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

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yezz
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2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post   12 Oct 2006, 16:49
2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. Find the product XYZ



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Re: 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post   12 Oct 2006, 17:09
8! ^4 is (8.7.6.5.4.3.2.1)^4
=(2^7.3^2.5.7)^4=>2^x.3^y.5^z
=xyz=7.2.1.4.=56

Answer is 56
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Re: 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post   12 Oct 2006, 17:35
vijay2001 wrote:
8! ^4 is (8.7.6.5.4.3.2.1)^4
=(2^7.3^2.5.7)^4=>2^x.3^y.5^z
=xyz=7.2.1.4.=56

Answer is 56


So I see that

2^7 => 2^x
3^2 => 3^y
5^1 => 5^z

but how did you get 7^1 => 4?

Thx.
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Re: 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post   12 Oct 2006, 18:48
No.

I got 896.

(8!)^4 = 2^28*3^8*5^4*7^4

so xyz = 28*8*4
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Re: 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post   12 Oct 2006, 18:56
tennis_ball wrote:
No.

I got 896.

(8!)^4 = 2^28*3^8*5^4*7^4

so xyz = 28*8*4


I understand now. he multiplied 7^1*4 but forgot to multiply powers of other factors by 4
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Re: 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post   12 Oct 2006, 19:04
Ans 896.

8! = (2^3)*7*(2*3)*5*(2^2)*3*2 = 7*5*(3^2)*(2*7)

So, (8!)^4= (2^28)*(3^8)*(5^4)*(7^4)

xyz = 28*8*4 = 896
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Re: 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post   13 Oct 2006, 14:37
Folks, i have already explained u earlier how to calculate the highest power of any prime number in a factorial .Please use that and save time in the exam.............

Anyway, i am repeating it again........
Let's calculate highest power of 11 in 274!

274/11 = 24 (don't worry about the remainder)
24/11 = 2
Add up now 24+2 = 26 will be the highest power of 11 in 274!

Let's look at one more example.
Highest power of 5 in 350!
350/5 = 70
70/5 = 14
14/5 = 2
So 70+14+2 = 86 is the highes power of 5 in 350!.
I guess the explanation is lucid

Given problem is 8!^4

Highest power of 2 in 8!
8/2 = 4
4/2 =2
2/2 =1
So highes power of 2 in 8! is 7

Similarly highest power of 3 is 8!
8/3 = 2
So highest power of 3 in 8! is 2
Similarly highes powr of 5 in 8! is1

So 8!^4 = (2^7)^4 x (3^2)^4 x(5^1)^4 x K
ie 8!^4 = 2^28 x 3^8 x 5^4 x K

So xyz = 28x8x4 = 896

This would be faster even if the given factorial is big........

ie
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Re: 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink] New post   13 Oct 2006, 19:24
cicerone wrote:
Let's look at one more example.
Highest power of 5 in 350!
350/5 = 70
70/5 = 14
14/5 = 2
So 70+14+2 = 86 is the highes power of 5 in 350!.
I guess the explanation is lucid

Thanks for the neat trick, Cicerone.



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Re: 2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4. [#permalink]
13 Oct 2006, 19:24
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