Question 1199210
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Prime factorization:
648000 = 2^6*3^4*5^3


Then rewrite the expression a bit.
648000 = 2^6*3^4*5^3
648000 = (2^3)^2*5^3*3^4
648000 = 8^2*5^3*3^4
It fits the format a^2*b^3*c^4 such that
a = 8
b = 5
c = 3
So a+b+c = 8+5+3 = 16


We can do a bit of rearranging like this
648000 = 2^6*3^4*5^3
648000 = 2^4*2^2*3^4*5^3
648000 = 2^4*(2*3^2)^2*5^3
648000 = 2^4*18^2*5^3
648000 = 18^2*5^3*2^4
Showing that
a+b+c = 18+5+2 = 25
which isn't smaller than the previous result (16) we computed.


Or we could do this
648000 = 2^6*3^4*5^3
648000 = 2^2*2^4*3^4*5^3
648000 = 2^2*(2*3)^4*5^3
648000 = 2^2*6^4*5^3
648000 = 2^2*5^3*6^4
So,
a+b+c = 2+5+6 = 13
which is smaller than the 16 computed earlier.


I used computer software to do a brute-force search of all possible triples a,b,c from 2 to 100. Those are the only three possible sums I could find. There may be others.



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Answer: 13
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