Integers
Every Data Sufficiency (DS) question can have 1 of 5 possible answers, and the possible answers are the same for every DS question. All DS questions are really two questions in one. You’re given a general problem, followed by two supplemental statements. The real problem is to evaluate each of the two statements independently to determine their individual sufficiency to solve the original general problem. As you determine the sufficiency of these statements, you are able to decide which of the following 5 answers to the Data Sufficiency question is the correct answer:
A. Statement 1 is sufficient, but statement 2 is not.
B. Statement 1 is not sufficient, but statement 2 is.
C. Neither statement 1 nor statement 2 is sufficient, but together they are.
D. Both statement 1 and statement 2 are individually sufficient.
E. Neither statement 1 nor statement 2 is sufficient, and together they are not sufficient.
The diagram above gives a pictorial representation of the 5 Data Sufficiency answers. The two statement numbers are shown in blue at the top of the first two columns. Note that green means sufficient, and red means not sufficient. As an example, if both statements 1 and 2 are individually sufficient, then the correct answer is D. For more information on the 5 DS answers, see our videos.
We illustrate these answers for integer problems with the following 12 examples. Each of the first three examples provides 5 different versions of supplementary statements for the same general problem, and each of these 5 versions yields a different one of the 5 possible answers. The answers to these 5 versions are explained immediately after the problem. After the first 3 examples, we then give 9 additional examples with just one version of the supplementary statements.
[Notice in the first 3 examples that the 5 versions of the supplementary statements provide examples of the 5 different answers, A – E. However, they are not given in A – E order.]
Example 1)
n is an odd, positive integer less than 200. What is its value?
Version I.
(1) n is divisible by 3.
(2) When divided by 7 or 11, n has a remainder of 5.
Version II.
(1) When divided by 7, n has a remainder of 5.
(2) When divided by 11, n has a remainder of 5.
Version III.
(1) When divided by 7 or 11, n has a remainder of 5.
(2) n is divisible by 3 and by the smallest prime number between 50 and 60.
Version IV.
(1) n is divisible by 3.
(2) When divided by 7, n has a remainder of 5.
Version V.
(1) When divided by 7 or 11, n has a remainder of 5.
(2) n is divisible by the smallest prime number between 50 and 60.
Answers: [ Version I: B Version II: C Version III: D Version IV: E Version V: A ]
Explanation:
Don’t forget to note the conditions on n mentioned in the statement of the problem: n is a positive integer which is odd and is less than 200. These combined with the information given in some of the statements will identify the value of n.
Version I: Answer B – Statement 1 is not sufficient, but statement 2 is.
(1) n is divisible by 3.
Partial information. There are many odd numbers less than 200 divisible by 3. Not sufficient.
(2) When divided by 7 or 11, n has a remainder of 5.
Full information. This establishes two equations: 7k + 5 = n and 11m + 5 = n. From these we get 7k + 5 = 11m + 5, so 7k = 11m. From this we know that 7k and 11m have the same prime factorization, so both 7k and 11m must contain 7 and 11 in their prime factorizations, and they’re equal to each other. The smallest number containing both 7 and 11 as factors is 77, and 5 more is 82. But 82 is even. The next number with both 7 and 11 as factors is 154, and 5 more is 159, which is odd. This is n. Sufficient.
Version II: Answer C – Neither statement 1 nor statement 2 is sufficient, but together they are.
(1) When divided by 7, n has a remainder of 5.
Partial information. There are many odd numbers less than 200 that leave a remainder of 5 when divided by 7. Examples are 19, 33, 47. Not sufficient.
(2) When divided by 11, n has a remainder of 5.
Partial information. There are many odd numbers less than 200 that leave a remainder of 5 when divided by 11. Examples are 27, 49, 71. Not sufficient.
Alone, neither of these statements is sufficient, but together they are sufficient, because these statements establish two equations: 7k + 5 = n and 11m + 5 = n. From these we get 7k + 5 = 11m + 5, so 7k = 11m. From this we know that 7k and 11m have the same prime factorization, so both 7k and 11m must contain 7 and 11 in their prime factorizations, and they’re equal to each other. The smallest number containing both 7 and 11 as factors is 77, and 5 more is 82. But 82 is even. The next number with both 7 and 11 as factors is 154, and 5 more is 159, which is odd. This is n.
Version III: Answer D – Both statement 1 and statement 2 are individually sufficient.
