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Find 4 consecutive odd integers where the product of the two smaller numbers is 64 less than the product of the two larger numbers.

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MattDid
Odd number is: (2n-1), (2n+1), (2n+3), (2n+x),... where x x changes every two
(2n-1)(2n+1)=(2n+3)(2n+5)-64
[tex]4n^2-1=4n^2+10n+6n+15-64[/tex]
16n=48|:16
n=3
Now we substitute to (2n-1), (2n+1), (2n+3), (2n+5):
2n-1 = 2*3-1=5
2n+1 = 2*3+1=7
2n+3 = 2*3+3=9
2n+5 = 2*3+5=11
5,7,9,11
AL2006id
If there are such numbers, then they can be written as 'x', (x + 2), (x + 4), and (x + 6).

Now, the problem says that  x(x+2) + 64 = (x+4) (x+6)

Expand each side:

x² + 2x + 64 = x² + 10x + 24

Subtract (x² + 24) from each side:

2x + 40 = 10x

Subtract 2x from each side:

40 = 8x

Divide each side by 8 :

x = 5

The numbers are 5, 7, 9, and 11.

(5 x 7) + 64 = 35 + 64 = 99  and  9 x 11 = 99 .    yay !

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