Find three consecutive positive integers such that the product of the first two is 11 more than five times the third.
We have three consecutive positive integers. Let's call them A, B, and C. One clue is that they are consecutive. That means, if we let A be the smallest of the integers and C to be the largest, that:
B = A + 1
C = A + 2
No matter what A is, the above mathematical statements or equations will be true.
The other piece of information that we have from the original problem is this: The product of the first two is 11 more than five times the third. We need to translate this piece of information into mathematical language. "Product" means the answer to a multiplication problem, so this English sentence is saying the following:
AB = 11 + 5C
We have three equations now with three unknowns. With any luck at all, these equations will be solvable.
I am going to take the third equation (AB = 11 + 5C) and substitute the values for B and C as defined in the first two equations to get the following:
B = A + 1
C = A + 2
AB = 11 + 5C
A(A + 1) = 11 + 5(A + 2)
Time to distribute to expand the equation:
A2 + A = 11 + 5A + 10
Let's move all the terms to the left side of the equal sign.
A2 + A = 11 + 5A + 10
–5A –11 –10 –11 –5A –10
A2 –4A – 21 = 0
Now we need to factor the polynomial expression on the left side of the equal sign.
(A – 7)(A + 3) = 0
This means that (A – 7) or (A + 3) = 0.
If (A – 7) is 0, then A = 7.
If (A + 3) is 0, then A = –3.
However, the original problem said that the integers (A, B, and C) were all positive. Thus, A is 7.
Since A is the smallest and C is the largest of our three consecutive positive integers, B is 8 and C is 9.
A = 7
B = 8 (=A + 1)
C = 9 (=A + 2)
To check, is it true that seven times eight equals 11 more than five times 9. Yes! 56 = 11 + 45.
