Question

The highest power of 2 by which the product of 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 answer:
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 actually find the number of 2 present in the multiplication i.e. the power of 2 in the prime factorisation of $100!$.
So, if there are 2 presents in the factorisation that means that 2 came from a number.
Now if also ${{2}^{2}}$ is present in the factorisation 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 value of ${{2}^{n}}$ can’t cross 100.
So, the maximum possible value of n will be 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 ${{2}^{n}},1\le n\le 6$ and find the box value to add them.
So, \[\left[ \dfrac{100}{{{2}^{1}}} \right]+\left[ \dfrac{100}{{{2}^{2}}} \right]+\left[ \dfrac{100}{{{2}^{3}}} \right]+\left[ \dfrac{100}{{{2}^{4}}} \right]+\left[ \dfrac{100}{{{2}^{5}}} \right]+\left[ \dfrac{100}{{{2}^{6}}} \right]=50+25+12+6+3+1=97\].

So, the correct answer is “Option A”.

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.