Is z even?
1) z/2 is even
2) 3z is even
(1) Z = 2 * even
=> z is even
Sufficient
(2) 3z = even
3 is odd
So z is even if z is an integer
But if z = 8/3
Not Sufficient
Answer  A
If x is an integer, is x/2 an even integer?
1) x is divisible by 2
2) x is divisible by 4
What is the ratio of number of odd integers to the number of even integers between 10.5 and 10.5?
Integers between 10.5 and 10.5 are
10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Odd integers are 9, 7, 5, 3, 1, 1, 3, 5, 7, 9 (Total 10)
Even integers are 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10 (Total 11)
Hence, ratio of number of odd integers to the number of even integers between 10.5 and 10.5 is 10/11
If m is an integer, is m odd?
1) m/2 is not an even integer
2) m – 3 is not an even integer
If n is a positive integer, then n (n + 1)(n + 2) is which of the following?
The problem is easiest to solve by substituting numbers for n. We’ll try an odd number first and then an even number.
For an odd number, let’s let n = 1: 1(1+1)(1+2) = 1(2)(3) = 6.
We see than choices A and C can’t be the correct choices. Choice A is false because, while the product is even, n is not even. Choice C is false because, while n is odd, the product is even.
For an even number, let’s let n = 2: 2(2+1)(2+2) = 2(3)(4) = 24.
We see than choices B and D can’t be the correct choices. Choice B is false because, while the product is even, n is not odd. Choice D is false because, while the product is divisible by 3, n is even.
Therefore, the only correct answer is choice E. When n is even, the product is indeed divisible by 4.
The product of integers x, y, and z is even. Is z even?
1) x/y = z
2) z = xy
The product of the units, tens, and hundreds digits of the positive 3digit integer x is 42. Is x even?
(1) x is less than 300. ?
(2) The tens digit of x is 7. ?
Correct Answer : d
Explanation : If the product of the three digits is 42 then the possible set of three digits is 2,3,7 or 6,7,1
Numbers from 2,3,7 are 237,273,327,372,723,732
Numbers from 6,7,1 are 167,176,617,671,716,761
Numbers less than 300 are 237,273,167,176 so x can be both even or odd Insufficient.
Tens digit is 7
Numbers can be 273,372,671,176.. x can be both even and odd Insufficient
Taking both the statements together
Number <300 and tens digit is 7 nos are 273,176 both even and odd possible so Insufficient
If m, n, and p are integers, is m+n odd??
(1) m = p^{2} + 4p + 4
(2) n = p^{2} + 2m + 1
So both are required
Is the positive integer p even?
(1) p^{2} + p is even.
(2) 4p + 2 is even.
If p and q are integers and p + q + p is odd, which of the following must be odd?
q must be odd, e.g. follwing patterns;
2+1+2 = 5
1+3+1 = 5
1+2+1 = 4
2+4+2 = 8
Hence all combinations with an even q would yield even results, thus q must be odd.
If a , b, and c are integers and ab^{2} / c is a positive even integer, which of the following must be true?
I. ab is even
II. ab > 0
III. c is even?
Given: = even > 0 ab^{2 }= c* even = even> either a is even or b or both.
I. ab is even > according to the above this must be true;
II. ab > 0 > not necessarily true, bb could be positive as well as negative (for example a=1, c = 1 and b =−2);
III. c is even > not necessarily true, see above example.
If k and y are integers, and 10k + y is odd, which of the following must be true?
Each digit in the twodigit number G is halved to form a new twodigit number H. Which of the following could be the sum of G and H? ?
If positive integers x and y are not both odd, which of the following must be even?
Positive integers x and y are NOT both odd, means that either both x and y are even or one is even and the other one is odd. In either case xy must be even.
If i and j are integers, is i + j even?
1) i < 10
2) i = j
If n is an integer, is n even?
1) n^{2} – 1 is an odd integer
2) 3n + 4 is an even integer
If x and y are both integers, is xy even?
1) x + y is odd
2) x is even
Correct Answer : C
Explanation : In order the product of two integers to be even either (or both) of them must be even. So, the question basically asks whether either x or y is even.
(1) x = y + 1. If x is odd then y is even and viseversa. Sufficient.
(2) x is even
Let x = 2, y= 3
x*y = 2*3 = 6 (Even), Sufficient
If x and y are integers, is xy even?
1) x = y + 1
2) x/y is an even integer
If either x or y is zero, then xy=0=even, because zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.
An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).
Is p^{2} an odd integer?
1) p is an odd integer
2) √p is an odd integer
If x and y are prime integers and x < y, which of the following cannot be true??
2X+Y===Even +odd= odd...Always False
(As x <y then x can be smallest prime 2 leaving only odd prime choices for y)
If x and y are positive integers, is the product xy even?
1) 5x  4y is even
2) 6x + 7y is even
1) 4y will always be even. Then we have 5x−even=even ,
For this to be the case, 5x must be even. Since 5 can't be even, then x must be even. Thus the product xy will be even. Sufficient.
2) 6x will always be even. Then we have even+7y=even.
Thus 7y is even, and y is even, and xy is even. Sufficient.
If x and y are integers, is x (y + 1) an even number?
1) x and y are prime numbers.
2) y > 7 ?
For all positive integers m, (m) = 3m when m is odd and (m) = (1⁄2) m when m is even, which of the following is equivalent to (9)*(6)?
Notice that [ ] is just some function such that "[m]=3m when m is odd and [m]=(1/2)*m when m is even".
So, [m]=3m when m is odd, [m]=(1/2)*m when m is even.
As 9 is odd then [9] equals to 3*9=27;
As 6 is even then [6] equals to 1/2*6=3;
So [9]*[6]=27*3=81. Note that numbers in the answer choices are also in boxes, so we have: [m]=81. m could be 27 (in this case as 27 is odd [27]=3*27=81) OR 162 (in this case as 162 is even [162]=162/2=81) > only [27] is in the answer choices.
If m and n are integers, is m odd??
1) m + n is odd
2) m + n = n^{2} + 5
(1) n + m is odd
The sum of two integers is odd only if one is odd and another is even, hence m may or may not be odd. Not sufficient.
(2) n + m = n^{2} + 5
> m−5=n^{2}−n
=> m−5=n(n−1)
either n or n−1 is even hence n(n−1)=even
=> n(n−1)=even
=> m−5=m−odd=even
> m=odd. Sufficient.
For nonnegative integers x, y, and z, is x^{z} odd?
1) xz = odd
2) x = 2^{y}
(1) The product xz is odd.
=> x, z are odd => x^{z }odd => SUF
(2) x = 2^{y}
=> x is even => x^{z} even => SUF
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