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tennis_ball

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jyotsnasarabu
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tennis_ball [#permalink] New post   01 Oct 2006, 21:13
Hi,

Can u help me solve the below mentioned prob:

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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Re: tennis_ball [#permalink] New post   02 Oct 2006, 01:11
Wow. I was frightened by the challenge.

but i got the answer C: 9.

Is it correct? :-D
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Re: tennis_ball [#permalink] New post   02 Oct 2006, 19:18
I went with E on this one...

I tried to come up w/ a test case to solve this.. I knew that the ones digits of the 4 consecutive digits had to add up to either 5 or 0 for it to be a multiple of 50..

I went w/ 61, 62, 63, and 64.. all consectuive..and add up to 250 ..

and thus the mean is greater than 10... so anything other than E has to be wrong...

tough one.
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Re: tennis_ball [#permalink] New post   02 Oct 2006, 19:43
treat the 4 numbers as x-1, x, x+1, x+2.
sum is 4x +2.

and the 3-digit 50 multiples are 100, 150, 200, ... 850, 900, 950.

then u get 4x + 2 = 50m
and x = (25m-1)/2.

x is an integer, then 25m-1 must be even. so 25m must be odd. then m must be odd.
among the multiples of 50:
2*50, 3*50, .... 18*50, 19*50,

there are 9 ms that are odd.
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Re: tennis_ball [#permalink] New post   02 Oct 2006, 20:37
but the mean of the numbers is (4x +2)/4 = 50m/4 = 12.5*m

but I guess mean can be anywhere from 12.5*1 to 12.5*9

so the answer is still 9
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Re: tennis_ball [#permalink] New post   02 Oct 2006, 21:24
Take the 4 consecutive integers as:

a - 3d, a --d, a+ d, a + 3d (where 2d = 1).

Sum them. It is 4a.

Now 4a = 100, 150, ....950.

Since a is an integer, you have to eliminate 150, 250,....etc. You have 100, 200,...,900 left. Count them. It is 9. So only 9 possibilities are there. Ans C.
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Re: tennis_ball [#permalink] New post   03 Oct 2006, 02:56
Zooroopa,

why do you say the mean must be an integer?

the question does not tell you that, don't think we can assume that

my answer is E because the mean does not have to be an integer
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Re: tennis_ball [#permalink] New post   03 Oct 2006, 05:24
4 consecutive integers cannot add up to 100, try it.

therefore we have 150,250,350 ... 950, so 9

some notes about consecutives:
the average of an even number of consecutives is never a whole number

the average of an odd number of consecutives is a whole number

the sum of an odd number of consecutives is divisible by the number of terms (1,2,3 = 6/3 = 2 or 10,11,12,13,14,15 = 65/5 = 13). this DOES not apply to an even number of terms.

the product of any n consecutives has n as a factor.
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Re: tennis_ball [#permalink] New post   03 Oct 2006, 05:49
How about this for a theory :lol: any care to prove it?

If, when N is even, the average of N consecutive numbers is always something ending in .5, then the sum of these integers cannot be divisible by N.

e.g

N(integer) + N(0.5) is not divisible by N

e.g.

ave {1,2,3,4} = 2.5
sum = 4(2) + 4(0.5) - NOT DIV BY 4

therefore in this case we are looking for integers not divisible by 4, which means all the factors of 100 are out. Leaving us with 9.

Anyone better at maths than me (i.e. most of you :lol:) care to comment on the above?

https://mathforum.org/library/drmath/view/55975.html
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Re: tennis_ball [#permalink] New post   03 Oct 2006, 06:02
tennis_ball wrote:
treat the 4 numbers as x-1, x, x+1, x+2.
sum is 4x +2.

and the 3-digit 50 multiples are 100, 150, 200, ... 850, 900, 950.

then u get 4x + 2 = 50m
and x = (25m-1)/2.

x is an integer, then 25m-1 must be even. so 25m must be odd. then m must be odd.
among the multiples of 50:
2*50, 3*50, .... 18*50, 19*50,

there are 9 ms that are odd.




the above solution sounds convincing



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Re: tennis_ball [#permalink]
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