Question 1199522: if  , where a,b and c are positive integers, and b < c, evaluate the smallest possible value of abc
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
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The variable 'a' is a positive integer {1,2,3,...}
The smallest possible value is a = 1.
If a = 1, then,
There are infinitely many solution pairings for b,c
Here's a small subset of possible integer solutions
b = 23, c = 8
b = 46, c = 17
b = 69, c = 26
b = 92, c = 35
The values of b are multiples of 23. The values of c are one less than a multiple of 9.
Unfortunately all of these b,c pairings do not satisfy the condition that b < c.
We'll ignore this value of 'a'.
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If a = 2, then,
leads to
follow similar steps shown in the previous section.
This time we have this subset of solutions
b = 14, c = 8
b = 28, c = 17
b = 42, c = 26
b = 56, c = 35
Like before the condition b < c isn't met.
We'll ignore this value of 'a'.
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If a = 3, then
leads to
Then we have this subset of solutions (out of infinitely many).
b = 5, c = 8
b = 10, c = 17
b = 15, c = 26
b = 20, c = 35
Finally the condition b < c has been met.
Go for the smallest b,c pairing so that we arrive at the smallest product.
a*b*c = 3*5*8 = 120
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If a = 4, then
leads to
Then that can be arranged into
b > 0 so 9b > 0 as well. The right hand side is positive.
But c > 0 leads to c+1 > 0 and -4(c+1) < 0
The left hand side is negative.
In short, we do not have any positive integer solutions when a = 4, b > 0 and c > 0.
We would need to allow b or c to be negative if we wanted solutions.
Similar situations happen when a = 5, a = 6, etc.
It allows us to ignore these values of 'a'.
Notice how  where 3 is the whole number part.
This is why a > 3 leads to b or c being negative.
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Answer: 120
This minimized product occurs when (a,b,c) = (3,5,8).
Answer by greenestamps(13200)  (Show Source):
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