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For positive integers n and m (n>m), is the average (arithmetic mean)

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For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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New post 25 Mar 2017, 14:41
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For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n-1}\), ......, and \(4*10^{n-m}\) an integer?

1) \(m<6\)
2) \(n=10\)

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Re: For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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New post 25 Mar 2017, 21:48
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ziyuen wrote:
For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n-1}\), ......, and \(4*10^{n-m}\) an integer?

1) \(m<6\)
2) \(n=10\)



Hi

Few important points to solve this Q..
1) n>m, so lowest term \(4*10^{n-m}\) will be ATLEAST 4*10 or 40
2) Therefore each term will be multiple of 40 or 2,4,5,8,10,20,40
3) Average of the terms is " INTEGER or Not" depends on number of elements.
4) number of elements depends on 'm' and is m+1
5) since each term multiple of 40, so if m is multiple of the factors of 40, answer is YES, otherwise it has to be checked


Let's see the statements:-
1) m<6
Therefore m can be 1 to 5
So number of terms can be 1 to 5+1..
We already know if number of elements are factors of 40, and is YES..
So 1,2,4,5 will have answer as YES..
Let's check for 3 and 6..

Each number in series consists of digits 4 and 0.... 40,400,4000 etc
So if there are 3 or 6 numbers, the SUM of digits of total of these numbers will be 3*4 or 6*4, which will be divisible by 3, thus we will have an integer here too..
Example:- 40+400+4000=4440.. sum of digits is 12, so number is div by 3
Or 400000+40000+4000=444000, again SUM of digits is 12

6 numbers will have total div by 3 and the numbers being EVEN, total will be div by 6, thus an integer again

Ans is YES always
Sufficient

2) n=10..
We are concerned about m only
Insufficient

A
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Re: For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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New post 26 Mar 2017, 12:23
Hi Chetan,

I did not understand your explanation.
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For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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New post 19 Aug 2018, 18:16
Ques: Arithmetic mean an integer?
we can simply take common from the expression 4∗10^n , 4∗10^n−1, ......, and 4∗10^n−m
=> 4*10^n (1+1/10 +1/10^2+......+1/10^m)---------(1)

so here we can see that the value of 'n' doesn't matter...it can be 40,400,40000.....
The only information you need is the value of 'm' , hence A sufficient !

Further: you can take the LCM of equation 1
4*10^n-m (10^m+10^m-1.....+1) => 4*10^n-m (11111.....m times)
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Re: For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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New post 12 Sep 2018, 00:06
How did we assume m to be between 1 to 5? if M< 6 then can't m be negative, like say m = -6?
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Re: For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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New post 08 Nov 2018, 06:32
Madhavi1990 wrote:
How did we assume m to be between 1 to 5? if M< 6 then can't m be negative, like say m = -6?

Stem mentions n and m are positive integers
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Re: For positive integers n and m (n>m), is the average (arithmetic mean) &nbs [#permalink] 08 Nov 2018, 06:32
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