If ((14!)^25 - (14!)^11)/((14!)^11 - (14!)^4) = X. What is the unit di

Question

If  \frac{(14!)^{25}-(14!)^{11}}{(14!)^{11}-(14!)^{4}}=X .

What is the unit digit of \frac{X}{(14!)^7}  ?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

EgmatQuantExpert · Accepted Answer

Solution

Given:  
 •  \frac{(14!)^{25}-(14!)^{11}}{(14!)^{11}-(14!)^{4}}=X  
 
  To find:  
 • The unit digit of  \frac{X}{(14!)^7}

Approach and Working:  
Assuming n = 14!, we can rewrite the whole expression as
 • X =  \frac{(n)^{25}-(n)^{11}}{(n)^{11}-(n)^{4}} 
=  \frac{(n)^{11} }{(n)^4 
=  (n)^7\frac{(n^7+1)(n^7-1)}{(n^7-1)} 
=  n^7 (n^7+1)

So, if we do  \frac{X}{(14!)^7} , we can write it as:
 •  \frac{X}{(14!)^7}  =  \frac{n^7 (n^7+1)}{n^7}  =  n^7 + 1

As n = 14!, n will always end with a 0.
 • Therefore, the unit digit of  n^7 + 1  = 0 + 1 = 1

Hence, the correct answer is option A.

Answer: A

Bunuel · Answer

Similar questions to practice: https://gmatclub.com/forum/if-12-16-12-8 ... 86818.html https://gmatclub.com/forum/baker-s-dozen ... l#p1057520 https://gmatclub.com/forum/if-a-13-1-6- ... 65186.html Hope this helps.

akshayvp · Answer

EgmatQuantExpert  : how you have assumed that 14! will always end with a 0. It is mugged up or any logic behind it.

EgmatQuantExpert · Answer

Let’s look at the factorial values of the following numbers:
 • 1! = 1
• 2! = 1 * 2 = 2 
• 3! = 1 * 2 * 3 = 6
• 4! = 1 * 2 * 3 * 4 = 24
• 5! = 1 * 2 * 3 * 4 * 5 = 120

Now, if you go beyond 5!, in every factorial value 2 and 5 will always be present.

The product of 2 and 5 will always end with a 0, hence, 5! onwards every factorial value should always end with at least one 0.

pikolo2510 · Answer

14! = 14 * 13 * 12 * 11 *   10   * .....

You have to multiply by 10 in the equation above. As you know, any integer when multiplied by 10 , you add a zero at the end. Thats the logic used in the question above

akshayvp · Answer

Thanks for the explanation  EgmatQuantExpert  and  pikolo2510

KanishkM · Answer

PLEASE  be careful while solving

So nothing you had to do a^2 - b^2 = (a-b) (a+b) & take the common out

\frac{(14!)^{25}-(14!)^{11}}{(14!)^{11}-(14!)^{4}}=X

take  (14!)^{11}  common from numerator and  (14!)^{4}  common from denominator

let 14! be x for sometime

x^11 - x^ 4 *   = X

x/ x^7 =

x/x^7 = x^7 + 1

put back 14! in the above expression

we will certainly have a zero in the product of 14!

0+1

UD = 1

A

NinetyFour · Answer

My reasoning: set  14! = a

Factor  a^{11}  out of the numerator and  a^4  out of the denominator. Since we're diving X by  a^7  we can cancel the  a^{11}  out of the numerator.

We're left with:  \frac{(a^{14} -1)}{(a^7-1)}

We can cancel the denominator out and we're left with:  a^7 +1

We know that the last digit in 14! will be 0 because there is a ten being multiplied. When we add one the final units digit will be 1. (answer choice A)

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If ((14!)^25 - (14!)^11)/((14!)^11 - (14!)^4) = X. What is the unit di

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If ((14!)^25 - (14!)^11)/((14!)^11 - (14!)^4) = X. What is the unit di

Prep Toolkit

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