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# N from 1-100 inclusive How many #'s such tha n*(n+1)

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CEO
Joined: 15 Aug 2003
Posts: 3469
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Kudos [?]: 703 [0], given: 781

N from 1-100 inclusive How many #'s such tha n*(n+1) [#permalink]  09 Oct 2003, 19:25
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N from 1-100 inclusive

How many #'s such tha n*(n+1) divided by 4.

25 mutliples of 4

25 numbers before 4 ( 3,4 ; 7,8 ; 11,12 ; 15,16 ...... 99,100)

24 numbers after 4 ( 4,5 ; 8,9 ; 12,13 ; 16,17 .........96,97)

So, a total of 74.

Is that correct?

thanks
praetorian
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 703 [0], given: 781

Re: PS : How many #'s [#permalink]  10 Oct 2003, 04:10
is my solution above correct?

thanks
praetorian
Manager
Joined: 26 Aug 2003
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Kudos [?]: 2 [0], given: 0

Re: PS : How many #'s [#permalink]  10 Oct 2003, 05:02
praetorian123 wrote:
N from 1-100 inclusive

How many #'s such tha n*(n+1) divided by 4.

25 mutliples of 4

25 numbers before 4 ( 3,4 ; 7,8 ; 11,12 ; 15,16 ...... 99,100)

24 numbers after 4 ( 4,5 ; 8,9 ; 12,13 ; 16,17 .........96,97)

So, a total of 74.

Is that correct?

thanks
praetorian

You can't take the original multiples of 4 in account because you're asked for n * (n+1). And don't forget the last number 10100. That makes the total of 50 numbers.
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Joined: 29 Aug 2003
Posts: 49
Location: Detroit, MI
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Kudos [?]: 2 [0], given: 0

Re: PS : How many #'s [#permalink]  10 Oct 2003, 06:03
I got 50 too. The second list should have 100, 101 combination i.e. N=100

Amar.
CEO
Joined: 15 Aug 2003
Posts: 3469
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Kudos [?]: 703 [0], given: 781

Re: PS : How many #'s [#permalink]  13 Oct 2003, 00:29
amarsesh wrote:
I got 50 too. The second list should have 100, 101 combination i.e. N=100

Amar.

thanks, i agree

i mixed it up..

a simpler way..

25 multiples of 4 and each number can occupy 2 positions...that gives us

50.
Re: PS : How many #'s   [#permalink] 13 Oct 2003, 00:29
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# N from 1-100 inclusive How many #'s such tha n*(n+1)

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