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If the sum of 4 consecutive positive integers is divisible b

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If the sum of 4 consecutive positive integers is divisible b [#permalink] New post 28 Sep 2009, 18:40
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If the sum of 4 consecutive positive integers is divisible by 5, which of the following must be true?

I. The smallest integer is odd

II. None of the integers is divisible by 5

III. One of the integers has 3 as its units digit

[Reveal] Spoiler:
Got this one right however was wondering if anybody knows how to proceed the algabraic way..
Maybe we can do something like this;
x+ x+1 + x+2 + x+3=5K
4x+6=5K ..?

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Re: sum of 4 consecutive positive integers [#permalink] New post 28 Sep 2009, 21:49
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4X+6=5k

1, x is odd, no hint, wrong

2, none of x-> x+3 is divisible by 5, If not

x is divisible by 5 then 4x+6 divides 5 remains 1 not 0
x+1 is ............5 then sum=4(x+1)+2 divides 5 remains 2 not 0
x+2 is ............5 then sum=4(x+2)-2 divides 5 remains 3 not 0
x+3 is.............5 then sum=4(x+3)-6 divides 5 remains 4 not 0

2 is right

3, 1 Number with the digit is 3, the other should be ended with
+ 0,1,2 then sum =10k+6 not divisible by 5
+ 1,2,4 then sum =10K+10 divisible by 5 these above enough to answer
+ 2,4,5 then
+ 4,5,6
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Re: sum of 4 consecutive positive integers [#permalink] New post 28 Sep 2009, 22:53
My logic is:

4x+6 = 5k
4x=5k-6
x=\frac{(5k-6)}{4}

Thus we start picking numbers for k

k=2 => x=1, numbers 1,2,3,4
k=3 => x=9/4
k=4 => x=14/4
k=5 => x=19/4
k=6 => x=6, numbers 6,7,8,9

If continue - get 11, 12, 13, 14, and so forth

1. False
2. True
3. False
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Re: sum of 4 consecutive positive integers [#permalink] New post 29 Sep 2009, 04:05
looks like ok, but need to try some more numbers.
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Re: sum of 4 consecutive positive integers [#permalink] New post 29 Sep 2009, 04:06
tejal777, do you have answer to this question?
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Re: sum of 4 consecutive positive integers [#permalink] New post 29 Sep 2009, 04:14
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tejal777 wrote:
If the sum of 4 consecutive positive integers is divisible by 5, which of the following must be true?

I. The smallest integer is odd

II. None of the integers is divisible by 5

III. One of the integers has 3 as its units digit

Got this one right however was wondering if anybody knows how to proceed the algabraic way..


I typically think in terms of remainders. If x = 5*k+y (where y = 0,1,2,3 or 4; sometimes I use y = -2,-1,0,1,2). In the above problem x can be replaced with y. it's pretty obvious that the neither of the 4 numbers can be divisible by 5. Otherwise, 4 options -2,-1,0,1 or -1,0,1,2 or 0,1,2,3 (-2) or -3(2), -2,-1,0. clearly, neither of those options work. necessary and sufficient condition is: neither of the 4 numbers is divisible by 5.

I. The smallest integer is odd -> false (6,7,8,9)

II. None of the integers is divisible by 5 -> true (from above)

III. One of the integers has 3 as its units digit -> false (6,7,8,9)
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Re: sum of 4 consecutive positive integers [#permalink] New post 29 Sep 2009, 04:19
It does work. K must equal 4n+2 to be divisible by 4 for any positive n.
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Re: sum of 4 consecutive positive integers [#permalink] New post 29 Sep 2009, 08:24
Let the 4 nos be a,a+1,a+2,a+3

4a + 6 = 5k.

a = 1,6,11,16...

so the series becomes 1,2,3,4 or 6,7,8,9 or 11,12,13,14.....etc

So from the above series only option 2 is correct
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Re: sum of 4 consecutive positive integers [#permalink] New post 29 Sep 2009, 09:36
Agreed guys, its working
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Re: sum of 4 consecutive positive integers [#permalink] New post 29 Sep 2009, 20:23
OA:
[Reveal] Spoiler:
only option 2 is correct

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Re: sum of 4 consecutive positive integers   [#permalink] 29 Sep 2009, 20:23
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