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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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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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New post Updated on: 01 Feb 2017, 08:07
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Question Stats:

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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.
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Re: If x+y+z = 6, 1+1+4=6 is possible?  [#permalink]

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New post 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
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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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New post 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.
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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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New post 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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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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New post 01 Feb 2017, 22:03
1
+1 B.

x=4
y=1
z=1

=> 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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New post 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
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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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New post 10 May 2017, 20:49
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1
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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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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New post 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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New post 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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New post 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.

\(x = 4\)
\(y = 1\)
\(z = 1\)

\(x+y+z = 4 + 1 + 1 = 6\)

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

Answer (B)...
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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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New post 17 Sep 2018, 10:35
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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] 17 Sep 2018, 10:35
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