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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2954
When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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Question Stats: 89% (00:55) correct 11% (01:52) wrong based on 363 sessions

### HideShow timer Statistics 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

Here is a fresh question from e-GMAT. Go ahead and give it a shot!

Regards,
The e-GMAT 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/number-properties-question-from-qr-2nd-edition-ps-96030.html

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Originally posted by EgmatQuantExpert on 05 May 2015, 06:19.
Last edited by EgmatQuantExpert on 07 Aug 2018, 07:02, edited 2 times in total.
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2954
Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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1
2
The correct answer is Option B

Dividend = Divisor*Quotient + Remainder

As per the question statement,

$$81 = z^{3}k+17$$, where quotient k is a non-negative 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 = 1
That 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

_________________

Originally posted by EgmatQuantExpert on 05 May 2015, 06:22.
Last edited by EgmatQuantExpert on 07 Aug 2018, 07:02, edited 2 times in total.
Manager  Joined: 07 Apr 2015
Posts: 162
Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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1
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.
Manager  Joined: 18 Nov 2013
Posts: 78
Concentration: General Management, Technology
GMAT 1: 690 Q49 V34 Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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1
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, 09:49.
Last edited by UJs on 05 May 2015, 20:54, edited 3 times in total.
Manager  Joined: 09 Jan 2013
Posts: 69
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Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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1
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.

Manager  Joined: 28 Jan 2015
Posts: 126
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38 Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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1
[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...
Target Test Prep Representative G
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2821
Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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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.

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Manager  S
Joined: 10 Jun 2014
Posts: 72
Location: India
Concentration: Operations, Finance
WE: Manufacturing and Production (Energy and Utilities)
Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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EgmatQuantExpert wrote:
The correct answer is Option B

Dividend = Divisor*Quotient + Remainder

As per the question statement,

$$81 = z^{3}k+17$$, where quotient k is a non-negative 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 = 1
That 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
_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 7763
Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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EgmatQuantExpert wrote:
The correct answer is Option B

Dividend = Divisor*Quotient + Remainder

As per the question statement,

$$81 = z^{3}k+17$$, where quotient k is a non-negative 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 = 1
That 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

Yes, you are correct. Z^3 will be between 17 and 81.
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Manager  G
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Location: India
Concentration: General Management, Strategy
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Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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1
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
Math Expert V
Joined: 02 Aug 2009
Posts: 7763
Re: When 81 is divided by the cube of positive integer z, the remainder is  [#permalink]

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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 device

Wherever 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
_________________ Re: When 81 is divided by the cube of positive integer z, the remainder is   [#permalink] 01 Sep 2018, 08:43
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