Question 1171176: If a^2 x b^3 x c^4 = 648 000, where a, b, and c are distinct positive integers greater than 1, what is the least possible value of a + b + c?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Find the prime factorization
648,000 = 2^6 * 3^4 * 5^3
Compare this to a^2 * b^3 * c^4
Note how 2^6 * 3^4 * 5^3 doesn't have a squared factor term, but it has cubic and fourth power terms.
We can fix this by pulling 2^4 out of 2^6 and having it pair with the 3^4 like so:
a^2 * b^3 * c^4 = 2^6 * 5^3 * 3^4
a^2 * b^3 * c^4 = (2^6) * 5^3 * 3^4
a^2 * b^3 * c^4 = (2^2*2^4) * 5^3 * 3^4
a^2 * b^3 * c^4 = 2^2 * 5^3 * (2^4*3^4)
a^2 * b^3 * c^4 = 2^2 * 5^3 * (2*3)^4
a^2 * b^3 * c^4 = 2^2 * 5^3 * 6^4
At this point, we can see
a^2 = 2^2
b^3 = 5^3
c^4 = 6^4
which leads to
a = 2
b = 5
c = 6
and finally,
a+b+c = 2+5+6 = 13
This is the smallest value of a+b+c possible. The proof is given in the next section.
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Here's a way to help prove we have the smallest a,b,c triple.
We know that a > 1 is some integer. Let's say 'a' is the smallest possible integer given that condition, so let's make a = 2.
a = 2
a^2 = 4
Divide like so
(648,000)/4 = 162,000
This must mean
b^3*c^4 = 162,000
c^4 = 162000/(b^3)
c = (162000/(b^3))^(1/4)
Let
c = (162000/(x^3))^(1/4)
so that implies that b = x
We can construct the function
f(x) = a+b+c
f(x) = 2+x+(162000/(x^3))^(1/4)
in which we want to minimize, such that x > 0 and x is an integer.
The use of differential calculus or a graphing calculator will find that the min of f(x) is located at approximately (4.71, 12.98)
But we can't have non-integer x values, so we could try f(4) and f(5) since 4.71 is between 4 and 5.
It turns out that f(4) = 13.09 approximately and f(5) = 13 exactly. Any other x value will lead to f(x) being larger. So f(5) = 13 leads to x = 5 and b = 5 and c = 6.
This fully proves that we have the smallest a,b,c triple possible. I skipped over the actual steps of using differential calculus, so let me know if you need me to go over that. The use of a graphing calculator is the preferred, quickest, and most efficient way to do this problem in my opinion.
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Answers:
a = 2
b = 5
c = 6
a+b+c = 13 is the smallest possible sum.
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