Question 883715: I am thinking of three counting numbers which have a product of 48. Also, one less than twice their sum is a perfect square. What are my numbers?
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! You are thinking of (2,3,8).
Their product is 48 and twice their sum less one is 25 - a perfect square.
I found the answer by generating a list of possible sums of the three counting numbers given by
(1) S = (n^2 + 1)/2
In order to get an integer sum, (n^2+1) must be even. Another constraint is that the sum must be greater than three and less than 51.
Chosing n = (1,2,3... 11) we get those values of n that satisfy these two conditions
(2) n = {3,5,7,9) giving the possible sum, S, of
(3) S = {5,13,25,41}
Now to select values of the three numbers a,b,c using the condition that their product is 48 and the their sum is an entry in (3).
First let a = 1, then
(4) b*c = 48 as given by the pairs
(5) (b,c) = {(1,48),(2,24),(3,16),(4,12),(6,8)} which when added to a = 1 gives the sums
(6) a+b+c = {50,27,20,17,15}
None of the values of the sum given by (5) are equal to any of the possible sums given by (3). So try the next value of a = 2. Then
(7) b*c = 24 as given by the pairs
(8) (b,c) = {(1,24),(2,12),(3,8),(4,6)} which when added to a = 2 gives the sums
(9) a+b+c = {27,16,13,12}
Note that the sum of 13 in (9) is also an entry in (3), thus we have a solution
(10) {a,b,c} = {2,3,8} QED
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