Jamierox4ev3r
  • Jamierox4ev3r
Find four consecutive odd integers whose sum is 232. I used the guess and check method, so I know the answer. However, I was wondering if anyone knew the formula. :)
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Four consecutive odd integers, eh? Let \(x\) be the first of these odd integers. The next odd integer is thus \(x+2\), and the next is \(x+4\), and the last is \(x+6\). The sum of all four of these is 232, so you have \[x+x+2+x+4+x+6=232\\ 4x+12=232\] solve for \(x\).
anonymous
  • anonymous
x can be even sith lord
Jamierox4ev3r
  • Jamierox4ev3r
Thank you @SithsAndGiggles . I will use the formula to see if it works; I already know the answer by guessing and checking (lol), thanks for providing an actual formula

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anonymous
  • anonymous
@thisSucks, in that case, let \(x=2n+1\). Replace all the x's, and solve for \(n\), then plug into \(x=2n+1\) and solve for \(x\).
anonymous
  • anonymous
yeah 2n+1 looks correct @SithsAndGiggles I am Lord Kaan

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