Is m +n^3 odd?

m+\(n^3\) will be odd when m and n are of opposite nature

Option A: m = odd and n- Even - serves the purpose - Hence Sufficient
Option B: m-n = odd
This means they are of opposite nature
Hence sufficient

Therefore Answer is D (Each one of them is individually sufficient)

Is m +n^3 odd?
1) m is odd and n is even
2) m-n is odd

Option A: m is odd and n is even
n^3 = even power raised 3 times = always even (how, 2^3=8 or 4^3=64, 6^3=216)
so m+n^3 = odd + even = odd

Option B: m-n is odd
this is possible when m is odd and n is even or m is even and n is odd ( how, m =5 & n=2 --> m-n=3 (odd) or m=6 & n =3--> m-n=3 (odd))
consider m being odd and n is even
this is similar to option A , so definitely m+N^3 is odd

if m is even and n is odd
odd ^3 will be always odd as well ( how 3^3=27, 5^3=125, 7^3=343)
so m+n^3= even + odd = odd
this option also looks good

Hence either of the two option is sufficient in itself
Option D( each statement is sufficient)

Is m +n^3 odd?
1) m is odd and n is even
2) m-n is odd


1 ) even ^(any integer) = even
and odd + even = odd
thus A is sufficient
2) m-n is odd
let m = 3 and n = 2
then 3 +8 = odd
but now consider
m = 3.5 and n = 0.5
m-n = 3.5-0.5= 3 (odd)
however 3.5 + (0.5)^3 = 3.5 + .125 = 3.625
neither odd nor even
hence not sufficient
thus A

1) m is odd and n is even. Sufficient

2) m-n is odd
When m is odd, n is even
When m is even, n is odd
Sufficient

IMO D

Is \(m +n^{3}\) odd?

(Statement1): m is odd and n is even
--> \(Odd+ (Even)^{3}= Odd +Even = Odd\) (Always YES)
Sufficient

(Statement2): m-n is odd
--> In order (m-n) to be Odd number, one of them should be Odd and the other one should be Even.

if m=Even and n=Odd, then
\(Even +(Odd)^{3}= Even +Odd = Odd\) (YES)

if m=Odd and n=Even, then
\(Odd+ (Even)^{3}= Odd +Even = Odd\) (YES)

—> but what if m= 3.5 and n= 0.5, then 3.5+(0.5)^{3} — not odd number. (NO)
Insufficient

The answer is A .

Answer : A

Solution posted in the picture attached.

Ans for this Question should be A.
For this type of Ques we should determine whether it is given that M and N are integers .
If not then we need to be careful.

With statement 1 since it clarifies that M and N are odd and even respectively thus we can say that they are sufficient
With statement 2 we cannot since M and N can be 4.2 and 1.2 respectively which in turn is not sufficient

#1
m is odd and n is even
so m +n3 ; shall be odd + even = odd always
sufficient
#2
m-n is odd
either of m & n has to be even or odd
in that case m +n3 = odd/even + even/odd ; ODD
Or m is 1 and n 0 no
Insufficient
IMO A

Is m +n3 odd?
1) m is odd and n is even
2) m-n is odd

Approach:

Important Concepts to be used here:
odd +- odd = even
even +- odd = odd
odd * odd = odd
odd * even = even

S1: m is odd and n is even
Sample cases:
m = 3, n = 2
\(m +n^3\) = 3+8 => 11 ... odd
m = -5, n = -4
\(m +n^3 \) = -5-64 => -69... odd
we get answers as odd in all cases. ............................Sufficient



S2: m-n is odd
for m-n to be odd, m and n should odd and even or vice versa, not same.
Sample cases:
m = 3, n = 2, m - n = 1
\(m +n^3\) = 3+8 => 11 ... odd
m = 2, n = -3, m - n = 5
\(m +n^3\) = 2-27 => -25 ... odd
we get answers as odd in all cases. ............................Sufficient

IMO Option D!

In the question it is not explicitly mentioned that m and n are integers

ST-1: m is odd, n is even . It is implicit here that m and n are integers
from st-1 we can determine that m+n^3 is odd

ST-2: we don't know if m and n are integers.for example value of m and n can be 3/2 and 1/2 respectively..we cannot that the value of m+n^3 is odd.
on the other hand, if m=1 and n=0 value is odd

Therefore only ST-1 is sufficient

Option a

Is m +n^3 odd?
1) m is odd and n is even
2) m-n is odd

Is m +n^3 odd = 1/3/5/7/9/-1/-3/-5/-7/-9??
Odd+even =odd or even+odd = odd, So, the equation to be odd, one of the variable has to be, either m or n^3, has to be odd and other one must be even.
1) m is odd and n is even
m = odd and n = even, so n^3 = even, So, statement 1 satisfies the required condition and it is sufficient to answer the problem. Let's try some numbers.
If, m= 1 and n=-2, m +n^3 = 1-8 = -7, which is odd
m=-5 and n=-2, m +n^3 = -5 -8 =-13, which is odd
m=1 and n=2, m +n^3 = 1+8 = 9, which is odd.
So, answer will be A or D.

2) m-n is odd
Odd-even = odd or even-odd = odd. So, Difference of two numbers is odd which means one has to be even and one has to be odd. So, statement 2 satisfies the required condition and it is sufficient to answer the problem. But still let's check with some numbers.
If, m= 3 and n=0, m-n=3, which is odd. So, m +n^3 = 3+0 = 3, which is odd.
m= 3 and n=2, m-n=1, which is odd. So, m +n^3 = 3+8 = 11, which is odd.
m= -1 and n=-2, m-n=1, which is odd. So, m +n^3 = -1-8 = -9, which is odd.
m= -6 and n=-1, m-n=-5, which is odd. So, m +n^3 = -6-1 = -7, which is odd.

Hence, Ans. is D.

option D .. is sufficient

Is m +n^3 odd?
1) m is odd and n is even --> sufficient: (odd)+(even)^3= odd+even=odd
2) m-n is odd --> insufficient: m-n = odd, so (m =odd & n= even) or (m=even & n= odd), so m+n^3=odd? yes, but if m = 3/2 & n = 1/2, then m+n^3 not equal to odd
Answer: A

1) Sufficient
Since m is odd and n is even, m+n^3=(odd)+(even)^3=(odd)+(even)=odd (Answer the question)
2) Insufficient
m and n can take on any values, even decimals. For example,
if m=5 and n=2 (m=n=3 which is odd), then m+n^3=13, which is an odd number.
if m=2.1 and n=1.1, m-n=1 is still an odd number. However, m+n^3=2.1+(1.1)^3 is clearly a decimal (no parity properties).
Answer: A

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