(1) When divided by 7 or 11, n has a remainder of 5.
Full information. This establishes two equations: 7k + 5 = n and 11m + 5 = n. From these we get 7k + 5 = 11m + 5, so 7k = 11m. From this we know that 7k and 11m have the same prime factorization, so both 7k and 11m must contain 7 and 11 in their prime factorizations, and they’re equal to each other. The smallest number containing both 7 and 11 as factors is 77, and 5 more is 82. But 82 is even. The next number with both 7 and 11 as factors is 154, and 5 more is 159, which is odd. This is n. Sufficient.
(2) n is divisible by 3 and by the smallest prime number between 50 and 60.
Full information. The smallest prime number between 50 and 60 is 53, and 3×53 = 159. This is less than 200 and is n. Sufficient.
Version IV: Answer E – Neither statement 1 nor statement 2 is sufficient, and together they are not sufficient.
(1) n is divisible by 3.
Partial information. There are many odd numbers less than 200 divisible by 3. Not sufficient.
(2) When divided by 7, n has a remainder of 5.
Partial information. There are many odd numbers less than 200 with a remainder of 5 when divided by 7. Examples are 19, 33, 47. Not sufficient.
Alone, neither of these statements is sufficient, and together they are not sufficient, because there are several odd numbers less than 200 which are both divisible by 3 and which have a remainder of 5 when divided by 7. These are 33, 75, 117, and 159.
Version V: Answer A – Statement 1 is sufficient, but statement 2 is not.
(1) When divided by 7 or 11, n has a remainder of 5.
Full information. This establishes two equations: 7k + 5 = n and 11m + 5 = n. From these we get 7k + 5 = 11m + 5, so 7k = 11m. From this we know that 7k and 11m have the same prime factorization, so both 7k and 11m must contain 7 and 11 in their prime factorizations, and they’re equal to each other. The smallest number containing both 7 and 11 as factors is 77, and 5 more is 82. But 82 is even. The next number with both 7 and 11 as factors is 154, and 5 more is 159, which is odd. This is n. Sufficient.
(2) n is divisible by the smallest prime number between 50 and 60.
Partial information. The smallest prime number between 50 and 60 is 53, but both 159 and 53 are odd, less than 200, and divisible by 53. Not sufficient.
To practice more of these types of problems, click here.
Example 2)
Besides yogurt and oatmeal, Susan likes to put bananas, oranges, and apples in her bircher muesli. If every 4 servings require 1 banana, 2 oranges, and an apple, does she have enough of these to serve 100 people at a picnic, assuming each person gets 1 serving?
Version I.
(1) She has 2 boxes of oranges, each with 30.
(2) She has twice as many oranges as apples and twice as many oranges as bananas.
Version II.
(1) She has 60 oranges, which is as many as she has apples and bananas combined.
(2) Altogether she has 120 oranges, apples, and bananas.
Version III.
(1) She has 2 boxes of oranges, each with 30, and she has twice as many oranges as apples and twice as many oranges as bananas.
(2) She has an equal number of apples and bananas, each of which is half as many as the 60 oranges she has.
Version IV.
(1) She has 2 boxes of oranges, each with 30, and she has twice as many oranges as apples and twice as many oranges as bananas.
(2) She has 60 oranges, which is as many as she has apples and bananas combined.
Version V.
(1) Altogether she has 120 oranges, apples, and bananas.
(2) She has an equal number of apples and bananas, each of which is half as many as the 60 oranges she has.
Answers: [ Version I: C Version II: E Version III: D Version IV: A Version V: B ]
Explanation:
Note that the numbers given are for 4 servings each. So to serve 100 people, we need 100 servings, and 25 times the numbers for apples, oranges, and bananas. So she needs at least 50 oranges, 25 apples, and 25 bananas.
Version I: Answer C – Neither statement 1 nor statement 2 is sufficient, but together they are.
(1) She has 2 boxes of oranges, each with 30.
Partial information. She has enough oranges, but what about apples and bananas? Not sufficient.
(2) She has twice as many oranges as apples and twice as many oranges as bananas.
Partial information. This is only relative information, and tells us nothing about the precise numbers of apples, oranges, and bananas. Not sufficient.
Alone, neither of these statements is sufficient, but together they are sufficient, because the first shows she has enough oranges, and the second that she has enough apples and bananas.
Version II: Answer E – Neither statement 1 nor statement 2 is sufficient, and together they are not sufficient.
