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# 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4.

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SVP
Joined: 03 Feb 2003
Posts: 1604
2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink]

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05 Jun 2003, 22:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

(2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. Find the product XYZ.
Founder
Joined: 04 Dec 2002
Posts: 15135
Location: United States (WA)
GMAT 1: 750 Q49 V42

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05 Jun 2003, 23:27
stolyar wrote:
(2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. Find the product XYZ.

Hm. I first came up with 64, but that seemed to have a mistake. A later revision gave me 14*4^3 close?
Founder
Joined: 04 Dec 2002
Posts: 15135
Location: United States (WA)
GMAT 1: 750 Q49 V42

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05 Jun 2003, 23:44
brstorewala wrote:
28*8*4

it is the same: 896; your and my results....
Manager
Joined: 28 Feb 2003
Posts: 100

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05 Jun 2003, 23:45
i realised it after i posted my answer
Founder
Joined: 04 Dec 2002
Posts: 15135
Location: United States (WA)
GMAT 1: 750 Q49 V42

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05 Jun 2003, 23:53
To tell the truth, it took me a minute to think about it. it seemed quite strange and unusual question - which makes it a good one. Of course, we might have approached it in a wrong way, but in any case, stolary is quite prolific.

Bogdan.
SVP
Joined: 03 Feb 2003
Posts: 1604

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06 Jun 2003, 01:55
This sort of problem is strange. Brstorewala knows about it. His approach is correct. BUT, it does not work when it comes to 10!*20!*30! When factorials are exceed 10, something strage happens... And I myself do not why.
Manager
Joined: 28 Feb 2003
Posts: 100

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06 Jun 2003, 03:36
stolyar,
i forgot to tell u about that 10!*20!*30! problem.....

try it out in the calculator of the computer (if u have patience) .........i think excel screws it up along the way........the answer with excel does not match with the answer from the calculator........
SVP
Joined: 03 Feb 2003
Posts: 1604

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06 Jun 2003, 03:40
brstorewala wrote:
stolyar,
i forgot to tell u about that 10!*20!*30! problem.....

try it out in the calculator of the computer (if u have patience) .........i think excel screws it up along the way........the answer with excel does not match with the answer from the calculator........

Yes, you are right. Excel may be overloaded with those factorials. Our approach should be OK.
Manager
Joined: 03 Jun 2003
Posts: 84
Location: Uruguay

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06 Jun 2003, 07:36
thank you
Founder
Joined: 04 Dec 2002
Posts: 15135
Location: United States (WA)
GMAT 1: 750 Q49 V42

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06 Jun 2003, 10:07
MBA04 wrote:
thank you

You probably don't want to be sexist in your replies. It is he or she - you never know here I or Stolyar may be a lady.... actually I am pretty sure that Stolyar is a cute russian girl... she is just afraid to admit it cause all the guys will start annoying her.

So...

as to the solution,

Get the factors of 8!^4 = 8^4*7^4*6^4*5^4*4^4*3^4*2^4*1^4

Then compile out of 2, 3, adn 5 the largest possible divisors - you will take care of all of them except 7. So, you will need 2^12 to take care of 8^4, and so on at the end you will get 28 power for 2's, 8th power for 3's and 4th power for 5.

Is it clear now?

STOLYAR!

Last edited by bb on 07 Jun 2003, 00:13, edited 1 time in total.
Manager
Joined: 25 May 2003
Posts: 54

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06 Jun 2003, 23:54
is not the greatest positive divisor of integer 'x' simply 'x'? ie. 32 is the largest positive divisor of 32 no?
Founder
Joined: 04 Dec 2002
Posts: 15135
Location: United States (WA)
GMAT 1: 750 Q49 V42

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07 Jun 2003, 00:06
skoper wrote:
is not the greatest positive divisor of integer 'x' simply 'x'? ie. 32 is the largest positive divisor of 32 no?

That's right.... go on... I see you have something in mind
Manager
Joined: 25 May 2003
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07 Jun 2003, 00:12
so then you shouldn't be allowed to ignore the 7^4. This is an amazing question btw
Founder
Joined: 04 Dec 2002
Posts: 15135
Location: United States (WA)
GMAT 1: 750 Q49 V42

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07 Jun 2003, 00:15
skoper wrote:
so then you shouldn't be allowed to ignore the 7^4. This is an amazing question btw

but how can you express it with 2,3, and 5?

At least I assumed that 7^4 is not divisble by any of them, which, just as well may not be true and you may be right.... but you will have to prove it
Manager
Joined: 25 May 2003
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07 Jun 2003, 00:19
agreed..you can't. Therefore 2^X * 3^Y * 5^Z cannot be the greatest divisor then because it does not equal (8!)^4, rather it equals (8!/7)^4
Founder
Joined: 04 Dec 2002
Posts: 15135
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07 Jun 2003, 00:22
skoper wrote:
agreed..you can't. Therefore 2^X * 3^Y * 5^Z cannot be the greatest divisor then because it does not equal (8!)^4, rather it equals (8!/7)^4

Yes, but we really dont' need to know what it equals to. Uncle Stoly took care of the shortcut and instead asked for the product of the powers...
Manager
Joined: 25 May 2003
Posts: 54

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07 Jun 2003, 00:27
uncle eh? so now who's being sexist?

so is it the greatest divisor or not?
Founder
Joined: 04 Dec 2002
Posts: 15135
Location: United States (WA)
GMAT 1: 750 Q49 V42

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07 Jun 2003, 00:29
skoper wrote:
uncle eh? so now who's being sexist?

so is it the greatest divisor or not?

You caught me!!!!

It is.
Manager
Joined: 25 May 2003
Posts: 54

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07 Jun 2003, 00:29
if it is, then x cannot be the greatest divisor of x
Founder
Joined: 04 Dec 2002
Posts: 15135
Location: United States (WA)
GMAT 1: 750 Q49 V42

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07 Jun 2003, 00:34
skoper wrote:
if it is, then x cannot be the greatest divisor of x

I am confused confused confused condlasjoqiweuzcf

OK, (8!/7)^4 is the greatest divisor of 8! with conditions provided above.
07 Jun 2003, 00:34

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