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When 81 is divided by the cube of positive integer z, the remainder is
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Updated on: 07 Aug 2018, 06:02
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When 81 is divided by the cube of positive integer z, the remainder is 17. Which of the following could be the value of z? I. 2 II. 4 III. 8
(A) I only (B) II only (C) III only (D) I and II only (E) I, II and IIIHere is a fresh question from eGMAT. Go ahead and give it a shot! Select your answer in the poll and provide your solution as a reply below. Regards, The eGMAT Quant Team P.S.: Solutions with clarity of thought and elegance will get kudos! Here is an easier official question which tests a similar concept: http://gmatclub.com/forum/numberpropertiesquestionfromqr2ndeditionps96030.htmlTo read all our articles: Must read articles to reach Q51
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Re: When 81 is divided by the cube of positive integer z, the remainder is
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Updated on: 07 Aug 2018, 06:02
The correct answer is Option BDividend = Divisor*Quotient + Remainder As per the question statement, \(81 = z^{3}k+17\), where quotient k is a nonnegative integer From this equation, we get: \(z^{3}k = 81 – 17\) > \(z^{3}k = 64\) > \(z^{3}k = 2^6\) . . . (1) We are given that z is a positive integer. And, since k is the quotient, it will be an integer as well. So, the following cases arise from Equation 1: Case 1: \(z^3 = 2^6\) and k = 1That is, z = \(2^2\) = 4 Case 2: \(z^3 = 2^3\) and \(k=2^3\)That is, z = 2 Case 3:\(z^3 = 1\) and \(k = 2^6\)That is, z = 1 So, by thinking through Equation 1, we have zeroed in on three possible values of z: {1, 2, 4}Now, we know that Remainder is always less than the divisor. So, from the question statement, we can write: \(17 < z^3\) That is, \(z^3 > 17\) By testing the 3 possible values of z for this inequality, we see that only z = 4 satisfies this inequality. Therefore, the only possible value of z is 4. Hope you enjoyed doing this question! Japinder
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Re: When 81 is divided by the cube of positive integer z, the remainder is
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05 May 2015, 06:11
81 divided by z^3 leaves a remainder of 17.
81 = q * 2^3 + 17 64 = q+z^3
Then I tried answer possibilities of 2, 4 and 8.
Only 2^3 = 64 fulfills the equation and yields a remainder of 17. So B.



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Re: When 81 is divided by the cube of positive integer z, the remainder is
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Updated on: 05 May 2015, 19:54
Q 81 divided by \(z^{3}\) leaves a remainder of 17.81 = \(z^{3}\) * k + 17 (where k is constant, some integer) 64 = \(z^{3}\) * k get prime factors of 64 = \(2^{6}\) now expression {64 = \(z^{3}\) * k} can be written as \(2^{6}\) = \(z^{3}\) * k
1> \(2^{3}\) * 8 (where k=8) but \(2^{3}\) = 8 > It is too small to give 17 as reminder
2> \(4^{3}\) * 1 (where k=1)
Ans : D (both I and II valid)Ahh .. saw my mistake Ans : B
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Originally posted by UJs on 05 May 2015, 08:49.
Last edited by UJs on 05 May 2015, 19:54, edited 3 times in total.



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Re: When 81 is divided by the cube of positive integer z, the remainder is
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05 May 2015, 18:22
When 81 divided by Z^3 leaves remainder 17. It means that the 17 < Z^3 < 81
The only option that fulfills this requirement is 4. You can workout 4^3 = 64. 81/64 leaves remainder 17.
Thus answer is B.



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Re: When 81 is divided by the cube of positive integer z, the remainder is
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05 May 2015, 19:28
[quote="EgmatQuantExpert"]When 81 is divided by the cube of positive integer z, the remainder is 17. Which of the following could be the value of z? I. 2 II. 4 III. 8
(A) I only (B) II only (C) III only (D) I and II only (E) I, II and III
I got B.
We are given 81/z^3 has a remainder of 17. We are given possible values for Z, so I just plug them in:
I) z = 2 gives 81/2^3 which leaves a remainder of 1 (81/8 = 10 r 1), not a remainder of 17 so not correct
II)z = 4 gives 81/4^3 = 81/64 = 1 r 17< this fits
III) I already know that 8^3 is too big to be divisible by 81 with a remainder, so I don't check.
II is the only correct answer...



