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If the sum of four consecutive positive integers a three

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If the sum of four consecutive positive integers a three [#permalink] New post 01 Oct 2006, 01:13
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A
B
C
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E

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If the sum of four consecutive positive integers a three digit multiple of 50, the mean of the these integers must be one of k possible values, where k=

(A) 7 (B) 8 (C) 9 (D) 10 (E) more than 10
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 [#permalink] New post 01 Oct 2006, 02:05
C is my answer.

Theare are 9 possible 3 digit numbers: 100, 200, ...900

9 means.

K will be 9.
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 [#permalink] New post 01 Oct 2006, 04:13
E more than 10

100,150,200,250,300,350...950
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 [#permalink] New post 01 Oct 2006, 06:05
thinking about this the multiples of 100 don't work. the number has to be not divisible by 4 for this to work as otherwise you cannot find 4 cons digits that add up.

so 150,250,350,450,550,650,750,850,950
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 [#permalink] New post 01 Oct 2006, 07:36
londonluddite wrote:
thinking about this the multiples of 100 don't work. the number has to be not divisible by 4 for this to work as otherwise you cannot find 4 cons digits that add up.

so 150,250,350,450,550,650,750,850,950


C

Yes , all these numbers can be put in the from 4n+ 6 .
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 [#permalink] New post 01 Oct 2006, 08:05
could you explain how you derived 4n+6 please? thanks
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 [#permalink] New post 01 Oct 2006, 17:51
londonluddite wrote:
could you explain how you derived 4n+6 please? thanks


sum of 4 consecutive numbers

n + n+1 + n+2 + n+3 = 4n + 6
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 [#permalink] New post 02 Oct 2006, 03:35
the word "integers" is important :beat
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 [#permalink] New post 02 Oct 2006, 06:39
guys,

but mutiples of 100 work for 4n+6 as well

4n+6=200, n=49
4n+6=300, n=64

and so on...

so all mutiples of 50 work for 4n+6
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Re: PS: Multiple of 50 [#permalink] New post 03 Oct 2006, 08:01
kevincan wrote:
If the sum of four consecutive positive integers a three digit multiple of 50, the mean of the these integers must be one of k possible values, where k=

(A) 7 (B) 8 (C) 9 (D) 10 (E) more than 10


Let the four integers be x,x+1,x+2 and x+3=4x+6

4x+6 has to be a multiple of 50 i.e 4x+6=50m m€{2,3...,19}

But x=(50m-6)/4= (25m-3)/2 must be an integer, so m must be odd.

Thus m can be any odd integer from 3 to 19 3=1+1*2 19=1+9*2

So there are 9 different values for m, x and 4x+6, as well as (4x+6)/4
Re: PS: Multiple of 50   [#permalink] 03 Oct 2006, 08:01
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