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sum of 4 consecutive positive integers [#permalink]
 28 Sep 2009, 19:40
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0% (00:00) correct
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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 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]
28 Sep 2009, 22:49
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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]
28 Sep 2009, 23:53
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]
29 Sep 2009, 05:05
looks like ok, but need to try some more numbers.
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Re: sum of 4 consecutive positive integers [#permalink]
29 Sep 2009, 05:06
tejal777, do you have answer to this question?
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Re: sum of 4 consecutive positive integers [#permalink]
29 Sep 2009, 05:14
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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]
29 Sep 2009, 05:19
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]
29 Sep 2009, 09:24
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]
29 Sep 2009, 10:36
Agreed guys, its working
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Re: sum of 4 consecutive positive integers [#permalink]
29 Sep 2009, 21:23
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Re: sum of 4 consecutive positive integers
[#permalink]
29 Sep 2009, 21:23
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