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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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If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an  [#permalink]

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17 Feb 2020, 01:01
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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  [#permalink]

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17 Feb 2020, 01:17
1
1
(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=odd-7=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 odd-even nature

2 is sufficient

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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an  [#permalink]

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17 Feb 2020, 02:05
1
If z is an integer, is z even?

(1) (z + 7)/3 = 2m + 1, where m is an integer.
Re-writing: 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  [#permalink]

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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  [#permalink]

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17 Feb 2020, 02:48
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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 + 1
since, 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 even
since, 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

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  [#permalink]

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17 Feb 2020, 04:17
1
(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

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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an  [#permalink]

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17 Feb 2020, 10:51
1
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.

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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an  [#permalink]

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17 Feb 2020, 13:48
1
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

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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an  [#permalink]

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17 Feb 2020, 17:17
1
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) sufic
z+7/3=2m+1
z+7/odd=even+odd=odd
z+7=odd*odd=odd
z+odd=odd
z=odd-odd=even

(2) sufic
z^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  [#permalink]

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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  [#permalink]

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17 Feb 2020, 22:07
1
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.

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Re: If z is an integer, is z even? (1) (z + 7)/3 = 2m + 1, where m is an  [#permalink]

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17 Feb 2020, 22:12
1
(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   [#permalink] 17 Feb 2020, 22:12
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