Question

The highest power of 2 by which the product of the first 100 counting numbers can be divided without any remainder is
A. 97
B. 96
C. 95
D. 94

Answer 1

Hint: We find the multiplication of the first 100 natural counting numbers as $100!$. We try to find the number of twos present in that multiplication. We try to break it with the powers of 2 and so we divide 100 with ${2^n}$ and take its box value where $1 \le n \le 6$. We add those numbers to get the solution of the problem.

Complete step-by-step solution:
We need to find the highest power of 2 by which the product of the first 100 counting numbers can be divided without any remainder. The first 100 numbers mean 1 t0 100.
We are taking multiplication of these 100 numbers which gives us $100!$.
So, we have to find the number of 2 present in the multiplication i.e., the power of 2 in the prime factorization of $100!$.
So, if there are 2 presents in the factorization that means that 2 came from a number.
Now if also ${2^2}$ is present in the factorization that means that ${2^2}$ came from a number.
So, we are trying to find which number and how many numbers are divisible by ${2^n}$.
Here, $n$ starts from 1 but to find the maximum value of n, we need to consider the boundary that the value of ${2^n}$ can’t cross 100.
So, the maximum possible value of $n$ will is 6 as ${2^6} = 64$ and ${2^7} = 128$ . There is no number in the first 100 numbers which can contain 7 or more numbers of twos.
Now we will divide the number 100 with $1 \le n \le 6$ and find the box value to add them. So,
$ \Rightarrow \left[ {\dfrac{{100}}{{{2^1}}}} \right] + \left[ {\dfrac{{100}}{{{2^2}}}} \right] + \left[ {\dfrac{{100}}{{{2^3}}}} \right] + \left[ {\dfrac{{100}}{{{2^4}}}} \right] + \left[ {\dfrac{{100}}{{{2^5}}}} \right] + \left[ {\dfrac{{100}}{{{2^6}}}} \right]$
On simplifying the terms, we get
$ \Rightarrow 50 + 25 + 12 + 6 + 3 + 1 = 97$

Hence, option (A) is the correct answer.

Note: We need to understand these 100 numbers are in multiplication which means all twos have been accumulated in one place. The number with a certain number of twos are responsible for bringing up that number. That’s why we divided every number with ${2^n}$ which gives us exact values of numbers that contain 2.