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If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 01:01
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Competition Mode Question If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an integer. (2) z^3 is even.
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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 01:17
(1) (z + 7)/3 = 2m + 1, where m is an integer.
z+7=3(2m+1)
(2m+1) is always odd and so 3(2m+1) is odd
z+7=odd z=odd7=even
1 is sufficient
(2) z^3 is even.
Since we are given that z is an integer, if z^3 is even then z is even since raising an integer to an integer power does not change the oddeven nature
2 is sufficient
Answer is D
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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 02:05
If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an integer. Rewriting: Z + 7 = 3 (2m + 1) ... Z = 3 (2m + 1)  7 We know 2m will always be EVEN irrespective of what 'm' is. 2m + 1 = Will be always odd (even + odd = odd) 3 (2m + 1) = Will be always odd (odd * odd = odd) 3 (2m + 1)  7 = will be always even (odd  odd = even)
Sufficient
A D / B C E(2) z^3 is even. From the number properties we know that even^3 = even and odd^3 = odd. Hence, Z will be always Even. Sufficient
'D' is the winner
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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 02:40
#1 (z + 7)/3 = 2m + 1, where m is an integer. LHS ; 2m+1 will be odd ; so now to get LHS as odd value of z has to be even ; sufficient #2 z^3 is even z has to even integer; sufficient IMO D
If z is an integer, is z even?
(1) (z + 7)/3 = 2m + 1, where m is an integer. (2) z^3 is even.



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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 02:48
Quote: If z is an integer, is z even?
(1) (z + 7)/3 = 2m + 1, where m is an integer. (2) z^3 is even Question: Is integer z even?Statement 1: (z + 7)/3 = 2m + 1since, m is an integer, (z + 7)/3 = 2m + 1 = odd i.e. (z + 7) =3* Odd i.e. z = 3* odd  7 = Odd  Odd = Even SUFFICIENT Statement 2: z^3 is evensince, z is an integer therefore, z^3 also will be an integer and since z^3 is even so z also must be even SUFFICIENT Answer: Option D P.S. The question would have been far more trickier if the question had not mentioned that z is an Integer because 2nd statement then would have been Insufficient
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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 04:17
(1) (z+7)/3= odd > z+7= 3*odd= odd > z is even SUFFICIENT (2) z^3 is even. If z^3 is even, then z is even SUFFICIENT
FINAL ANSWER IS (D)
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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 10:51
If z is an integer, is z even? (1) \(\frac{(z + 7)}{3} = 2m + 1\), where m is an integer. 2m + 1 = odd since m is an integer. \(\frac{(z + 7)}{3} = 2m + 1\) = \(\frac{e + o}{o} = o\) z is even. SUFFICIENT. (2) \(z^3\) is even. \(even^3\) = even z is even. SUFFICIENT. Answer D.
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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 13:48
If z is an integer, is z even?
(Statement1): \(\frac{(z + 7)}{3}= 2m + 1\), where m is an integer. —> z+7 = 6m+ 3 z= 6m —4 = 2( 3m—2) Z—even Sufficient
(Statement2): \(z^{3}\) is even. Since \(z^{3}\) is even, z must be even (Even if z is equal to zero) Sufficient
The answer is D
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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 17:17
Quote: If z is an integer, is z even?
(1) (z + 7)/3 = 2m + 1, where m is an integer. (2) z^3 is even. z is integer (1) suficz+7/3=2m+1 z+7/odd=even+odd=odd z+7=odd*odd=odd z+odd=odd z=oddodd=even (2) suficz^3=even then z=even, because if z=odd then an odd^3=odd Ans (D)



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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 20:49
If z is an integer, is z even?
(1) \(\frac{z + 7}{3} = 2m + 1\), where m is an integer. => z+7 = 6m+3 => z = 6m  4 => z is even => Suff
(2) \(z^3\) is even. For example z^3 = 8 => z=2 => z is even but \(z^3 = 6 => z = \sqrt[3]6\) => z is not an integer. => Not suff
=> Choice A



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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 22:07
If z is an integer, is z even?
(1) (z + 7)/3 = 2m + 1, where m is an integer. (2) Z^3 is even.
From statement (1), (z + 7)/3 = 2m + 1, where m is an integer. Or, (z + 7) = 3(2m + 1) Or, z = 3(2m + 1) – 7 Or, z = 3( 2 *even/Odd + 1) 7 Or, z = 3(even + 1) – 7 Or, z = Odd *Odd 7 Or z = Odd –odd = Even, Sufficient.
From statement (2), (2) Z^3 is even. Z is even. Sufficient.
Answer: D



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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 22:12
(1) (z + 7)/3 = 2m + 1, where m is an integer.
z = 2(3m) + 3  7 z = 2(3m) 4 z = even  even z is even > sufficient
(2) z^3 is even. z^3 = Z * Z * Z
if z is odd, then Z^3 > (( Z * Z ) * Z) is ((odd) * odd) = odd if z is even, then Z^3 > (( Z * Z ) * Z) is ((even) * even) = even > Sufficient




Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an
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17 Feb 2020, 22:12






