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The product of a list of (not necessarily distinct) positive integers is 120. What is the least possible sum of the numbers in the list? What if the product were 144000? What is the least possible sum in this case?

 Oct 19, 2017
 #1
avatar+349 
0

Not sure, but... for the 120 case, I guess it's 239? {120, 1, 1..., 1, 1}? Anyway, ask CPhill.

 Oct 19, 2017
 #2
avatar+128408 
+1

120  factors as    2^3 * 3 * 5  and the sum of these is

 

3(2)  + 3 + 5  =

 

6 + 8

 

14

 

144000  factors as  2^7 * 3^2 * 5^3  and the sum of these is

 

7(2) + 2(3) + 3(5)  =   

 

14 + 6 + 15  =

 

35

 

 

cool cool cool

 Oct 19, 2017
 #3
avatar+118608 
0

Tom..riddle asked me to reopen this one. 

I think she would like you to explain more how you got your answer Chris.

(I have not looked at it at all as yet)

 Aug 27, 2019
edited by Melody  Aug 27, 2019
 #4
avatar+128408 
+1

First one

 

2* 2 * 2 * 3 * 5    =   8 * 3 * 5   =  24 * 5   = 120

 

Sum

2,2,2,3,5

2 + 2 + 2 + 3 + 5   =   14

 

Second one

 

2^7 * 3^2 * 5^3  =  144000

 

Sum 

2,2,2,2,2,2,2,3,3,5,5,5

2 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 5 + 5 + 5  =  35

 

 

cool cool cool

 Aug 27, 2019
 #5
avatar+1712 
0

Oh thank you.

tommarvoloriddle  Aug 27, 2019
 #6
avatar+1712 
0

I just didn't realize it... but heres the gist of my way thats basically the same as CPHILL I didn't get it at first because instead of 3^2 I thought it was 32...

 

The prime factorization of 144000 is2^7*3^2*5^3 , so the minimum sum in this case is 35 .

 

Thanks Melody and Cphill and mathhemath and waffles!

 Aug 27, 2019
edited by tommarvoloriddle  Aug 27, 2019

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