When 81 is divided by the cube of positive integer z, the remainder is

Question

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

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Here is an easier official question which tests a similar concept: https://gmatclub.com/forum/number-properties-question-from-qr-2nd-edition-ps-96030.html

To read all our articles: Must read articles to reach Q51

Here is an easier official question which tests a similar concept: https://gmatclub.com/forum/number-properties-question-from-qr-2nd-edition-ps-96030.html To read all our articles: Must read articles to reach Q51 https://e-gmat.com/blogs/wp-content/uploads/2018/08/Free-Session-NP-e1533292073470.jpg

EgmatQuantExpert · Accepted Answer

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

Japinder https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-NP-e1533292144892.png

akhilbajaj · Answer

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.

noTh1ng · Answer

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.

UJs · Answer

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

sabineodf · Answer

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...

JeffTargetTestPrep · Answer

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

janadipesh · Answer

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

chetan2u · Answer

Yes, you are correct. Z^3 will be between 17 and 81.

saurabh9gupta · Answer

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

chetan2u · Answer

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

Abhishek009 · Answer

Plug in and check the options -

I.  z = 2  , So  z^3 = 8  , Hence  \frac{81}{8}  = remainder  1

II.  z = 4  , So  z^3 = 64  , Hence  \frac{81}{64}  = remainder  17

III.  z = 8  , So  z^3 = 512  , Hence  \frac{81}{512}  = remainder  81

Hence, Answer must be (B) II Only.

100mitra · Answer

Correct option : B (4)
Only 4^3 = 64 can be avialable between 17 to 81

Sushen · Answer

D 81=z^3 k+17 z^3 k=81-17=64=2^6

Possibilities:
k=1 z=2^2=4
k=2^3 z = 2
k=2^6 z = 1
so z∈{1,2,4}

Restrictions:
The remainder is always less than the divisor. SO… The divisor was z^3 
The remainder was 17 SO… z^3<17
 z<∛17 z is an integer
 SO z<3
 
Refined, z≠4 z∈{1,2}

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When 81 is divided by the cube of positive integer z, the remainder is

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