Ps: divisor

Question

what is XYZ if (2^X)*(3^Y)*(5^Z) is the greatest positive divisor of (8!)^4?

jaynayak · Answer

8! can be written as 2^7*3^2*5*7

Hence GCD
28*8*4

GMATT73 · Answer

Why isn`t the answer 7*2*1 :?:

ywilfred · Answer

(8!)^4 

= (8*7*6*5*4*3*2*1)^4
= (2^3 * 7 * 2*3 * 5 * 2^2 * 3 * 2 * 1)^4
= (2^7 * 7 * 5 * 3^2 * 1)^4
= 2^28 * 7^4 * 5*4 * 3*8

TO divide (8!)^4 evenly, x = 28, y = 8, z = 4

MA · Answer

you missed "greatest +ve divisor".

= (8!)^4
= (8x7x6x5x4x3x2x1)^4
= (2^3 x 7 x 2 x 3 x 5 x 2 x 2 x 3 x 2 x 1)^4
= (2^7 x 7 x 3^2 x 5)^4
= 2^28 x 7^4 x 3^8 x 5^4
so x = 28, y = 8 and z = 4
xyz = 28 x 8 x 4

haas_mba07 · Answer

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

As (2^x)x(3^y)x(5^z) is GCD of (8!)^4,

(8!)^4/(2^x)x(3^y)x(5^z) = k

or (2^28)x(3^8)x(5^4)x(7^4) = k x(2^28)x(3^8)x(5^4)x(7^4)

=> x = 28, y = 8, z= 4

xyz = 28 x 8 x 4 = 28 x 32

Am I even going the right way with this?? :roll:

Professor · Answer

thanx everybody, you all hit the ball into the post.

goal = 28 x 8 x 4

GMATT73 · Answer

Crystal clear explanation! Thanks MA. :-D

Ps: divisor

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