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A convex polygon has 629 distinct diagonals. How many sides

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kevincan
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A convex polygon has 629 distinct diagonals. How many sides [#permalink] New post   14 Jul 2006, 04:09
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A convex polygon has 629 distinct diagonals. How many sides does it have?
A) 29 B) 33 C) 37 D) 43 E) 51



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Re: A convex polygon has 629 distinct diagonals. How many sides [#permalink] New post   14 Jul 2006, 05:10
yes, using the formula n*(n-3)/2 for the number of diagonals.
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Re: A convex polygon has 629 distinct diagonals. How many sides [#permalink] New post   14 Jul 2006, 09:32
Answer : C) 37

Am I expected to remember that the square root of 5041 is 71 for the GMAT?? :?:
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Re: A convex polygon has 629 distinct diagonals. How many sides [#permalink] New post   14 Jul 2006, 11:38
I did not know the formula but I created one using this approach:

For a pentagon there are a total of 10 (4 + 3 +2 + 1 i.e sum of first n-1 integers) lines out of which 5 are outer line not the diagnols.
So the formula will be = (sum of first n-1 integers) -n

Total diagnols = (n-1)(1+n-1)/2 -n = n(n-1)/2 -n

629 = n(n-1)/2 -n i.e 629 = n^2-n/2 - n
i.e n^2 - 3n -1258 = 0
i.e (n-37)(n+34) = 0
So n = 37
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Re: A convex polygon has 629 distinct diagonals. How many sides [#permalink] New post   14 Jul 2006, 13:03
That formula would've helped me immensely.
I faced:
"Which of the following could not be the number of diagonals in a polygon"
when it counted most & I totally bricked on it.
Also keep in mind that the number of diagonals in polygon follow a recursive series:
0,2,5,9,14,20,27....

where each term a[n] = a[n-1] + (a[n-1] - a[n-2]) + 1
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Re: A convex polygon has 629 distinct diagonals. How many sides [#permalink] New post   14 Jul 2006, 13:46
ps_dahiya wrote:
I did not know the formula but I created one using this approach:

For a pentagon there are a total of 10 (4 + 3 +2 + 1 i.e sum of first n-1 integers) lines out of which 5 are outer line not the diagnols.
So the formula will be = (sum of first n-1 integers) -n

Total diagnols = (n-1)(1+n-1)/2 -n = n(n-1)/2 -n

629 = n(n-1)/2 -n i.e 629 = n^2-n/2 - n
i.e n^2 - 3n -1258 = 0
i.e (n-37)(n+34) = 0
So n = 37


Excellent! And if we notice that n(n-3)=1258, we can cheat a little and reason that n must end have a units digit of 7 or 6!



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Re: A convex polygon has 629 distinct diagonals. How many sides [#permalink]
14 Jul 2006, 13:46
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