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

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Question Stats: 18% (02:11) correct 82% (02:34) wrong based on 62 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

Weekly Quant Quiz #3 Question No 8

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Originally posted by gmatbusters on 06 Oct 2018, 10:34.
Last edited by gmatbusters on 07 Oct 2018, 06:39, edited 2 times in total.
Renamed the topic and edited the question.
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GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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

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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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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

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

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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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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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GMAT 1: 540 Q49 V16 GMAT 2: 680 Q49 V33 Re: For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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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

AnS E
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GMAT 1: 640 Q48 V29 Re: For positive integers n and m (n>m), is the average (arithmetic mean)  [#permalink]

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2
gmatbusters 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

Weekly Quant Quiz #3 Question No 8

the answwer is defincaty A ,

easy loigic is

AMean is integer for all odd numbers, so we need to worry for 2,4 ( values of m).

for a series with 2 or 4 values that are multiple of 4 . the AM is always an integer.

gmatbuster any views Re: For positive integers n and m (n>m), is the average (arithmetic mean)   [#permalink] 13 Sep 2019, 03:49
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