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The product of three positive integers is 300. If one of them is 5

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The product of three positive integers is 300. If one of them is 5 [#permalink]

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New post 20 Nov 2017, 11:08
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The product of three positive integers is 300. If one of them is 5, what is the least possible value of the sum of the other two?

A. 16
B. 17
C. 19
D. 23
E. 32
[Reveal] Spoiler: OA

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Re: The product of three positive integers is 300. If one of them is 5 [#permalink]

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New post 20 Nov 2017, 11:14
we know that the product of the two remaining numbers is 60. the smallest number in the list is 16 which satisfies out condition by multiplying 10 and 6.
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Re: The product of three positive integers is 300. If one of them is 5 [#permalink]

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New post 20 Nov 2017, 11:18
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Bunuel wrote:
The product of three positive integers is 300. If one of them is 5, what is the least possible value of the sum of the other two?

A. 16
B. 17
C. 19
D. 23
E. 32

\(\frac{300}{5}= 60\). The closest factors of 60 are 10 and 6 and their sum is 16. A
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Re: The product of three positive integers is 300. If one of them is 5 [#permalink]

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New post 21 Nov 2017, 02:49
Let a,b and c be the no.s
Given abc=500 and one of the nos is 5. Let a = 5
We know 500=5*60
For the sum of b and c to be minimum, their values have to be minimum too. Let's see -
Product Sum
60 = 15*4 19
30*2 32
10*6 16

Answer - A
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Re: The product of three positive integers is 300. If one of them is 5 [#permalink]

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New post 21 Nov 2017, 08:11
Bunuel wrote:
The product of three positive integers is 300. If one of them is 5, what is the least possible value of the sum of the other two?

A. 16
B. 17
C. 19
D. 23
E. 32

\(5*x*y = 300\)

So, \(xy = 60\)

Possible sets of x,y will be -

(1*60) , (2*30) , (3*20) , (4*15) , (5*12) , (6*10) , (10*6) , (12*5)................

Thus, the least possible value of the sum of the other two will be 10 + 6 = 16, answer must be (A)

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Re: The product of three positive integers is 300. If one of them is 5 [#permalink]

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New post 21 Nov 2017, 09:46
Bunuel wrote:
The product of three positive integers is 300. If one of them is 5, what is the least possible value of the sum of the other two?

A. 16
B. 17
C. 19
D. 23
E. 32


Hi..
The logic behind it is that the factors closest would give least sum and max product.
Example \(16=4^2=4*4\)
So sum =4+4=8
Rest will keep increasing as we move outwards 2*8, so 2+8=10
1*16, so 1+16=17

Let's see the solution now..
300/5=60..
60~8^2
So look for factors close to 8.... 6*10
Sum =6+10=16

A
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Re: The product of three positive integers is 300. If one of them is 5 [#permalink]

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New post 22 Nov 2017, 13:15
Bunuel wrote:
The product of three positive integers is 300. If one of them is 5, what is the least possible value of the sum of the other two?

A. 16
B. 17
C. 19
D. 23
E. 32


Since 300/5 = 60, we need to determine two numbers that multiply to 60. The possibilities are 1 x 60, 2 x 30, 3 x 20, 4 x 15, 5 x 12 and 6 x 10.

Of these products, we see that if we add each pair of factors, the sum that is a minimum is 10 + 6 = 16.

Answer: A
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Re: The product of three positive integers is 300. If one of them is 5   [#permalink] 22 Nov 2017, 13:15
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