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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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20 Nov 2017, 10:08
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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

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

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20 Nov 2017, 10:14
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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20 Nov 2017, 10:18
1
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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21 Nov 2017, 01:49
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

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

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21 Nov 2017, 07:11
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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21 Nov 2017, 08:46
1
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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22 Nov 2017, 12:15
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.

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

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12 Jul 2018, 00:45
The product of three positive integers is 300

Let the three positive integers be a, b and c

=> $$abc = 300$$

Given one of the integers is 5 => a = 5

=> $$bc = \frac{300}{a}$$

=> $$bc = \frac{300}{5} = 60$$

=> $$bc = 60$$

Now writing all the factor pairs of 60 we have

1 and 60 sum to 61
2 and 30 sum to 32
3 and 20 sum to 23
4 and 15 sum to 19
5 and 12 sum to 17
6 and 10 sum to 16

The least possible sum of b and c is 60

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

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12 Jul 2018, 00:49
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

As this a question about a product, we'll first need to use factorization.
This is a Precise approach.

300 = 5 * 60 = 5 * 6 * 10 = 5 * 3 * 2 * 5 * 2.
So our two numbers have (together) the factors 2,2,3 and 5.
Let's try our answers, starting from the smallest:
Can we get (A) 16?
2*3 = 6 and 2*5 = 10.
Since 6 + 10 = 16, then yes, we can.

*Note, in general, if two positive numbers a,b have a constant product then a+b is minimized when a = b = sqrt(ab). So a good guess is to try making a,b, as close to each other as possible, in this case by picking one small factor (2) and one large factor (3 or 5) for each.
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Re: The product of three positive integers is 300. If one of them is 5  [#permalink]

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12 Jul 2018, 04:16
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

a*b*5=300
ab=60
The closer a and b are to 60/2 the lesser will be the sum
a=10 and b =6
least possible sum =16
Re: The product of three positive integers is 300. If one of them is 5 &nbs [#permalink] 12 Jul 2018, 04:16
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# The product of three positive integers is 300. If one of them is 5

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