The factorial of a whole number is the function that multiplies the number by every natural number below it. Symbolically, a factorial can be represented by using the symbol “!”. So, ” \(n\) factorial” is the product of the first \(n\) natural numbers and is represented as \(n\) !

The formula to find the factorial of a number is
\(
n !=n \times(n-1) \times(n-2) \times(n-3) \times \ldots . . \times 3 \times 2 \times 1
\)

For an integer \(n \geq 1\), the factorial representation in terms of pi product notation is:
\(
n !=\prod_{i=1}^n i
\)

From the above formulas, the recurrence relation for the factorial of a number is defined as the product of the factorial number and the factorial of that number minus 1 . It is given by:
\(
n !=n \cdot(n-1) !
\)

For example, \(5 !\) can be written as: \(5 !=5 \times 4 \times 3 \times 2 \times 1=120\)

Note: \( 0 !=1\)

Prime Factorization

Since \(n !\) is the product of all positive integers not exceeding \(n\), it is clear that it is divisible by all primes \(p \leq n\), and not divisible by any prime \(p>n\). But what is the power of a prime \(p \leq n\) in the prime factorization of \(n\) !? We can find it as the sum of powers of \(p\) in all the factors \(1,2, \ldots, n\); but rather than counting the power of \(p\) in each factor, we shall count the number of factors divisible by a given power of \(p\). Among the numbers \(1,2, \ldots, n\), exactly \(\left\lfloor\frac{n}{p^k}\right\rfloor\) are divisible by \(p^k\) (here \(\lfloor\cdot\rfloor\) is the floor function). The ones divisible by \(p\) give one power of \(p\). The ones divisible by \(p^2\) give another power of \(p\). Those divisible by \(p^3\) give yet another power of \(p\). Continuing in this manner gives
\(
\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\ldots
\)
for the power of \(p\) in the prime factorization of \(n\) !. The series is formally infinite, but the terms converge to 0 rapidly, as it is the reciprocal of an exponential function.

For example, the power of 7 in 100 ! is just \(\left\lfloor\frac{100}{7}\right\rfloor+\left\lfloor\frac{100}{49}\right\rfloor=14+2=16\left(7^3=343\right.\) is already greater than 100\()\).

Example 1: \(\text { Simplify the factorial expression } \frac{5 !}{2 ! 3 !} \text {. }\)

Solution:

Expanding these factorials, we have:
\(
\frac{5 !}{2 ! 3 !}=\frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)}
\)
We can simplify the numbers from 1 to 3. We can also simplify the 4 with the 2 :
\(
\frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)}=10
\)

Example 2: \(\text { Simplify the expression } \frac{n !}{(n-2) !} \text {. }\)

Solution:

The factorial expression in the numerator is greater than the factorial expression in the denominator, so we expand \(n\) ! partially until the expression \((n-2) !\) appears:
\(
\frac{n !}{(n-2) !}=\frac{n(n-1)(n-2) !}{(n-2) !}
\)
Now, we can cancel the common factors:
\(
\frac{n(n-1)(n-2) !}{(n-2) !}=n(n-1)
\)
We can apply the distributive property to simplify:
\(
n(n-1)=n^2-n
\)

Example 3: Prove that 0 ! is 1

Solution:
Let us write down the factorial formula
\(
n !=n(n-1) !
\)
Let us find the value when \(n=1\)
\(
\begin{aligned}
& 1 !=1(1-1) ! \\
& 1=1 .(0 !) \\
& 1=0 !
\end{aligned}
\)
Hence, \(0 !=1\)

Example 4: If \((n+1) !=5 n !\), solve for \(n\).

Solution:
Given \((n+1) !=5 n\) !
We know that by the factorial formula, \(n !=n ! \times(n-1) !\)
Therefore \((n+1) !=(n+1) \times(n+1-1) !(\) Because here \(n=n+1)\)
But \((n+1) !=5 n !\)
Substituting
\(
\begin{aligned}
& 5 n !=(n+1) \times(n+1-1) ! \\
& 5 n !=(n+1) \times(n) !
\end{aligned}
\)
\(\mathrm{n}\) ! Gets canceled at both LHS and RHS
\(
\begin{aligned}
& 5=n+1 \\
& 5-1=n \\
& n=4
\end{aligned}
\)

Example 5: How many prime divisors does 3! have?

