Answer:
n = 10
Step-by-step explanation:
Factorial is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.
Therefore, n! represents the product of all positive integers from 1 to n.
[tex]\boxed{n!=n \times(n-1) \times(n-2) \times ... \times 1}[/tex]
Given expression:
[tex]n! = (2^8)(3^4)(5^2)(7)[/tex]
The expression for n! has been given as the product of prime factors.
As n! represents the product of all positive integers from 1 to n, begin by writing out the positive integers from 1 in ascending order as the product of primes (using exponents where possible):
[tex]\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\cline{1-14}\vphantom{\dfrac12}n&1&2&3&4&5&6&7&8&9&10&11&12&13\\\cline{1-14}\vphantom{\dfrac12}\sf Product\;of\;primes&1&2&3&2^2&5&3\cdot 2&7&2^3&3^2&5 \cdot 2&11&2^2\cdot 3&13\\\cline{1-14}\end{aligned}\;\;\sf etc.[/tex]
If we examine the prime products of the given expression, we can see that largest prime number 7 appears only once. Therefore, n must be less than 14, since the next time 7 appears as a prime factor is when 2 · 7 = 14.
The prime number 5 appears twice in the given expression.
From the above table, we can see that the first two times the number 5 is present is (1) on its own, and (2) as a factor of 10. Therefore, n must be equal to or more than 10.
The prime number 3 appears four times in the given expression.
From the above table, we can see that the first four times the number 3 is present is (1) on its own, (2) as a factor of 6, (3) & (4) as both factors of 9.
The 5th time prime number 3 is present is as a prime factor of 12. Therefore, n must be less than 12, else 2⁵ would be a factor of n!.
Therefore, we have determined that 10 ≤ n < 12.
As 11 is a prime number and does not appear in the given expression for n!, we can conclude that n = 10.
We can check this by calculating the given expression and 10!:
[tex]\begin{aligned}n! &= (2^8)(3^4)(5^2)(7)\\&=256 \cdot 81\cdot25\cdot7\\&=20738\cdot25\cdot7\\&=518400\cdot7\\&=3628800\end{aligned}[/tex]
[tex]\begin{aligned}10!&=10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\&=90 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\&=720 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\&=5040 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\&=30240 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\&=151200 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\&=604800 \cdot 3 \cdot 2 \cdot 1\\&=1814400 \cdot 2 \cdot 1\\&=3628800 \cdot 1\\&=3628800\end{aligned}[/tex]
Therefore, this proves that n = 10.
