If 6^y is a factor of (10!)^2, What is the greatest possible

Question

If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?

A. 2
B. 4
C. 6
D. 8
E. 10

Bunuel · Accepted Answer

6=2*3. Now, there will be obviously less 3's than 2's in (10!)^2, so maximum power of 3 will be limiting factor for maximum power of 6, which is y. Finding the maximum powers of a prime number 3, in 10!: \frac{10}{3}+\frac{10}{3^2}=3+1=4 (take only quotients into account). So, we have that the maximum power of 3 in 10! is 4, thus maximum power of 3 in (10!)^2 will be 8: (3^4)^2=3^8 . As discussed 8 is the maximum power of 6 as well. Answer: D. For more on this subject check: everything-about-factorials-on-the-gmat-85592.html (explanation of this concept in details). Similar questions to practice: p-and-q-are-integers-if-p-is-divisible-by-10-q-and-cannot-109038.html question-about-p-prime-in-to-n-factorial-108086.html how-many-zeros-does-100-end-with-100599.html if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html what-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html find-the-number-of-trailing-zeros-in-the-product-of-108248.html find-the-number-of-trailing-zeros-in-the-expansion-of-108249.html if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html if-m-is-the-product-of-all-integers-from-1-to-40-inclusive-108971.html if-10-2-5-2-is-divisible-by-10-n-what-is-the-greatest-106060.html Hope it helps.

vomhorizon · Answer

Factorial of 10 can be written as 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

or 2 x 5 x 3 x 3 x 2 x 2 x 2 x 7 x 2 x 3 x 5 x 2 x 2 x 3 x 2 x 1

So in 10 factorial we have 08 2's and 4 three's ... Square of 10 factorial will give us 16 2's and 8 three's

to get six we know that we would have to take one 2 and one three ..the maximum number of three's we can take is 8 therefore 8 different 6's can be formed therefore max possible value of y can be 6 ..

2 x 3
2 x 3 
2 x 3
2 x 3
2 x 3
2 x 3
2 x 3
2 x 3

D..

pappueshwar · Answer

hi bunuel,

i did nt understand how did u get 3's more than 2's in 10!.

i am of the view that 10! = 10*9*8*7*6*5*4*3*2*1

if we expand this we get =  2*5   *3*3   *2*2*2  * 7  * 3*2 * 5*  2*2 *  3 *  2  *1

so there are 8 2's and 4 3's in the expansion above.

so how to interpret this and proceed

Bunuel · Answer

It's: "there will be obviously LESS 3's than 2's in (10!)^2, so  maximum power of 3 will be limiting factor  for maximum power of 6, which is y."

essarr · Answer

wow, that link really helped thanks soooo much; it's so much simpler now that I understand the concept

hellscream · Answer

Thank Bunuel for very clear and concise answer.

PareshGmat · Answer

(10!)^2 = 10^2 *  9^2  * 8^2 * 7^2 *  6^2  * 5^2 * 4^2 *  3^2  * 2^2

Just concentrate on the power of 3 (Power of 2's would be more as compared to 3; so it can be ignored)

9^2  =  3^4

6^2  =  3^2  * 2^2

3^2  =  3^2

Total powers of 3 = 8

Answer = 8

chetan2u · Answer

Hi,

Since you already know there are 8 3s and 14 2s, I'll start from thereon...
Each 6 is composed of 3 and 2...
So it requires equal number of 3 and 2..

But (14-8) that is 6 of 2s do not have a 3 to make a 6..
Therefore the number of 6s will depend on the integer (3or2) with lowest value thus answer is 8..

EgmatQuantExpert · Answer

Hey, PFB the solution. • 6^y can be written as 2^y * 3^y • To find the greatest possible value of y, we need to find out how many 3 's are there in (10!)^2 • Now 10! = 1 * 2 *3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 • Which can be written as - o 10! = 2^8 * 3^4 * 5^2 * 7^1 • Therefore (10!)^2 = 2^{16} * 3^8 * 5^4 * 7^2 • As we can see there are 16 2's but only 8 3's • But to make a 6 we need both one 2 and one 3 . • Therefore, the maximum number of 6 's that we can make is 8 . • Please note : that out of the 16 2's we can use only 8 of them and the rest 8 cannot be clubbed with any 3's, as there aren't any left. • Hence, the value of y = 8 . Thanks, Saquib Quant Expert e-GMAT Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

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Akriti_Khetawat · Answer

Hey what's a limiting factor and why are we considering only that number to find the answer?

Bunuel · Answer

There will be obviously less 3's than 2's in (10!)^2, so the power of 3 in (10!)^2 will determine the power of 6 in (10!)^2. Please follow the links given in that post for more.

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If 6^y is a factor of (10!)^2, What is the greatest possible

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If 6^y is a factor of (10!)^2, What is the greatest possible

Prep Toolkit

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