(1) She has 60 oranges, which is as many as she has apples and bananas combined.
Partial information. But this doesn’t mean she has 30 apples and 30 bananas. For example, she may have 40 apples and 20 bananas. Not sufficient.
(2) Altogether she has 120 oranges, apples, and bananas.
Partial information. But this doesn’t mean she has at least 50 oranges, 25 apples, and 25 bananas. For example, she may have 60 oranges, 40 apples and 20 bananas. Not sufficient.
Alone, neither of these statements is sufficient, and together they are not sufficient, because although the first statement tells us we have enough oranges, the second doesn’t tell us anything specific about the number of apples and bananas.
Version III: Answer D – Both statement 1 and statement 2 are individually sufficient.
(1) She has 2 boxes of oranges, each with 30, and she has twice as many oranges as apples and twice as many oranges as bananas.
Full information. She needs 50 oranges, and she has 60. This is twice the number both of apples and of bananas, each of which must therefore be 30. Sufficient.
(2) She has an equal number of apples and bananas, each of which is half as many as the 60 oranges she has.
Full information. 50 oranges is what she needs, and she has 60. The number of apples and bananas is 1/2 of 60 or 30, each. So she has enough of those, also. Sufficient.
Version IV: Answer A – Statement 1 is sufficient, but statement 2 is not.
(1) She has 2 boxes of oranges, each with 30, and she has twice as many oranges as apples and twice as many oranges as bananas.
Full information. She needs 50 oranges, and she has 60. This is twice the number both of apples and of bananas, each of which must therefore be 30. Sufficient.
(2) She has 60 oranges, which is as many as she has apples and bananas combined.
Partial information. But this doesn’t mean she has 30 apples and 30 bananas. For example, she may have 40 apples and 20 bananas. Not sufficient.
Version V: Answer B – Statement 1 is not sufficient, but statement 2 is.
(1) Altogether she has 120 oranges, apples, and bananas.
Partial information. But this doesn’t mean she has at least 50 oranges, 25 apples, and 25 bananas. For example, she may have 60 oranges, 40 apples and 20 bananas. Not sufficient.
(2) She has an equal number of apples and bananas, each of which is half as many as the 60 oranges she has.
Full information. 50 oranges is what she needs, and she has 60. The number of apples and bananas is 1/2 of 60 or 30, each. So she has enough of those, also. Sufficient.
To practice more of these types of problems, click here.
Example 3)
Is the sum of 27 distinct, positive integers odd or even?
Version I.
(1) More than half the numbers are even.
(2) One number is even, and the other 26 can be placed in odd-even pairs – that is, each pair consists of two numbers, one odd and the other even.
Version II.
(1) More than half the numbers are even.
(2) The smallest 20 numbers are between 15 and 40.
Version III.
(1) Exactly 14 of the numbers are even.
(2) The smallest 20 numbers are between 15 and 40.
Version IV.
(1) 26 of the numbers can be placed in odd-even pairs – that is, each pair consists of two numbers, one odd and the other even.
(2) Over half the numbers are even.
Version V.
(1) One number is even, and the other 26 can be placed in odd-even pairs – that is, each pair consists of two numbers, one odd and the other even.
(2) Exactly 14 of the numbers are even.
Answers: [ Version I: B Version II: E Version III: A Version IV: C Version V: D ]
Explanation:
The sum of two odds or two evens is always even, and the sum of an odd and an even is odd. If an odd number of the 27 integers in this problem are odd, then their sum is odd, and since all the other numbers are even, then the sum of all 27 will be odd.
Version I: Answer B – Statement 1 is not sufficient, but statement 2 is.
(1) More than half the numbers are even.
Partial information. Without an exact number to work with we can’t tell how many are odd and how many even. Not sufficient.
(2) One number is even, and the other 26 can be placed in odd-even pairs – that is, each pair consists of two numbers, one odd and the other even.
Full information. Since 1/2 of 26 is 13, then this says that 13 of the integers are odd, so their sum is odd. The other 14 are even, so their sum is even. The sum of the 27 equals the sum of the 13 odds and the 14 evens, which will be the sum of an odd and an even. Therefore, the sum of the 27 is odd. Sufficient.
Version II: Answer E – Neither statement 1 nor statement 2 is sufficient, and together they are not sufficient.
(1) More than half the numbers are even.
Partial information. Without an exact number to work with we can’t tell how many are odd and how many even. Not sufficient.
