Question 1045107
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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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{{{n*(n+1)}}} = {{{(n+2)^2 - 19}}}.


{{{n^2 + n}}} = {{{n^2 + 4n + 4 - 19}}},

n = 4n - 15,

15 = 3n,

n = 5.

The three consecutive numbers are  5,  6,  and 7.
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