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GMAT Instructor
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If the sum of four consecutive positive integers a three [#permalink]
01 Oct 2006, 02:13
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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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VP
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C is my answer.
Theare are 9 possible 3 digit numbers: 100, 200, ...900
9 means.
K will be 9.
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The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.
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Senior Manager
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E more than 10
100,150,200,250,300,350...950
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Senior Manager
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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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Senior Manager
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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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Senior Manager
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could you explain how you derived 4n+6 please? thanks
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Senior Manager
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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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VP
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the word "integers" is important
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Manager
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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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GMAT Instructor
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Re: PS: Multiple of 50 [#permalink]
03 Oct 2006, 09:01
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
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Re: PS: Multiple of 50
[#permalink]
03 Oct 2006, 09:01
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