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

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Re: For positive integers n and m (n>m), is the average (arithmetic mean) [#permalink]
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]
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]
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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Re: For positive integers n and m (n>m), is the average (arithmetic mean) [#permalink]
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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Re: For positive integers n and m (n>m), is the average (arithmetic mean) [#permalink]
2
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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]
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