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Re: When 81 is divided by the cube of positive integer z, the remainder is
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24 Jun 2017, 07:36
EgmatQuantExpert wrote: When 81 is divided by the cube of positive integer z, the remainder is 17. Which of the following could be the value of z? I. 2 II. 4 III. 8
(A) I only (B) II only (C) III only (D) I and II only (E) I, II and III We are given that when 81 is divided by the cube of positive integer z, the remainder is 17. Let’s test each Roman numeral to determine which could be z. I. 2 2^3 = 8 Since a number divided by 8 cannot have a remainder that is greater than 7, z cannot be 2. II. 4 4^3 = 64 81/64 = 1 remainder 17 z can be 4. III. 8 8^3 = 512 81/512 = 0 remainder 81 z cannot be 8. Answer: B
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Re: When 81 is divided by the cube of positive integer z, the remainder is
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01 Sep 2018, 02:36
EgmatQuantExpert wrote: The correct answer is Option BDividend = Divisor*Quotient + Remainder As per the question statement, \(81 = z^{3}k+17\), where quotient k is a nonnegative integer From this equation, we get: \(z^{3}k = 81 – 17\) > \(z^{3}k = 64\) > \(z^{3}k = 2^6\) . . . (1) We are given that z is a positive integer. And, since k is the quotient, it will be an integer as well. So, the following cases arise from Equation 1: Case 1: \(z^3 = 2^6\) and k = 1That is, z = \(2^2\) = 4 Case 2: \(z^3 = 2^3\) and \(k=2^3\)That is, z = 2 Case 3:\(z^3 = 1\) and \(k = 2^6\)That is, z = 1 So, by thinking through Equation 1, we have zeroed in on three possible values of z: {1, 2, 4}Now, we know that Remainder is always less than the divisor. So, from the question statement, we can write: \(17 < z^3\) That is, \(z^3 > 17\) By testing the 3 possible values of z for this inequality, we see that only z = 4 satisfies this inequality. Therefore, the only possible value of z is 4. Hope you enjoyed doing this question! Japinder 81=Z^3 *k +17 Z^3 must be greater tha 17 CAN I SAY Z^3 MUST BE LESS THAN 81 then out of 2,4 & 8 only 4 satisfy the conditions If I am not wrong anywhere then it is much easier way to choose the ans ...please help experts Bunuel chetan2u



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Re: When 81 is divided by the cube of positive integer z, the remainder is
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01 Sep 2018, 02:53
janadipesh wrote: EgmatQuantExpert wrote: The correct answer is Option BDividend = Divisor*Quotient + Remainder As per the question statement, \(81 = z^{3}k+17\), where quotient k is a nonnegative integer From this equation, we get: \(z^{3}k = 81 – 17\) > \(z^{3}k = 64\) > \(z^{3}k = 2^6\) . . . (1) We are given that z is a positive integer. And, since k is the quotient, it will be an integer as well. So, the following cases arise from Equation 1: Case 1: \(z^3 = 2^6\) and k = 1That is, z = \(2^2\) = 4 Case 2: \(z^3 = 2^3\) and \(k=2^3\)That is, z = 2 Case 3:\(z^3 = 1\) and \(k = 2^6\)That is, z = 1 So, by thinking through Equation 1, we have zeroed in on three possible values of z: {1, 2, 4}Now, we know that Remainder is always less than the divisor. So, from the question statement, we can write: \(17 < z^3\) That is, \(z^3 > 17\) By testing the 3 possible values of z for this inequality, we see that only z = 4 satisfies this inequality. Therefore, the only possible value of z is 4. Hope you enjoyed doing this question! Japinder 81=Z^3 *k +17 Z^3 must be greater tha 17 CAN I SAY Z^3 MUST BE LESS THAN 81 then out of 2,4 & 8 only 4 satisfy the conditions If I am not wrong anywhere then it is much easier way to choose the ans ...please help experts Bunuel chetan2uYes, you are correct. Z^3 will be between 17 and 81.
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Re: When 81 is divided by the cube of positive integer z, the remainder is
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01 Sep 2018, 04:52
I merely substituted the values in the options to solve. chetan2u egmat can there be any other pitfall in such type of questions Posted from my mobile device



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Re: When 81 is divided by the cube of positive integer z, the remainder is
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01 Sep 2018, 07:43
saurabh9gupta wrote: I merely substituted the values in the options to solve. chetan2u egmat can there be any other pitfall in such type of questions Posted from my mobile deviceWherever you are looking for a value of a variable and the choices give you different values as in this, you can surely substitute the choices and get your answer
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Re: When 81 is divided by the cube of positive integer z, the remainder is &nbs
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