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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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Joined: 30 Jun 2012
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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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01 Jul 2012, 15:40
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45% (medium)

Question Stats:

70% (01:38) correct 30% (01:40) wrong based on 259 sessions

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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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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

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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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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

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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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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.

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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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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 $$x*y*z = 3*2*2 = 12$$
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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