Solution:

Given \(3 !\)
\(
3 !=3 \times 2 \times 1=6
\)
3 ! has 2 prime divisors, which are 3 and 2.

Example 6: \(\text { Find the sum of the factors of (5!). }\)

Solution:

Begin with the prime factorization of (5!): \(5 !=5 \cdot 4 \cdot 3 \cdot 2=2^3 \cdot 3 \cdot 5\), so the sum of the factors is \((2^0+2^1+2^2+2^3)(3^0+3^1)(5^0+5^1)=360\)

Example 7: \(\text { Find the largest power of } 3 \text { that divides (9!) }\)

Solution:

We are looking for the power of 3 in the prime factorization of (9!). Instead of trying to write out the entire prime factorization, we only need to look for powers of 3: There are three numbers in \(9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2\) which contribute 3 as a factor: 3,6, and 9 . 3 and 6 each contribute one 3, while 9 contributes two 3 ‘s for a total of four 3’s. Our answer is, therefore \(3^4\).

Example 8: \(\text { Find the largest power of } 3 \text { that divides ( } 99 \text { !). }\)

Solution:

Every multiple of 3 contributes one three, but every multiple of 9 contributes two 3 ‘s, every multiple of 27 contributes three 3 ‘s, and 81 contributes four 3 ‘s. The counting is easier than it looks. There are 33 multiples of 3 (that’s thirty-three 3’s). There are 11 multiples of 9. This adds 11 more threes. Do you see why this is not 22 more? We have already counted one of the 3 ‘s in each multiple of 9. There are three multiples of 27 (three more 3’s) and one multiple of 81 (one more 3) for a total of \(33+11+3+1=48\) threes or \(3^{48}\).

Example 9: What is the largest possible value of \(n\) for which \(n\) ! is not divisible by 1,024?

Solution:

\(1024=2^{10}\), so we are looking for \(n!\) where there are fewer than 10 two’s in the prime factorization. 2 . 4 . 6 . 8 .10 . 12 is divisible by \(2^{10}\), so \(n\) must be less than 12. \(11!\) is not divisible by 1024 (but \(12! \) is).

Example 10: \(\text { Find the power of } 7 \text { in the prime factorization of (343). }\)

Solution:

We use the floor function here

\(
\left\lfloor\frac{343}{7}\right\rfloor+\left\lfloor\frac{343}{7^2}\right\rfloor+\left\lfloor\frac{343}{7^3}\right\rfloor=49+7+1=57\), or

\(7^{57}\).

Example 11: The year 1849 was the last calendar year that had exactly 3 factors. What is the next calendar year which will have exactly 3 factors?

Solution:

\(1849=43^2\), we are looking for the square of the next prime which is \(47^2=2209\)

Example 12: What is the greatest number of factors that any three-digit number has?

Solution:

\(720=2^4 \times 3^2 \times 5\) has 5 .3 .2 = 30 factors. The smallest integer with 32 factors is \(2^3 \times 3 \times 5 \times 7=840\). The smallest integer with 36 factors is \(2^2 \times 3^2 \times 5 \times 7=1260\). This shows 32 factors is the maximum for any 3-digit integer.

Example 13: \(\text { What is the smallest prime factor of } 1,517?\)

Solution:

When we look for prime factors of 1,517, we check factors up to \(\sqrt{1,517}\) (about 39). 37 would typically) be the last prime factor we check and we find that \(1,517=\mathbf{3 7} \cdot 41\).

Example 14: How many of the factors of 12,321 are perfect squares?

Solution:

\(12,321=111^2=3^2 \cdot 37^2\). Perfect square factors have only even exponents (including zero): \(3^0 \cdot 37^{\circ}=1\), \(3^2 \cdot 37^{\circ}=9,3^0 \cdot 37^2=1,369\), and \(3^2 \cdot 37^2=12,321\) for a total of 4 perfect square factors.

Example 15: What is the sum of all three-digit multiples of 11 which have exactly 10 factors?

Solution:

For a multiple of 11 to have 10 factors, its prime factorization must be in the form \(11^9, 11 \cdot a^4\), or \(a\). \(11^4\). We can rule out \(11^9\) and \(a \cdot 11^4\), as both are larger than 999, so we look for values of a which make \(11 \cdot a^4<1,000\). Only two values work for \(a\) : 2 and 3. \(11 \cdot 2^4=176\) and \(11 \cdot 3^4=891\), so we have \(176+891=\mathbf{1 , 0 6 7}\).