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If 2^x*3^y*7^z is divisible by 168 and 441 What is the least

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If 2^x*3^y*7^z is divisible by 168 and 441 What is the least  [#permalink]

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New post 01 Jul 2012, 15:40
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Question Stats:

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If 2^x*3^y*7^z is divisible by 168 and 441. What is the least value of x*y*z if we consider x, y and z integers?

A. 3
B. 4
C. 5
D. 6
E. 12
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Re: What is the least value of x.y.z...  [#permalink]

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New post 01 Jul 2012, 20:44
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ferrarih wrote:
If \(2^x.3^y.7^z\) is divisible by 168 and 441. What is the least value of \(x.y.z\) if we consider x, y and z integers?

a) 3
b) 4
c) 5
d) 6
e) 12

Hi,

\(168 = 2^33^17^1\)
\(441 = 3^27^2\)
So, for \(2^x.3^y.7^z\) to be divisible by both the numbers x=3, y=2, z=2
xyz=12

Answer (E).

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Re: If 2^x*3^y*7^z is divisible by 168 and 441 What is the least  [#permalink]

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New post 26 Aug 2014, 00:34
This question typically asks LCM of 168 & 441 in terms of powers of 2, 3 & 7

\(168 = 2^3 * 3^1 * 7^1\)

\(441 = 21^2 = 3^2 * 7^2\)

LCM\(= 2^3 * 3^2 * 7^2\)

Multiplication of powers = 3*2*2 = 12

Answer = E
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Re: If 2^x*3^y*7^z is divisible by 168 and 441 What is the least  [#permalink]

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New post 26 Sep 2017, 15:38
ferrarih wrote:
If 2^x*3^y*7^z is divisible by 168 and 441. What is the least value of x*y*z if we consider x, y and z integers?

A. 3
B. 4
C. 5
D. 6
E. 12


Let’s factor 168 and 441 into primes:

168 = 24 x 7 = 2^3 x 3^1 x 7^1

441 = 9 x 49 = 3^2 x 7^2

The least value that is divisible by 168 and 441 is the LCM of 168 and 441, which is 2^3 x 3^2 x 7^2. Thus, the least value of x * y * z is 3 x 2 x 2 = 12.

Answer: E
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Re: If 2^x*3^y*7^z is divisible by 168 and 441 What is the least  [#permalink]

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New post 25 Oct 2018, 07:28
ferrarih wrote:
If 2^x*3^y*7^z is divisible by 168 and 441. What is the least value of x*y*z if we consider x, y and z integers?

A. 3
B. 4
C. 5
D. 6
E. 12


OA: E

If \(2^x*3^y*7^z\) is divisible by \(168\) i.e. \((2^3*3*7)\),then \(x\geq{3}, y\geq{1}\) and \(z\geq{1}\).
If \(2^x*3^y*7^z\) is divisible by \(441\) i.e. \((2^0*3^2*7^2)\),then \(x\geq{0}, y\geq{2}\) and \(z\geq{2}\).

So If \(2^x*3^y*7^z\) is divisible by \(168\) and \(441\), Minimum Value of \(x\) should be \(3\), \(y\) should be \(2\) and \(z\) should be \(2\).
The least value of \(x*y*z = 3*2*2 = 12\)
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Re: If 2^x*3^y*7^z is divisible by 168 and 441 What is the least &nbs [#permalink] 25 Oct 2018, 07:28
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If 2^x*3^y*7^z is divisible by 168 and 441 What is the least

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