Question 1045089
Let {{{n-1}}}, {{{n}}}, and {{{n+1}}} be the 3 consecutive numbers.
{{{(n-1)n}}}= the product of the smaller two numbers
{{{(n+1)^2}}}= the square of the largest number
{{{(n+1)^2-19}}}= 19 less than the square of the largest number
The problem says that
{{{(n-1)n=(n+1)^2-19}}} .
 
Solving that equation:
{{{(n-1)n=(n+1)^2-19}}}
{{{n^2-n=n^2+2n+1-19}}}
{{{-n=2n-18}}}
{{{-n+n=2n-18+n}}}
{{{0=3n-18}}}
{{{18=3n}}}
{{{18/3=n}}}
{{{n=6}}}
The 3 consecutive numbers are
{{{n-1=6-1=highlight(5)}}} , {{{n=highlight(6)}}} , and {{{n+1=6+1=highlight(7)}}} .
 
Verification:
The product of the smaller two numbers is {{{5*6=30}}}
19 less than the square of the largest number is
{{{7^2-19=49-19=30}}}