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If 6^y is a factor of (10!)^2, What is the greatest possible
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19 Mar 2012, 22:15
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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
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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19 Mar 2012, 22:47
essarr wrote: 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
... I was hoping to get a better explanation, as I'm still confused about the explanation provided. Thanks! 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: everythingaboutfactorialsonthegmat85592.html (explanation of this concept in details). Similar questions to practice: pandqareintegersifpisdivisibleby10qandcannot109038.htmlquestionaboutpprimeintonfactorial108086.htmlhowmanyzerosdoes100endwith100599.htmlifnistheproductofintegersfrom1to20inclusive106289.htmlwhatisthegreatestvalueofmsuchthat4misafactorof105746.htmlfindthenumberoftrailingzerosintheproductof108248.htmlfindthenumberoftrailingzerosintheexpansionof108249.htmlifdisapositiveintegerandfistheproductofthefirst126692.htmlifmistheproductofallintegersfrom1to40inclusive108971.htmlif10252isdivisibleby10nwhatisthegreatest106060.htmlHope it helps.
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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23 Mar 2012, 01:45
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
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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23 Mar 2012, 01:49



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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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23 Mar 2012, 21:44
wow, that link really helped thanks soooo much; it's so much simpler now that I understand the concept



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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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11 Apr 2012, 01:19
essarr wrote: 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
... I was hoping to get a better explanation, as I'm still confused about the explanation provided. Thanks! 10!= 10*9*8*7*6*5*4*3*2*1 = 2*5*3*3*2*2*2*7*2*3*5*2*2*3*2 = 2^8*3^4*5^2*7 6= 2*3 Therefore only the exponents of 2 and 3 are relevant, 2^8 or 3^4 > higher number counts = 8 > Answer Choice D



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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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11 Apr 2012, 01:22
Thank Bunuel for very clear and concise answer.
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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01 Oct 2012, 05:13
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..
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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01 Oct 2012, 05:13
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 8 .. 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..
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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15 Oct 2012, 07:21
The tricky thing is limiting factor : 3... Bunuel you made it look simple.. But is it really sub 600?



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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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10 Jul 2014, 00:06
\((10!)^2 = 10^2 * 9^2 * 8^2 * 7^2 * 6^2 * 5^2 * 4^2 * 3^2 * 2^2\) 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 \(3^2\) = 3^2Total powers of 3 = 8 Answer = 8
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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15 Apr 2015, 05:33
H:
I was doing trailing zero questions and I got the concept but I have some difficulty in understanding a concept in which sometimes we factorize the denominator but sometimes we don't factorize. For example
If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?. We are not factorizing 6 as 2 and 3 because higher power of 3 is sufficient but in this question... If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer? we are factoring 18 as 2^2 and 3.
Could you please help me to understand when we have to factorize and when we do not.
I shall be looking for your reply.



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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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15 Apr 2015, 05:41
harrisadiq wrote: H:
I was doing trailing zero questions and I got the concept but I have some difficulty in understanding a concept in which sometimes we factorize the denominator but sometimes we don't factorize. For example
If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?. We are not factorizing 6 as 2 and 3 because higher power of 3 is sufficient but in this question... If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer? we are factoring 18 as 2^2 and 3.
Could you please help me to understand when we have to factorize and when we do not.
I shall be looking for your reply. Aren't we make prime factorization in both cases? 6=2*3 and 18=2*3^2. Sorry, but your question is not very clear...
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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15 Apr 2015, 05:45
in question number 1, we are not considering the power of 2's but in q2 we are considering the power of 2 and 3 both. I am unable to understand when we consider all the bases as we did in q2 but not in q1



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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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17 Jan 2016, 04:42
Hi Bunuel, Is it ok to square the 10 before computing or did I just get lucky? \(\frac{100}{6}+ \frac{100}{36}= 6 + 2\)
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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17 Jan 2016, 05:22
Bunuel wrote: Icecream87 wrote: Hi Bunuel,
Is it ok to square the 10 before computing or did I just get lucky? \(\frac{100}{6}+ \frac{100}{36}= 6 + 2\) Why do you want to square? What do you square? How is 10/3+10/3^2 squared equal to \(\frac{100}{6}+ \frac{100}{36}\)? How do you get \(\frac{100}{6}= 6\)? Good question. I somehow managed to omit the 1 (from 16) on 100/6 to only maintain 6 to my liking. And I also just squared the 10 initially not the whole fraction and I dind't get why 3 was used instead of 6. Anyway, I am having trouble understanding all the powers. Does this mean that if we had 15 instead of 6 we would have taken the biggest of the primes of 15 to find the powers: \(\frac{10!^2}{15^x}\)then we would use \(\frac{10}{5}= 2*2\) so as to get \(15^4\) ? Thanks
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible
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24 Jan 2017, 19:18
1) The problem is asking us to find the number of 2*3 prime factors in the sequence of consecutive integers 2) 10/3=3; 10/9=1; 3+1=4 3) Since the sequence is squared 4*2=8
The greatest possible value of y is 8




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