in the above equation.
⇒(1+100)100−1=1100+(100C1)1001+(100C2)1002+(100C3)1003+........+(100C100)100100−1\Rightarrow {\left( {1 + 100} \right)^{100}} - 1 = {1^{100}} + \left( {{}^{100}{C_1}} \right){100^1} + \left( {{}^{100}{C_2}} \right){100^2} + \left( {{}^{100}{C_3}} \right){100^3} + ........ + \left( {{}^{100}{C_{100}}} \right){100^{100}} - 1
Now we had to solve the RHS of the above equation.
As the first term of the RHS of the equation is 1 and last term is -1. So, cancelling both the terms.
⇒(100C1)1001+(100C2)1002+(100C3)1003+........+(100C100)100100\Rightarrow \left( {{}^{100}{C_1}} \right){100^1} + \left( {{}^{100}{C_2}} \right){100^2} + \left( {{}^{100}{C_3}} \right){100^3} + ........ + \left( {{}^{100}{C_{100}}} \right){100^{100}} (1)
Now as we know by the above stated formula to find the value of nCr{}^n{C_r}, that 100C1=(100)!(1)!(99)!=100{}^{100}{C_1} = \dfrac{{\left( {100} \right)!}}{{\left( 1 \right)!\left( {99} \right)!}} = 100
So, putting the value of 100C1{}^{100}{C_1} in equation 1.
⇒100×100+(100C2)1002+(100C3)1003+........+(100C100)100100\Rightarrow 100 \times 100 + \left( {{}^{100}{C_2}} \right){100^2} + \left( {{}^{100}{C_3}} \right){100^3} + ........ + \left( {{}^{100}{C_{100}}} \right){100^{100}}
⇒1002+(100C2)1002+(100C3)1003+........+(100C100)100100\Rightarrow {100^2} + \left( {{}^{100}{C_2}} \right){100^2} + \left( {{}^{100}{C_3}} \right){100^3} + ........ + \left( {{}^{100}{C_{100}}} \right){100^{100}}
Now as we know that in each term of the above equation the power of 100 is more than or equal to 2.
So, we can take 1002{100^2} (i.e. 10000) form the above equation.
So, (101)100−1=10000[1+(100C2)1002+(100C3)1003+........+(100C100)100100]{\left( {101} \right)^{100}} - 1 = 10000\left[ {1 + \left( {{}^{100}{C_2}} \right){{100}^2} + \left( {{}^{100}{C_3}} \right){{100}^3} + ........ + \left( {{}^{100}{C_{100}}} \right){{100}^{100}}} \right]
So, 101100−1{101^{100}} - 1 will be divisible by 10000.
So, if 101100−1{101^{100}} - 1 is divisible by 10000, then it must be divisible by 100 and 1000, because 10000 is divisible by 100 and 1000.
So, if this will be a multi correct question then the correct options will be A, B and C.
But if the question is a single correct question then the best option will be C.

Note: Whenever we face such a type of question then we have to recall the method of binomial expansion to solve these types of questions. Note that the binomial expansion was done to eliminate the extra non divisible part (i.e. -1) of the question which is the key trick to remember in this particular question. And after that we had to expand the first term using formula nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} . And then take the maximum common part outside. This will be the easiest and efficient way to find the solution of the problem.