SSAT Upper Level Quantitative Quiz: Prime And Composite Numbers

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Prime And Composite NumbersQuestion 1 of 13

If $p$ is prime, how many positive factors does $p^2$ have?

SSAT Upper Level Quantitative Quiz

SSAT Upper Level Quantitative Quiz: Prime And Composite Numbers

Practice Prime And Composite Numbers in SSAT Upper Level Quantitative with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Prime And Composite Numbers, giving you a quick way to practice the rules, question types, and explanations that matter most for SSAT Upper Level Quantitative.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If $p$ is prime, how many positive factors does $p^2$ have?

2
3 (correct answer)
4
5

Explanation: A prime number p has exactly two positive factors: 1 and p. Squaring it adds p^2 as a third factor, so p^2 has exactly 1, p, and p^2. The tempting wrong answer is 2, which forgets to count 1 as a factor.

Question 2

If $p$ and $q$ are distinct primes, which statement must be true?

$p × q$ is odd.
$p + q$ is prime.
$pq$ is composite. (correct answer)
$p - q$ is even.

Explanation: Since p and q are distinct primes, pq has four distinct positive divisors: 1, p, q, and pq. That makes pq composite, not prime. The tempting wrong choice is p times q is odd, but that fails when one of the primes is 2, because 2 times an odd prime is even.

Question 3

For a prime $n$ greater than 3, which expression can never be prime?

$n + 2$
$n - 2$
$2n + 1$
$n^2 + 1$ (correct answer)

Explanation: Any prime greater than 3 is odd. Squaring an odd number gives an odd number, so n^2 + 1 is even. Since n > 3, n^2 + 1 is greater than 2, so it cannot be prime. The tempting wrong choice is 2n + 1, but it can be prime; for example, n = 5 gives 11.

Question 4

If a and b are different primes and $a + b = 21$ , what is $ab$ ?

34
38 (correct answer)
42
46

Explanation: Since 21 is odd, one of the two primes must be 2. That means the other prime is 21 - 2 = 19. Multiplying gives 2 * 19 = 38. The tempting wrong answer is 34, which comes from 2 and 17, but those sum to 19, not 21.

Question 5

Which number is composite?

119 (correct answer)
127
131
137

Explanation: 119 = 7 × 17, so it has factors other than 1 and itself, making it composite. The other choices are prime: 127, 131, and 137 have no divisors besides 1 and themselves. 127 is the tempting trap, but it is prime, not composite.

Question 6

How many prime numbers are there between $2^6$ and $2^6 + 20$ , inclusive?

2
3
4
5 (correct answer)

Explanation: First,

2^6 = 64

, so we need primes between 64 and 84, inclusive. Let's check each number systematically: 67 (prime), 71 (prime), 73 (prime), 79 (prime), 83 (prime). All other numbers in this range are composite. For example: 64 =

2^6

, 65 = 5×13, 69 = 3×23, 77 = 7×11, 81 =

3^4

. Therefore, there are exactly 5 prime numbers in this range.

Question 7

The sum of two prime numbers is 30. If one of the primes is greater than 20, how many such pairs of primes are possible?

1 (correct answer)
2
3
4

Explanation: Let the two primes be

p

and

q

where

p + q = 30

and one prime is greater than 20. The primes greater than 20 and less than 30 are: 23, 29. If

p = 23

, then

q = 30 - 23 = 7

, and 7 is prime. If

p = 29

, then

q = 30 - 29 = 1

, but 1 is not considered prime. Therefore, there is only one pair: (23, 7). Choice B would suggest there are 2 pairs, but the second possibility doesn't work. Choices C and D overcount non-existent pairs.

Question 8

If $n$ is the smallest positive integer such that $2n + 1$ is composite, what is the value of $2n + 1$ ?

9 (correct answer)
15
21
25

Explanation: We need to find the smallest positive integer

n

such that

2n + 1

is composite. Let's check values: For

n = 1

2(1) + 1 = 3

(prime). For

n = 2

2(2) + 1 = 5

(prime). For

n = 3

2(3) + 1 = 7

(prime). For

n = 4

2(4) + 1 = 9 = 3^2

(composite). So the smallest value of

n

is 4, making

2n + 1 = 9

. Choice B (15) corresponds to

n = 7

, choice C (21) to

n = 10

, and choice D (25) to

n = 12

, all of which are larger than

n = 4

Question 9

A positive integer $k$ has the property that both $k$ and $k + 2$ are prime. If $k > 3$ , how many possible values of $k$ are there less than 50?