(2) The smallest 20 numbers are between 15 and 40.
No information. This statement gives us no information as to whether the numbers are odd or even. Not sufficient.
Alone, neither of these statements is sufficient, and together they are not sufficient, because the first gives us nothing precise, and the second gives us nothing at all.
Version III: Answer A – Statement 1 is sufficient, but statement 2 is not.
(1) Exactly 14 of the numbers are even.
Full information. This says that 13 of the integers are odd, so their sum is odd. The other 14 are even, so their sum is even. The sum of the 27 equals the sum of the 13 odds and the 14 evens, which will be the sum of an odd and an even. Therefore, the sum of the 27 is odd. Sufficient.
(2) The smallest 20 numbers are between 15 and 40.
No information. This statement gives us no information as to whether the numbers are odd or even. Not sufficient.
Version IV: Answer C – Neither statement 1 nor statement 2 is sufficient, but together they are.
(1) 26 of the numbers can be placed in odd-even pairs – that is, each pair consists of two numbers, one odd and the other even.
Partial information. This tells us that half the numbers are odd and half are even. Because 13 are odd, the sum of these 26 will be odd, but we can’t tell about the sum of the 27, because we don’t know whether the 27th is odd or even. Not sufficient.
(2) Over half the numbers are even.
Partial information. Without an exact number to work with we can’t tell how many are odd and how many even. Not sufficient.
Alone, neither of these statements is sufficient, but together they are sufficient, because, since 1/2 of 26 is 13, then statement 1 says that 13 of the integers are odd, so their sum is odd. Statement 2 says over half the numbers are even, and since we know from statement 1 that 13 are odd, then the other 14 are even, so their sum is even. The sum of the 27 thus equals the sum of the 13 odds and the 14 evens, which will be the sum of an odd and an even. Therefore, the sum of the 27 is odd.
Version V: Answer D – Both statement 1 and statement 2 are individually sufficient.
(1) One number is even, and the other 26 can be placed in odd-even pairs – that is, each pair consists of two numbers, one odd and the other even.
Full information. Since 1/2 of 26 is 13, then this says that 13 of the integers are odd, so their sum is odd. The other 14 are even, so their sum is even. The sum of the 27 equals the sum of the 13 odds and the 14 evens, which will be the sum of an odd and an even. Therefore, the sum of the 27 is odd. Sufficient.
(2) Exactly 14 of the numbers are even.
Full information. This says that 13 of the integers are odd, so their sum is odd. The other 14 are even, so their sum is even. The sum of the 27 equals the sum of the 13 odds and the 14 evens, which will be the sum of an odd and an even. Therefore, the sum of the 27 is odd. Sufficient.
To practice more of these types of problems, click here.
Example 4)
The bleachers in a certain high school’s gymnasium has r rows, and each row can seat s people. With 1000 spectators expected for tonight’s big game, are there enough seats in the bleachers?
(1) 72 ≤ s and r ≤ 18
(2) 72 ≤ s ≤ 75, and 14 ≤ r ≤ 18
Answer: B – Statement 1 is not sufficient, but statement 2 is.
Explanation:
The total number of spectator seats available is equal to rs. So, unless there are some other adjustments, the entire problem is to determine whether rs ≥ 1000.
(1) 72 ≤ s and r ≤ 18
Partial information. This is deceiving, since we’re tempted to multiply 18 and 72 and get 1296, which is more than enough seats. But we don’t know that we have 18 rows. In fact, the number of rows may be only 10, which would give us only 720 seats. Not sufficient.
(2) 72 ≤ s ≤ 75, and 14 ≤ r ≤ 18
Full information. We look at the minimum values for r and s to see if rs is large enough. These are 72 for s, and 14 for r, and for these values rs = 1008. Sufficient.
Example 5)
A PE class of 54 students is divided into two major groups – those who want to play tennis, and those who want to play basketball. Those playing tennis are placed in groups of 2 people each, and the basketball teams have 5. How many basketball teams are there?
(1) There are more than 4 but fewer than 10 tennis pairs.
(2) There are more than 7 basketball teams.
Answer: A – Statement 1 is sufficient, but statement 2 is not.
Explanation:
We’re working with products of 2 and of 5, and sums of these various products. Since there are 5 on a basketball team, the number of players on basketball teams must be some multiple of 5. If the number of basketball teams is even, the number of basketball players will be even, but if the number of these teams is odd, then the number of basketball players will be odd.
