The prime decomposition of the number 1215000 is


    1215000 = ...


Comparing it with  , you see that


    either  a= 3^5,  b= 5,    c= 2,      with  a+b+c = 3^5 + 5 + 2   = 250,

    or      a= 3^2,  b= 5,    c = 2*3,   with  a+b+c = 3^2 + 5 + 2*3 =  20,

    or      a = 3,   b= 5*3,  c = 2,     with  a+b+c = 3   + 5*3 + 2 =  20.


So, a+b+c is maximal and equal to 250  at  a= 3^5,  b= 5  and c= 2.    ANSWER

Prime factors of 1,215,000: 

FACTS: 1) "a" MUST have the largest value

       2) The order of the other 2 prime factors that come after "a" doesn't 

          MATTER, which means that we can either have:  

This then leaves us with: 

Thus, greatest possible value of  

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!
 


The prime factorization of 1215000 is (2^3)*(3^5)*(5^4).  This is to be expressed as (a)(b^4)(c^3), with the sum a+b+c as large as possible.


Since in that form b and c are small numbers, and since we want the sum a+b+c to be as large as possible, we want a to contain as many large factors as possible.


So we want a to contain all 4 factors of 5.


There are only 3 prime factors of 2 in the number, so b can't be 2.  So b should be 3, with c = 2.


Then (b^4)(c^3) = (3^4)(2^3); and then a is made up of the remaining factors of the number: (3^1)(5^4).


So


a = 3*5^4 = 3*625 = 1875

b = 3

c = 2


The maximum sum a+b+c is 1875+3+2 = 1880.


 

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