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If n is an integer and n^3 is between 1 and 100, inclusive [#permalink]

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19 Oct 2010, 11:25

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If n is an integer and n^3 is between 1 and 100, inclusive, what is the value of n?

(1) n = 2k+1, where k is an integer. (2) n is a prime number.

Need help on this question which is explained in the Number Properties book by Manhattan. According to Manhattan, the OA is C, but to me, the OA should be A.

We know that n>0 because n^3 is between 1 and 100, so n is a positive integer. The value of n could be 1, 2, 3, and 4.

Statement 1: n = 2k+1. So n is an odd number. So n is either 1 or 3. According to MGMAT, statement 1 is not sufficient to answer the question because n could be either 1 or 3 and so there is not a unique value of n as it yields two possible values. But this is where I beg to differ. According to me, the value of n could be only 3 (n = 2*1+1).

Need help on this question which is explained in the Number Properties book by Manhattan. According to Manhattan, the OA is C, but to me, the OA should be A.

If n is an integer and n^3 is between 1 and 100, inclusive, what is the value of n? (1). n = 2k+1, where k is an integer. (2). n is a prime number. We know that n>0 because n^3 is between 1 and 100, so n is a positive integer. The value of n could be 1, 2, 3, and 4.

Statement 1: n = 2k+1. So n is an odd number. So n is either 1 or 3. According to MGMAT, statement 1 is not sufficient to answer the question because n could be either 1 or 3 and so there is not a unique value of n as it yields two possible values. But this is where I beg to differ. According to me, the value of n could be only 3 (n = 2*1+1). Tell me where I am wrong.

If n is an integer and n^3 is between 1 and 100, inclusive, what is the value of n?

\(n\) is an integer and \(n^3\) is between 1 and 100, inclusive, means that \(n\) could be 1, 2, 3 or 4 (but not 5 or more since 5^3=125>100).

(1) n = 2k+1, where k is an integer --> \(n\) is an odd number --> \(n\) could be 1 or 3. Not sufficient.

(2) n is a prime number --> \(n\) could be 2 or 3. Not sufficient.

(1)+(2) \(n\) could be only 3. Sufficient.

Answer: C.

As for your doubt: \(n=2k+1\), where \(k\) is an integer is a fromula of odd number so you can get ANY odd number with it, including 1: if \(k=0\) then \(n=2k+1=1\).

Thanks. It cleared my doubt, especially the last part in your solution where you talk about the value of the integer k. I simply forgot the fact that 0 is and integer and so n could be 1 or 3. thats the reason why I was getting 3 as the only value of n from the 1st statement. +1 to you.

so K can take ----- -4 -3 -2 -1 0 1 2 3 4 ---- but if i take any negative it wont be between 1 and 100 as a power cube.. so negative numbers out. from n = 2K + 1 ===> k should be either 0 or 2 if it is 3 then n = so i should consider only 0,1,2,3,.....

but n = 2k + 1 and n^3 is between 1 and 100 inclusive... there could be a chance of 2 values for n

if k = 0, then n = 1 ===> n^3 =1 <= 100 K = 1, then n= 3 ===> n^3 = 27 <= 100 k = 2 then n = 5 ====> n^3 = 5^3 = 125 > 100

So from 1 only 2 possible values 1 and 3 for the N

Re: doubt on question from manhattangmat book number properties [#permalink]

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05 Jun 2012, 14:40

kashishh wrote:

given n*n*n = inclusive 1 - 100 St.1 ) n = 2k+1 implies n is odd and n*n*n = 1 - 100, n= 1, 3 - NOT sufficient

St.2) n=prime no. implies n= 2,3 (as again n*n*n = 1- 100) NOT sufficient

St.1) and St. 2) = 3, sufficient.

Answer - C

thanks a lot......i need 1 more help....while solving quant section,most of my questions on inequalities,probablity,standard deviation etc are always gettin wrong....ny idea from where should i study these......

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Re: If n is an integer and n^3 is between 1 and 100, inclusive [#permalink]

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22 Jul 2014, 04:54

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