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

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Question Stats: 16% (02:29) correct 84% (02:28) wrong based on 220 sessions

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

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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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Hi Chetan,

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

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

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

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hazelnut 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) $$n=10$$

Asked: 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?
Number of terms = n - (n-m) +1 = m+1
the average (arithmetic mean) of $$4*10^n$$, $$4*10^{n-1}$$, ......, and $$4*10^{n-m}$$ = $$\frac{4*10^n + 4*10^{n-1} + ......+4*10^{n-m}}{(m+1)}$$

1) $$m<6$$
Let us take m=0
$$4*10^n/1 = 4*10^n$$ Integer
Let us take m=1
$$\frac{4*10^n + 4*10^{n-1}}{2} = \frac{4*10^{n-1} ( 1+10)}{2} =\frac{4*10^{n-1} *11}{2}$$ Integer
Let us take m=2
$$\frac{4*10^n + 4*10^{n-1} + 4*10^{n-2}}{3} = \frac{4*10^{n-2} ( 1+10+100)}{3} =\frac{4*10^{n-2} *111}{3}$$ Integer
Let us take m=3
$$\frac{4*10^n + 4*10^{n-1} + 4*10^{n-2} + 4*10^{n-3}}{4} = \frac{4*10^{n-3} ( 1+10+100+1000)}{4} = \frac{4*10^{n-3} *1111}{4}$$ Integer
Let us take m=4
$$\frac{4*10^n + 4*10^{n-1} + 4*10^{n-2} + 4*10^{n-3} + 4*10^{n-4}}{5} = \frac{4*10^{n-4} ( 1+10+100+1000+10000)}{5} = \frac{4*10^{n-4} *11111}{5}$$ Integer
Let us take m=5
$$\frac{4*10^n + 4*10^{n-1} + 4*10^{n-2} + 4*10^{n-3} + 4*10^{n-4} + 4*10^{n-5}}{6} = \frac{4*10^{n-5} ( 1+10+100+1000+10000 +10^5)}{6} =\frac{4*10^{n-5} *111111}{6}$$ Integer
SUFFICIENT

2) $$n=10$$
Since if we take n=10 and m=6
The expression will not be divisible by 7
NOT SUFFICIENT

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

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hazelnut 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.....

Not convinced with the OA..... Let us consider n=10, m=3, which gives Avg as an Integer. However, if we consider n=10, m=5, its Avg will not be an integer. I had opted for Optn E.... Please clarify.

Bunuel chetan2u request your help Re: For positive integers n and m (n>m), is the average (arithmetic mean)   [#permalink] 28 Aug 2019, 20:16
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