Question

What is the highest power of 2 that divides $20!$ completely?

Answer 1

Hint: We start solving the problem by recalling the definition of factorial$(!)$ of a positive number n. Using this definition, we find the value of the given factorial $20!$. We then take the product of even numbers and odd numbers separately. We then factorize the even numbers to get each power of 2. After factorizing every even number, we use the law $a^m.a^n=a^{m+n}$ to get the total power 2 which will be the required highest power.

Complete step-by-step solution:
According to the problem, we need to find the highest power of 2 that divides $20!$ completely.
Let us first recall about the definition of factorial$(!)$. We know that the factorial of a positive number n is denoted by $n!$, is the product of all positive numbers that is less than or equal to n.
i.e., $n!=n\times (n-1)\times (n-2)\times ......\times 2\times 1$. We use this definition to find the value of $20!$.
So, we have $20!=20\times 19\times 18\times 17\times 16\times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$.
Let us take the even product of even numbers and product of odd numbers, as all the even numbers are divisible by 2.
$\Rightarrow 20!=(20\times 18\times 16\times 14\times 12\times 10\times 8\times 6\times 4\times 2)\times (19\times 17\times 15\times 13\times 11\times 9\times 7\times 5\times 3\times 1)$.
Let us factorize the even numbers present in the factorial.
$\Rightarrow 20!=((2^2\times 5)\times (2\times 9)\times 2^4\times (2\times 7)\times (2^2\times 3)\times (2\times 5)\times 2^3\times (2\times 3)\times 2^2\times 2)\times (19\times 17\times 15\times 13\times 11\times 9\times 7\times 5\times 3\times 1)$.
Let us take the product of exponents of 2 separately.
$\Rightarrow 20!=(2^2\times 2\times 2^4\times 2\times 2^2\times 2\times 2^3\times 2\times 2^2\times 2)\times (5\times 9\times 7\times 3\times 5\times 3)\times (19\times 17\times 15\times 13\times 11\times 9\times 7\times 5\times 3\times 1)$.
From law of exponents, we know that $a^m.a^n=a^{m+n}$.
$\Rightarrow 20!=\left(2^{2+1+4+1+2+1+3+1+2+1}\right)\times (5\times 9\times 7\times 3\times 5\times 3)\times (19\times 17\times 15\times 13\times 11\times 9\times 7\times 5\times 3\times 1)$.
$\Rightarrow 20!=\left(2^{18}\right)\times (5\times 9\times 7\times 3\times 5\times 3)\times (19\times 17\times 15\times 13\times 11\times 9\times 7\times 5\times 3\times 1)$.
Let us assume the product other than the exponent of 2 be ‘d’.
$\Rightarrow 20!=\left(2^{18}\right)\times d$ ---(1).
So, we have found that $2^{18}$ divides $20!$ completely from equation (1), which makes 18 is the highest power of 2 that divides $20!$ completely.
$\therefore$ The highest power of 2 that divides $20!$ completely is 18.

Note: We can also solve this by applying step function for the division of 20 and every number which is a power of 2. The process of solving can be seen as detailed as follows:
$\Rightarrow$ Highest power of 2 that divides $20!$ = $\left[{\displaystyle \frac{20}{2}}\right]+\left[{\displaystyle \frac{20}{2^2}}\right]+\left[{\displaystyle \frac{20}{2^3}}\right]+\left[{\displaystyle \frac{20}{2^4}}\right]+\left[{\displaystyle \frac{20}{2^5}}\right]+......$.
$\Rightarrow$ Highest power of 2 that divides $20!$ = $\left[{\displaystyle \frac{20}{2}}\right]+\left[{\displaystyle \frac{20}{4}}\right]+\left[{\displaystyle \frac{20}{8}}\right]+\left[{\displaystyle \frac{20}{16}}\right]+\left[{\displaystyle \frac{20}{32}}\right]+......$.
$\Rightarrow$ Highest power of 2 that divides $20!$ = $\left[10\right]+\left[5\right]+\left[2.5\right]+\left[1.25\right]+\left[0.625\right]+......$.
We know that step function takes the value of the integer that is less than or equal to the number present inside the function.
$\Rightarrow$ Highest power of 2 that divides $20!$ = $10+5+2+1+0+0......$.
$\Rightarrow$ Highest power of 2 that divides $20!$ = 18.
We neglected other terms as we are getting the step function as 0.