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# p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos

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Intern
Joined: 01 Feb 2017
Posts: 1
p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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Updated on: 01 Feb 2017, 08:07
1
4
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Difficulty:

55% (hard)

Question Stats:

54% (01:03) correct 46% (01:09) wrong based on 247 sessions

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p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are positive integers and x+y+z = 6, what is the smallest possible value of p?

(A) 64
(B) 240
(C) 360
(D) 640
(E) 900

Hi, I was working on this questions below and I assumed that x, y and z had to be 3 different numbers. But apparently I was wrong. I haven't been using math for a long time, but I remember we would always use different letters to refer to different values. So I thought: 1+2+3=6 was correct.
However, based on the final answer, it seems that the correct assumption is x+y+z = 6 = 1+1+4=6. So x= 1 and y also =1

I would be wrong then if I assume in other math problems that different letters refer to different values?
Thanks!

Originally posted by MaggieSG on 01 Feb 2017, 07:24.
Last edited by Bunuel on 01 Feb 2017, 08:07, edited 1 time in total.
Renamed the topic and edited the question.
Intern
Joined: 20 Jan 2017
Posts: 1
Re: If x+y+z = 6, 1+1+4=6 is possible?  [#permalink]

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01 Feb 2017, 08:35
x, z, and y can have the same value. I got 240 2^4+3*5, since y and z must be > than 0
Intern
Joined: 13 Dec 2016
Posts: 43
Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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01 Feb 2017, 09:12
MaggieSG, Thank you for posting this question. I made a mistake of considering x =6, y=0 and z=0 only to realize that the question clearly states POSITIVE INTEGERS and 0 is neither positive nor negative. Hence, lesson learnt. Need to keep this fact in mind while solving questions such as this.
Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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01 Feb 2017, 11:34
MaggieSG wrote:
p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are positive integers and x+y+z = 6, what is the smallest possible value of p?

(A) 64
(B) 240
(C) 360
(D) 640
(E) 900

Hi, I was working on this questions below and I assumed that x, y and z had to be 3 different numbers. But apparently I was wrong. I haven't been using math for a long time, but I remember we would always use different letters to refer to different values. So I thought: 1+2+3=6 was correct.
However, based on the final answer, it seems that the correct assumption is x+y+z = 6 = 1+1+4=6. So x= 1 and y also =1

I would be wrong then if I assume in other math problems that different letters refer to different values?
Thanks!

Unless it is explicitly stated otherwise, different variables CAN represent the same number.
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Manager
Joined: 04 Dec 2016
Posts: 57
Location: India
GPA: 3.8
WE: Operations (Other)
Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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01 Feb 2017, 22:03
+1 B.

x=4
y=1
z=1

=> 2^x*3^y*5^z = 2^4 * 3^1 * 5^1 = 16*3*5 = 240
Joined: 30 May 2015
Posts: 40
Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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04 Feb 2017, 11:13
1
x + y + Z = 6
we can chose bigger value for x , as it is power of 2 which is smallest number out of given three values

so x =4 ; y =1 ; z = 1

then 2^x 3^y 5^z => 16*3*5 = 240
Manager
Joined: 02 Aug 2013
Posts: 65
Location: India
WE: Programming (Consulting)
Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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10 May 2017, 20:49
2
I took a little different approach to solve this question.

p = $$2^x * 3^y * 5^z$$

So, p = multiple of 2*3*5 = 30.

Smallest multiple in answer chose is 240.

Hence, ans: Option B.
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Posts: 87
Location: India
Schools: IIMA , IIMA PGPX"18
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Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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19 May 2017, 02:58
MaggieSG wrote:
p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are positive integers and x+y+z = 6, what is the smallest possible value of p?

(A) 64
(B) 240
(C) 360
(D) 640
(E) 900

Hi, I was working on this questions below and I assumed that x, y and z had to be 3 different numbers. But apparently I was wrong. I haven't been using math for a long time, but I remember we would always use different letters to refer to different values. So I thought: 1+2+3=6 was correct.
However, based on the final answer, it seems that the correct assumption is x+y+z = 6 = 1+1+4=6. So x= 1 and y also =1

I would be wrong then if I assume in other math problems that different letters refer to different values?
Thanks!

As we want to make number as small as possible, so we will take power of smallest number i.e 2, as greater as we can take.
so we take,
x=4
y=1
z=1

x+y+z=6

=> 2^x*3^y*5^z = 2^4 * 3^1 * 5^1 = 16*3*5 = 240
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Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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13 Aug 2017, 04:31
Remember words play important role. since x, y, and z are positive they cannot be 0.
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Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos  [#permalink]

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13 Aug 2017, 22:13
MaggieSG wrote:
p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are positive integers and x+y+z = 6, what is the smallest possible value of p?

(A) 64
(B) 240
(C) 360
(D) 640
(E) 900

$$p = (2^x )( 3^y)(5^z)$$

$$x + y + z = 6$$

$$x, y,$$ and $$z$$ are positive integers.

Smallest value of '$$p$$' would be when $$x, y , z$$ would be smallest positive integers.

$$x = 4$$
$$y = 1$$
$$z = 1$$

$$x+y+z = 4 + 1 + 1 = 6$$

$$p = (2^4)( 3^1)(5^1) = (16)(3)(5) = 240$$

Re: p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are pos &nbs [#permalink] 13 Aug 2017, 22:13
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