Question 1199210: If  , where a, b and c are distinct positive integers greater than 1, what is the least possible value of a+b+c?
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
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Ignore this response; it is faulty. Some of the prime factors can be combined to give a smaller value of a+b+c than shown in my response.
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Find the prime factorization of 648000. Since doing that doesn't give you practice in any particularly useful skills, I would use an online calculator.
648000 = (2^6)(3^4)(5^3)
So
(a^2)(b^3)(c^4) = (2^6)(3^4)(5^3)
Because that is a prime factorization, each exponent on the right must be a multiple of an exponent on the left. That means
(1) 5^3 = b^3, so b is 5; and
(2) 3^4 = c^4, so c is 4
That leaves a^2 = 2^6 = (2^3)^2, so a is 2^3 = 8
So for this particular problem there is only one value of a+b+c with a, b, and c distinct positive integers greater than 1.
ANSWER: a+b+c = 8+5+4 = 17
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
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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