But since the tennis players are in pairs there must always be an even number of tennis players. And since the total number of students (54) is even, then the number of basketball players and the number of basketball teams must also be even. This means that the number of basketball players must be 0, 10, 20, 30, 40, or 50. And correspondingly, the number of tennis players must be 4, 14, 24, 34, 44, or 54.
(1) There are more than 4 but fewer than 10 tennis pairs.
Full information. This means there are more than 8 but fewer than 20 tennis players. The only number of tennis players from the general discussion above that is between 8 and 20 is 14. So there are 14 tennis players and 40 basketball players, meaning there are 8 basketball teams. Sufficient.
(2) There are more than 7 basketball teams.
Partial information. The number of basketball teams is an even number, so there can be 8 or 10 teams. But we need one definite answer, not two possible answers. Not sufficient.
Example 6)
The object in a Kakuro puzzle is to fill a two dimensional grid with numbers from 1 to 9 whose sums equal the number indicated on the left of the row, or at the top of the column, to which the number applies. Suppose there is a row of just three numbers, selected from the digits 1 through 9, but without any duplicates. If the indicated sum of these three numbers is 10, is 2 one of the numbers in the row?
(1) 6 is not in the row.
(2) 7 is not in the row.
Answer: E – Neither statement 1 nor statement 2 is sufficient, and together they are not sufficient.
Explanation:
There are 4 possible combinations of three positive integers, with no duplicates, adding up to 10:
1, 2, 7
1, 3, 6
1, 4, 5
2, 3, 5
Notice that two of the combinations contain the number 2. Aspects of these and the other two combinations will be used in the clarifying statements.
(1) 6 is not in the row.
Partial information. Three rows do not have 6. Two of them have 2, and 1 does not. Not sufficient.
(2) 7 is not in the row.
Partial information. Three rows do not have 7. One of them has 2, and the other 2 do not. Not sufficient.
Alone, neither of these statements is sufficient, and together they are not sufficient, because these two statements eliminate all but two rows, one of which has 2 and the other of which does not.
Example 7)
Both f and g are integers which have a remainder of 1 when divided by 3. Is fg odd?
(1) f – g is even.
(2) 2f – g is odd.
Answer: C – Neither statement 1 nor statement 2 is sufficient, but together they are.
Explanation:
The product fg will be odd if and only if both f and g are odd. If either or both are even, the product will be even. Integers with a remainder of 1 on division by 3 may be even or odd. For example, here is a list of even integers with a remainder of 1 on division by 3: 4, 10, 16, etc. And here’s one of odd integers: 7, 13, 19, 25, etc. Notice that the differences (or sums) of any two of the even integers listed is always even, and the differences (or sums) of the odd integers listed is always even. But the difference (or sum) of an odd integer and an even integer on the two lists is always odd. For example, 19 – 10 = 9, which is odd, or 16 + 13 = 29, which is also odd. So if f – g (or f + g) is odd, then either f or g is even, and the other one is odd. Keep these facts in mind in looking at the statements (1) and (2).
(1) f – g is even.
Partial information. This just says that f and g are both even, OR they are both odd. If both are even, then fg is even. If both odd, then fg is odd. Not sufficient.
(2) 2f – g is odd.
Partial information. This statement tells us that, since 2f is always even, g must be odd, but f could be either even or odd. Not sufficient.
Alone, neither of these statements is sufficient, but together they are sufficient, because the second statement tells us that, since 2f is always even, g must be odd. The first tells us that f and g are both even, OR they are both odd. Therefore, since g is odd, both f and g are odd, and fg is odd.
Example 8)
b and a are positive integers such that b + a = x. What is x?
(1) \(\begin{align}b^2 + a^2\end{align}\) = 106
(2) b – a = 4
Answer: A – Statement 1 is sufficient, but statement 2 is not.
Explanation:
(1) \(\begin{align}b^2 + a^2\end{align}\) = 106
Full information. The easiest way to solve this problem is to plug in small integers and see if we can find a (b, a) pair that works. We’ll plug values into a. It’s easy to see that for a = 1, 2, 3, and 4, there is no b that works. But for a = 5, b = 9 will work. Moreover, none of the larger values of a = 6, 7, 8, or 10 will work. For a = 9, b = 5, and is the same solution we got before. Nothing larger than 10 will work, because the square of any larger number is bigger than 106. Sufficient.
(2) b – a = 4
Partial information. Many integer pairs have a difference of 4. We need more information to nail down a and b. Not sufficient.
