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10 Jan 2015, 14:05
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
48% (01:22) correct
52%
(01:01)
wrong
based on 180
sessions
History
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Is Z an integer?
(1) Z^3 is an integer
(2) 3Z is an integer
Below is a number plugin approach but wanted to know an algebra approach for the same. Please help.
.
(1) Z^3 is an integer
(2) 3Z is an integer
Below is a number plugin approach but wanted to know an algebra approach for the same. Please help.
Statement 1:
if z^3 is a PERFECT CUBE, such as 1, 8, or 27, then z will be an integer.
if z^3 is NOT a perfect cube, such as 2, 3, 4, etc., then z will NOT be an integer.
therefore, INSUFFICIENT.
Statement 2:
if Z = 5/3 then 3 (5/3) is an integer bur Z is not an integer. Hence insufficient.
Combining 1 & 2:
Consider all the numbers that satisfy statement (2):
1/3, 2/3, 1, 4/3, 5/3, 2, etc. of these, the only ones that satisfy statement (1) as well are 1, 2, 3, ...
(all the fractional ones will still be fractions when you cube them)
since these – the numbers that satisfy BOTH statements – are all integers, Z is an Integer.
Hence answer is C
if z^3 is a PERFECT CUBE, such as 1, 8, or 27, then z will be an integer.
if z^3 is NOT a perfect cube, such as 2, 3, 4, etc., then z will NOT be an integer.
therefore, INSUFFICIENT.
Statement 2:
if Z = 5/3 then 3 (5/3) is an integer bur Z is not an integer. Hence insufficient.
Combining 1 & 2:
Consider all the numbers that satisfy statement (2):
1/3, 2/3, 1, 4/3, 5/3, 2, etc. of these, the only ones that satisfy statement (1) as well are 1, 2, 3, ...
(all the fractional ones will still be fractions when you cube them)
since these – the numbers that satisfy BOTH statements – are all integers, Z is an Integer.
Hence answer is C
26 Mar 2015, 09:03
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mash7777
Is Z an integer?
(1) Z^3 is an integer --> z can be any integer or an irrational number, cube root from an integer, for example, \(\sqrt[3]{2}\). Not sufficient.
(2) 3Z is an integer --> z can be any integer or reduced fraction, integer/3 (2/3, 4/3, 5/3, ...). Not sufficient.
(1)+(2) Since an irrational number cannot equal to integer/3 (rational number), then z must be an integer. Sufficient.
Answer: C.
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General Discussion
11 Jan 2015, 05:34
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AjChakravarthy
Hi, plugin approach is the best way to solve this question, but let's just look at the algebraic approach as well.
st.1
z^3= I, here I is an integer and can take both positive as well as negative values.
z= (I)^1/3
now z can be integer depending upon the value of I. if I = -3, then z = (-3)^1/3, which is clearly not an integer. if I =8, then (8)^1/3 =2, which is an integer. hence st. 1 alone is not sufficient
st.2
3z = k, here k is an integer, which can take both positive as well as negative values.
z = k/3
now depending upon the value of k, z can either integral or non-integral values.
e.g. if k=5 then z is not an integer
if k=-3, then z is an integer.
st.1 + st.2
z= (I)^1/3
z= k/3
therefore we have (I)^1/3 = k/3
3(I)^1/3 = k
right hand side = k which is an integer. therefore left hand side must also be an integer. i.e.3(I)^1/3 must be an integer. which is possible only if (I)^1/3 is an integer.
if (I)^1/3 is an integer, then z is also an integer.
hence z is an integer.
therefore C.
