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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 Updated on: 07 Oct 2018, 05:39
1
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A
B
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Question Stats:

21% (01:15) correct 79% (01:42) wrong based on 39 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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Originally posted by gmatbusters on 06 Oct 2018, 09:34.
Last edited by gmatbusters on 07 Oct 2018, 05:39, edited 2 times in total.
Renamed the topic and edited the question.
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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 06 Oct 2018, 09:53
The answer should be A.

Please find soln attached.

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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 06 Oct 2018, 09:43
E.

No information is given about n in Statement 1. No information is given about m in Statement 2. Combining the statements, it can be determined that the sum of all entries amount to an integer but because we don't know if there is an odd or even number of entries and, thus, if sum of all entries divided by the number of entries will be an integer or not.
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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 06 Oct 2018, 09:46
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


av= 4 (10^n +10^n-1 + 10 ^ n-2 +......10^(n-m)/(m+1)

= (4 *10^(n) (1+1/10 +1/10^2 +.....+1/10^m))/(m+1)

= (4* 10^n * (1* (1/10)^m-1 - 1)/ (1-1/10). )/(m+1)

so if m<6 we cant say for m+1 in denominator as if m=2 m+1 becomes 3 ...
hence not sufficient
Option B is not required

E is the answer
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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 06 Oct 2018, 09:50
E.
statment 1 is not sufficient. if we know m is less than 6 then we could have at least 6 terms in total. but we know that the numerator will not be divisible by 6. however if m is 3 then we have 4 terms and the numerator is divisible by 4

statment 2 is not sufficient because it doesnt say anything about m

both these statments won't solve the question because we dont know what m is
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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 06 Oct 2018, 09:57
St 1 suggests m<6

Since m will be in the denominator (m decides the number of factors) if we want to calculate the average, if we have m has 5 or 4 or 3 or 2, we have numerator as 4x10^n-m(1 + 10 + .... + 10^m) which are divisible by those numbers regardless of the value of n (as long as n>m)

Sufficient

St 2 gives us a value of n. But no value of m is specified (m = 1,2....11) which means when we have m in the denominator, we are not sure if it divides numerator or not)

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 06 Oct 2018, 09:57
Correct answer is 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 06 Oct 2018, 10:18
avg of 4(10^n), 4(10^(n-1)), ......, and 4(10^(n-m)) is integer when n-m=1, 2, 3 ,4, 5 ,7,8,10......etc but not for 6 & as 1111111/7 gives a non terminating number.

1) not suff..... no info about n and there could or could not be a possibility of n-m EQ & NE 6
2)not suff..... no info about M and there could or could not be a possibility of n-m EQ & NE 6

1+2) not suff........ as there could or could not be a possibility of n-m EQ & NE 6

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Re: For positive integers n and m (n>m), is the average (arithmetic mean) &nbs [#permalink] 06 Oct 2018, 10:18
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