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If the sum of four consecutive positive integers a three [#permalink]
01 Oct 2006, 01:13

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 1 sessions

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=

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.

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 .

Re: PS: Multiple of 50 [#permalink]
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 mâ‚¬{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

gmatclubot

Re: PS: Multiple of 50
[#permalink]
03 Oct 2006, 08:01

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...