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Caas
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Product of 2 consecutive odd integers is 783. Find their 21 May 2007, 11:49
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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Product of 2 consecutive odd integers is 783. Find their sum.

A) 44
B) -56
C) -76
D) 64
E) -36



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KillerSquirrel
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Product of 2 consecutive odd integers is 783. Find their 21 May 2007, 12:49
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x*(x+2) = 783 (two consecutive numbers i.e x=3,x+2=5)

x^2+2x = 783

x^2+2x-783 = 0

factoring 783

783:3
261:3
87:3
29

3*3*3*29 = 29*27

(x+29)(x-27) = 0

x can be either -29,27

if x = 27 then x+2 = 29 and 27+29 = 56

if x = -29 then x+2 = -27 and -29+(-27) = -56

the only possible answer is (B)

:-D
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Product of 2 consecutive odd integers is 783. Find their 23 May 2007, 00:21
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factorize 783

783 = 3 x 3 x 3 x 29

hence all factors - 1,783,3,9,27,29,87,261

only two consecutive odds = 27,29

Hence B
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Juaz
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Product of 2 consecutive odd integers is 783. Find their 23 May 2007, 01:03
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Caas
Product of 2 consecutive odd integers is 783. Find their sum.

A) 44
B) -56
C) -76
D) 64
E) -36
Show more


As two integers are next to each other, one quick way of sloving this is look for square root of 783 = 28 appx (as 28*28 =784).

so integers could be 27 & 29 OR -27 & -29.
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KillerSquirrel
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Product of 2 consecutive odd integers is 783. Find their 23 May 2007, 01:36
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Juaz
Caas
Product of 2 consecutive odd integers is 783. Find their sum.

A) 44
B) -56
C) -76
D) 64
E) -36

As two integers are next to each other, one quick way of sloving this is look for square root of 783 = 28 appx (as 28*28 =784).

so integers could be 27 & 29 OR -27 & -29.
Show more


Nice thanks ! a real time saver !

:-D
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Caas
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Product of 2 consecutive odd integers is 783. Find their 23 May 2007, 09:01
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Thank you for the discussion guys :)
You are all right. OA is B indeed.
For those who wonder - another way of solving it
(2n-1)*(2n+1)=783
4n^2-1=783
n^2=784/4
n=+/- 14
2n=+/- 28 ----> so the consecutive odd integers are +/- 27 and +/- 29 and their sum is +/- 56



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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