The value of the expression (20!)(10!) / (2)(5!)(5!) is an integer

Question

The value of the expression \frac{(20!)(10!) }{ (2)(5!)(5!)} is an integer that ends in how many zeros? (A) 0 (B) 1 (C) 3 (D) 4 (E) 5 or more Screenshot 2023-12-31 151757.png

Bunuel · Accepted Answer

\frac{(20!)(10!) }{ (2)(5!)(5!)}=\frac{(20!)(6*7*8*9*10) }{ (2)(5!)}=\frac{(20!)(6*7*8*9*2) }{ (2)(4!)} .

After reducing to an integer, the only source of factors of 5 in the expression comes from 20!, and there are certainly enough factors of 2 to pair with them. The highest power of 5 in 20! is 20/5 = 4. Therefore,  \frac{(20!)(10!) }{ (2)(5!)(5!)}  ends with 4 trailing zeros.

Answer: D.

P.S. The exact value is 306,545,653,030,256,640,000

astor900 · Answer

Can we just directly search for the highest power of 5 in each factorial then do the operation like written which give : 4*2/2*1*1 =4

Bunuel · Answer

It would work for this question, however if it were 20!10!/(2^20*5!5!) instead, according to your logic the number of trailing zeros would still be 4, wherease in reality 20!10!/(2^20*5!5!) does not end with 0 (it's 584689432201875). So, we need to make sure that we also have enoguhg 2's to pair with 5's.

Diwen2000 · Answer

Bunuel 
Could you please explicate how you convert "an integer ends in how many zeros" to identifying the quantity of 2*5?

Bunuel · Answer

For theory check:  Number Properties

For questions check the following topics from our  Special Questions Directory :

12.  Trailing Zeros 
13.  Power of a number in a factorial

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

anish777 · Answer

Can we solve this sum like this? The highest power of 5 in 20! and 10! are 4 and 2 respectively. Adding 'em, we get 6. The highest power in 5! is 1, adding both of 'em in denominator gives 2. Hence, 6-2=4 ( as we are dividing here).

asingh91 · Answer

Bunuel

Hi Bunuel , Thanks for illustrating
but in the eg you used, numerator as 26 factors of 2 and denominator has 20 factors. that would leave us with 6 powers of 2 remaining right?

Bunuel · Answer

Not sure I follow what you mean with your doubt. The trailing zeros come from pairing 2s and 5s. There are plenty of 2s, so only the number of 5s matters. That is why the count comes from 20!/5 = 4.

shakedAF · Answer

Zeros come from factors of 10, which are formed from pairs of 2 and 5. Since factorials always contain more 2’s than 5’s, the number of trailing zeros is determined only by the number of 5’s in the product.

Simplify the expression: 
  \frac{ (20!*10!)}{(2*5!*5!) }=\frac{(20!*10*9*8*7*6)}{(2*5*4*3*2)}=20!*7*6*3   
The extra factor  7*6*3  has no 5’s, so only 20! matters. 
The number of factors of 5 in 20!  \frac{20}{5}=4  so the result of 20! ends with  4 zeros .

In general, finding the number of zeros of n!, by using the formula:  https://gmatclub.com/forum/download/file.php?id=85488  rounded to the lower integer. For cases when n<25  \frac{n}{5}  is suffiecient.

Answer D

MegB07 · Answer

I approached this by identifying number forming pairs 10s. So 20, 15, 18, 10, 5, 2, 10, 5, 2. divided by 5x2 and 5x2 and got 4 as number of 0s. Please tell if approach is right

Bunuel · Answer

It's not clear what you are doing there. Please elaborate.

MegB07 · Answer

So 20! would give us 20, 15, 10, 2, 5, any even integer which multiplied by 15 would give 0 at the end (18). Similarly for 10! 10, 5,2. When dividing by 5x2, 5x2. This would leave us with 4 0s. The remaining numbers would cancelled with each other

Bunuel

It's not clear what you are doing there. Please elaborate.

Sakha · Answer

Bunuel , What I did is isolate the 0 causing numbers in Numerator and Denominator:

(2 * 5 * 10 * 15 * 20 * 2 * 5 * 10) / ( 2*5*2*5)

Got 40,000 from it, thus selected 4 Zeroes.

Is this method accurate?

The value of the expression (20!)(10!) / (2)(5!)(5!) is an integer

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The value of the expression (20!)(10!) / (2)(5!)(5!) is an integer

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