Question 858210
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Find 3 odd consecutive integers such that the product of the first and third minus the middle integer is 338.
Ans:
Let the 3 odd integers be n-2,n and n+2
Product of 1st and 3rd = {{{(n - 2)*(n + 2) = n^2 - 4}}}
Subtracting middle integer
{{{n^2 - 4 - n = 338}}}
Or
{{{n^2 -n - 342 = 0}}}
This is a simple quadratic equation which can be solved by factorizing.
{{{n^2 + 18*n - 19*n - 342 = 0}}}
{{{(n + 18) * (n - 19) = 0}}}
So possible solutions for n are {{{n = -18}}} or {{{n = 19}}}

Correspondingly, the 3 numbers are (-20,-18,-16) or (17,19,21)

Since is is not mentioned that the integers are positive, both solutions are valid.

Check: {{{(-20)*(-16) - (-18) = 320 + 18 = 338}}}
{{{17*21 - 19 = 357 - 19 = 338}}}
Both are correct!

:)
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