Question 1150079
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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.<br>
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.<br>
So we want a to contain all 4 factors of 5.<br>
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.<br>
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).<br>
So<br>
a = 3*5^4 = 3*625 = 1875
b = 3
c = 2<br>
The maximum sum a+b+c is 1875+3+2 = 1880.<br>