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If x, y and z are largest positive integers for which 2^x3^y5^z is a f

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Bunuel
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If x, y and z are largest positive integers for which 2^x3^y5^z is a f [#permalink] New post   03 Oct 2019, 03:33
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If x, y and z are largest positive integers for which \(2^x3^y5^z\) is a factor of \((7!)^2\), what is the value of \(x + y + z\) ?

A. 7
B. 12
C. 13
D. 14
E. 16


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Bunuel
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Re: If x, y and z are largest positive integers for which 2^x3^y5^z is a f [#permalink] New post   06 Dec 2021, 07:11
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Official Solution:


If \(x\), \(y\) and \(z\) are largest positive integers for which \(2^x3^y5^z\) is a factor of \((7!)^2\), what is the value of \(x + y + z\) ?


A. \(7\)
B. \(12\)
C. \(13\)
D. \(14\)
E. \(16\)


To answer the question we need to find out the largest integer powers of 2, 3, and 5 in \((7!)^2\).

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

So, \(x_{max}=8\), \(y_{max}=4\), \(z_{max}=2\), and thus, \(x + y + z=8+4+2=14\).


Answer: D
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Kanika3agg
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Re: If x, y and z are largest positive integers for which 2^x3^y5^z is a f [#permalink] New post   03 Oct 2019, 09:43
Ans - D

Applied the divisibility formula which is applied in finding trailing zeroes.
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Re: If x, y and z are largest positive integers for which 2^x3^y5^z is a f [#permalink] New post   03 Oct 2019, 11:27
power of 2 in 7! -- \(\frac{7}{2}\) - \(\frac{3}{2}\) - 1 -- 3+1 = 4 -- in \((7!)^2\) = 4*2 = 8

power of 3 in 7! -- \(\frac{7}{3}\) - 2 -- 2 -- in \((7!)^2\) = 2*2 = 4

power of 5 in 7! -- \(\frac{7}{5}\) - 1 -- 1 -- in \((7!)^2\) = 1*2 = 2

x = 8, y = 4, and z = 2

x+y+z = 8+4+2 = 14

D is the answer
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Re: If x, y and z are largest positive integers for which 2^x3^y5^z is a f [#permalink] New post   03 Oct 2019, 11:56
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7!^2 = (1*2*3*4*5*6*7)^2 = (2^4 * 3^2 *5^1*7^1)^2

so x= 8
y= 4
z = 2
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Re: If x, y and z are largest positive integers for which 2^x3^y5^z is a f [#permalink] New post   07 Oct 2019, 11:26
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Bunuel wrote:
If x, y and z are largest positive integers for which \(2^x3^y5^z\) is a factor of \((7!)^2\), what is the value of \(x + y + z\) ?

A. 7
B. 12
C. 13
D. 14
E. 16


Since 7! = 7 x 6 x 5 x 4 x 3 x 2 = 7 x 2 x 3 x 5 x 2^2 x 3 x 2 = 2^4 x 3^2 x 5 x 7, then we see that (7!)^2 = 2^8 x 3^4 x 5^2 x 7^2. Therefore, x = 8, y = 4, z = 2 and hence x + y + z = 14.

Answer: D
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Re: If x, y and z are largest positive integers for which 2^x3^y5^z is a f [#permalink] New post   01 Feb 2023, 08:57
Solving it by writing out 7! makes sense to me, but I'm wondering why I got the wrong answer when I used the formula for prime factors of a factorial.


(7/2)+(7/4)+(7/8) (throw out remainder) = 3+1+0 = 4
(7/3)+(7/9) (throw out remainder) = 2+0 = 2
(7/5) + (7/25) (throw out remainder) = 1 + 0 = 1

4+2+1 = 7

So I got 7, but the answer should be 14? I'm confused.


Edit: I realized my error. The question says (7!)^2... I reread it so many times and didn't even notice the ^2... ugh...
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Re: If x, y and z are largest positive integers for which 2^x3^y5^z is a f [#permalink]
01 Feb 2023, 08:57
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