3
4
5 (correct answer)
6

Explanation: We're looking for twin prime pairs where both

k

and

k + 2

are prime, with

3 < k < 50

. The twin prime pairs in this range are: (5,7), (11,13), (17,19), (29,31), and (41,43). Therefore, the possible values of

k

are: 5, 11, 17, 29, and 41. This gives us exactly 5 possible values.

Question 10

If $m = 2^3 \cdot 3^2 \cdot 5$ and $n = 2^2 \cdot 3 \cdot 7$ , what can be concluded about $m + n$ ?

$m + n$ is always prime regardless of the values
$m + n$ is always composite since both $m$ and $n$ are composite
$m + n$ is composite because $m + n = 444$ , which has multiple factors (correct answer)
$m + n$ could be either prime or composite depending on other factors

Explanation: First, let's calculate

m

and

n

m = 2^3 \cdot 3^2 \cdot 5 = 8 \cdot 9 \cdot 5 = 360

and

n = 2^2 \cdot 3 \cdot 7 = 4 \cdot 3 \cdot 7 = 84

. Therefore,

m + n = 360 + 84 = 444

. Since

444 = 4 \cdot 111 = 4 \cdot 3 \cdot 37 = 12 \cdot 37

, we see that 444 is composite. Choice A is wrong because we have specific values, not a general statement. Choice B is incorrect because the sum of composite numbers isn't necessarily composite. Choice D is wrong because we can determine the exact nature of this specific sum.

Question 11

Which of the following statements about the number 91 is true?

91 is prime because it is not divisible by 2, 3, 5, or 7
91 is composite because it equals 7 × 13 (correct answer)
91 is prime because it is not divisible by any integer from 2 to 9
91 is composite because it is divisible by 3

Explanation: To determine if 91 is prime or composite, we need to check for factors.

91 = 7 \times 13

, so 91 is composite. Choice A is incorrect because checking divisibility by 2, 3, 5, 7 is insufficient - we need to check all primes up to

\sqrt{91} \approx 9.5

, and 91 is divisible by 7. Choice C is wrong because 91 IS divisible by 7 (and 7 is between 2 and 9). Choice D is incorrect because 91 is not divisible by 3 (sum of digits: 9 + 1 = 10, which is not divisible by 3). The correct answer is B because

7 \times 13 = 91

Question 12

A number $N$ has exactly 6 positive divisors. If $N = 2^a \cdot 3^b$ where $a$ and $b$ are non-negative integers, and $N > 12$ , what is the smallest possible value of $N$ ?

18 (correct answer)
20
32
36

Explanation: If

N = 2^a \cdot 3^b

, then the number of divisors is

(a+1)(b+1) = 6

. The factor pairs of 6 are (1,6), (2,3), (3,2), and (6,1), giving us

(a,b) = (0,5), (1,2), (2,1), (5,0)

. The corresponding values are:

N = 3^5 = 243

N = 2^1 \cdot 3^2 = 18

N = 2^2 \cdot 3^1 = 12

, and

N = 2^5 = 32

. Since

N > 12

, we exclude

N = 12

. Among the remaining candidates (18, 32, 243), the smallest is 18.

Question 13

If $n$ is a positive integer such that $n^2 + n + 41$ is composite, what is the smallest possible value of $n$ ?

40 (correct answer)
41
80
81

Explanation: We need to find when

n^2 + n + 41

is composite. Let's test values systematically. When

n = 40

40^2 + 40 + 41 = 1600 + 40 + 41 = 1681 = 41^2

, which is composite. We can verify that for smaller values like

n = 39

39^2 + 39 + 41 = 1521 + 39 + 41 = 1601

, which is prime. Choice B (41) gives

41^2 + 41 + 41 = 41(41 + 1 + 1) = 41 imes 43

, which is also composite, but 41 > 40. Choice C (80) gives a composite number but is larger than 40. Choice D (81) also works but is larger.