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2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink]
 05 Jun 2003, 22:45
(2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. Find the product XYZ. 
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Re: product number two [#permalink]
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?
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brstorewala wrote: 28*8*4
it is the same: 896; your and my results....
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i realised it after i posted my answer 
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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.
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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. 
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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........ 
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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.
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Can someone explain his answer
thank you
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MBA04 wrote: Can someone explain his answer
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.
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is not the greatest positive divisor of integer 'x' simply 'x'? ie. 32 is the largest positive divisor of 32 no?
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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 
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so then you shouldn't be allowed to ignore the 7^4. This is an amazing question btw
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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
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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
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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... 
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uncle eh? so now who's being sexist? 
so is it the greatest divisor or not?
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skoper wrote: uncle eh? so now who's being sexist? so is it the greatest divisor or not? You caught me!!!! 
It is.
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if it is, then x cannot be the greatest divisor of x
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Founder
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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.
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close
