SOLUTION: Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.

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Question 1045089: Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.
Answer by KMST(5328) About Me  (Show Source):
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Let n-1, n, and n%2B1 be the 3 consecutive numbers.
%28n-1%29n= the product of the smaller two numbers
%28n%2B1%29%5E2= the square of the largest number
%28n%2B1%29%5E2-19= 19 less than the square of the largest number
The problem says that
%28n-1%29n=%28n%2B1%29%5E2-19 .

Solving that equation:
%28n-1%29n=%28n%2B1%29%5E2-19
n%5E2-n=n%5E2%2B2n%2B1-19
-n=2n-18
-n%2Bn=2n-18%2Bn
0=3n-18
18=3n
18%2F3=n
n=6
The 3 consecutive numbers are
n-1=6-1=highlight%285%29 , n=highlight%286%29 , and n%2B1=6%2B1=highlight%287%29 .

Verification:
The product of the smaller two numbers is 5%2A6=30
19 less than the square of the largest number is
7%5E2-19=49-19=30