Example 9)
The Smalltown Theater-in- the-Round sells tickets for $10 and $15. How many tickets did they sell today for next week’s play?
(1) On sales of $375, they sold fewer than twenty $15 tickets and fewer than ten $10 tickets.
(2) They sold 10 more $15 tickets than $10 tickets, and had sales of over $350 and less than $400.
Answer: D – Both statement 1 and statement 2 are individually sufficient.
Explanation:
(1) On sales of $375, they sold fewer than twenty $15 tickets and fewer than ten $10 tickets.
Full information. The maximum number of $15 and $10 tickets allowed by this condition is 19 and 9, respectively. If they sold 19 $15 tickets and 9 $10 tickets they would have exactly $375 in revenue. If they sold any number less of either type of ticket, they would have less than $375. So they sold 19 and 9. Sufficient.
(2) They sold 10 more $15 tickets than $10 tickets, and had sales of over $350 and less than $400.
Full information. If they sold 18 $15 tickets and 8 $10 tickets they would have revenue of $350. If they sold 20 $15 tickets and 10 $10 tickets their revenue would be $400. Therefore, they must have sold 19 $15 tickets and 9 $10 tickets. Sufficient.
Example 10)
Let a, b, and c, no two of which are equal, be the ages in ascending order of 3 siblings. Is the sum of their combined ages (a + b + c) over 44?
(1) The combined ages of the two oldest siblings is 36.
(2) The youngest of the three is at least 15.
Answer: B – Statement 1 is not sufficient, but statement 2 is.
Explanation:
(1) The combined ages of the two oldest siblings is 36.
Partial information. The youngest could be 7, and the combined ages of all three would be 43, or the youngest could be 10, and their combined ages 46. Not sufficient.
(2) The youngest of the three is at least 15.
Full information. If the youngest is 15, then the other two must be at least 15 as well, and 15 + 15 + 15 = 45 > 44. Sufficient.
Example 11)
Thaddeus has x pens and y pencils in holders on his desk. What is x + y?
(1) The total number of pens and pencils is between 16 and 27, with 1/3 more pens than pencils.
(2) He has 3 more pens than pencils, and the number of pencils is 3/4 the number of pens.
Answer: D – Both statement 1 and statement 2 are individually sufficient.
Explanation:
It’s important to remember in problems like these that we don’t deal in fractions of pens and pencils, so all the numbers we’re working with are integers. Thus, when we divide, we need the division to have no remainder.
(1) The total number of pens and pencils is between 16 and 27, with 1/3 more pens than pencils.
Full information. The easiest way to solve this problem is simply to try some numbers and see what works. Since there are 1/3 more pens than pencils, then the number of pencils, y, must be divisible by 3. Here are some candidate values:
y = 6, x = 8 x + y = 14
y = 9, x = 12 x + y = 21
y = 12, x = 16 x + y = 28
It’s easy to see that y = 9 and x = 12 is the only pair of values that satisfies the requirement that the total number of pencils and pens be between 16 and 27.
Sufficient.
(2) He has 3 more pens than pencils, and the number of pencils is 3/4 the number of pens.
Full information. This gives us two equations:
x = y + 3, and
¾ x = y, or 3x = 4y
Substituting y + 3 for x in this last equation, we get 3(y + 3) = 4y, so y = 9. Thus, he has 9 pencils and 12 pens. Sufficient.
Example 12)
k is a positive integer. Is k odd?
(1) k(k – 1) is even.
(2) 2k + 1 is odd.
Answer: E – Neither statement 1 nor statement 2 is sufficient, and together they are not sufficient.
Explanation:
Review of the math on even/odd integer operations:
For any integer n, 2n + 1 is odd, and 2n is even.
The sum (or difference) of two evens or two odds is even.
The sum (or difference) of an odd and an even is odd.
The product of an even with an odd or with an even is even.
The product of an odd with an odd is odd.
(1) k(k – 1) is even.
No information. If k is even, k – 1 is odd, and if k is odd, then k – 1 is even. Either way, their product will be even. So this tells us nothing about k. Not sufficient.
(2) 2k + 1 is odd.
No information. Whether k is even or odd, 2k + 1 is always odd. Not sufficient.
Alone, neither of these statements is sufficient, and together they are not sufficient, because neither of these statements tell us anything about k, since they’re true for all integers.
To practice more of these types of problems, click here.