\(10! – (2)(5!)^2 = 5!*(6*7*8*9*10 - 2*120) = 5!*240*(7*2*9-1) = 5^2*24*48*(125) = 5^5*24*48\)



i.e. if \(5^n\) is divisor of \(10!-2*5!^2\) then,

\(n_{max} = 5 \)

Answer: Option D
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\(10!- 2(5!)^2\) = \(5! [(10*9*8*7*6)- (2*5!)]\)

= \(5! [(2*5*3*3*2*4*7*2*3)- (2*5!)]\)

= \(5! [(2*5*4*3*2*2*7*3*3)- (2*5!)]\)

= \(5![(2*5!*126) - 2(5!)]\)

=\(2*5!*5!(126-1)\)

=\(2*5!*5!*5^3\)

= \(2*4!*5*4!*5*5^3\)

n= 1+1+3=5

\(10! = 5!* 6*7*8*9*10 = 5!*(2*3)*7*(2*4)*9(2*5)= (5!)^{2}* 252\)

--> \(10! – (2)(5!)^2= (5!)^{2}* 252 - (2)(5!)^2= (5!)^2 (252-2)= 250*(5!)^2= 2*5^{3}*(5!)^{2}\)
\(5^{n} =5^{5}\)
--> \(n =5 \)

Answer (D).

\(10! – (2)(5!)^2=5!(10*9*8*7*6-2*5!)\)

\(5!(2*5*3*3*2*2*2*7*2*3-2*2*3*2*2*5)=5!(2^5*3^3*5*7-2^4*3*5)=\)

\(=5!*2^4*3*5(2*3^2*7-1)=5!*2^4*3*5(126-1)=\)

\(5*4!*2^4*3*5*125=2^4*3*5^{1+1+3}*4!\)

So we have five 5s in the terms, and the maximum value of n is 5

D
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10! - 2(5!)(5!)

10*9*8*7*6*5! - 2(5!)(5!)

5! [10*9*8*7*6 - 2(5*4*3*2)]

5! [10*9*8*7*6 - 10*8*3]

5! [80 (9*7*6) - 80(3)]

5! [80(375)]

Now, 5! has one 5; 80 has one 5; 375 has three 5s.

So, the max number of 5s are 5 i.e. n = 5

Kudos = Thank you.

Alternate Method:

If you know the values up to 10!, you can solve this question using the actual values as follows:
10! = 3628800 & 5!= 120.
Now the expression is 10!–(2)(5!)^2
We need 2 * Square of (5!) = 2* Square of 120 = 2* 14400= 28800

So, expression becomes 3628800-28800= 36,00,000 Since there are five zeroes, the maximum power of 5 that can divide it will be five.
And hence n=5.

The question is basically asking us to determine how many 5's we can factor out of \(10! – (2)(5!)^2\)

This calls for some prime factorization!
--------------ASIDE---------------------
Here's a quick technique for finding the prime factorization of a factorial.

For this example, we'll find the prime factorization of 10!

First, I know that the prime numbers "hiding" in 10! will range from 2 to 7 (i.e., all primes less than or equal to 10), which means we can write: \(10! = (2^?)(3^?)(5^?)(7^?)\)
From here, I'll focus on one prime number at a time, starting with 2.
In 10!, we have:
1 2 hiding in 2
2 2's hiding in 4 (since 4 = 2x2)
1 2 hiding in 6 (since 6 = 2x3)
3 2's hiding in 8 (since 8 = 2x2x2)
1 2 hiding in 10 (since 10 = 2x5)
So, there's a total of 8 2's hiding in 10!, which means we can now write: \(10! = (2^8)(3^?)(5^?)(7^?)\)

Now let's focus on how many 3's are hiding in 10!.
We have:
1 3 hiding in 3
1 3 hiding in 6 (since 6 = 2x3)
2 3's hiding in 9 (since 9 = 3x3)
So, there's a total of 4 3's hiding in 10!, which means we can write: \(10! = (2^8)(3^4)(5^?)(7^?)\)

Onto the 5's:
1 5 hiding in 5
1 5 hiding in 10
So, there's a total of 2 5's hiding in 10!, which means we can write: \(10! = (2^8)(3^4)(5^2)(7^?)\)

Finally, we can see that there's exactly 1 7 hiding in 10!, which means we can write: \(10! = (2^8)(3^4)(5^2)(7^1)\)
-------BACK TO THE QUESTION----------------------------

Following the same procedure for \(5!\), we get: \(10! – (2)(5!)^2 \) \(= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 - 2(2^3 \cdot 3 \cdot 5)^2\)

Simplify the right side: \(= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 - 2(2^6 \cdot 3^2 \cdot 5^2)\)

Simplify more: \(= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 - 2^7 \cdot 3^2 \cdot 5^2\)

Factor: \(= 2^7 \cdot 3^2 \cdot 5^2(2 \cdot 3^2 \cdot 7 - 1)\)

Evaluate the part in parentheses: \(= 2^7 \cdot 3^2 \cdot 5^2(125)\)

Rewrite \(125\) as a power of \(5\) to get: \(= 2^7 \cdot 3^2 \cdot 5^2(5^3)\)

Simplify one last time: \(= 2^7 \cdot 3^2 \cdot 5^5\)

So, in total, we can factor five 5's out of \(10! – (2)(5!)^2\)

Answer: D

Cheers,
Brent
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