SSAT Middle Level Quiz: Prime And Composite Numbers

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Prime And Composite NumbersQuestion 1 of 20

A prime has exactly two factors; a composite has more than two. Small examples: 2, 3, 5, 7 are prime; 4 and 6 are composite. Identify the composite numbers in the following set: 18, 19, 20, 21.

18, 20, and 21

19 and 21

18 and 19

Only 19

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SSAT Middle Level Quiz

SSAT Middle Level Quiz: Prime And Composite Numbers

Practice Prime And Composite Numbers in SSAT Middle Level 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 Middle Level.

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

A prime has exactly two factors; a composite has more than two. Small examples: 2, 3, 5, 7 are prime; 4 and 6 are composite. Identify the composite numbers in the following set: 18, 19, 20, 21.

18, 20, and 21 (correct answer)
19 and 21
18 and 19
Only 19

Explanation: This question tests middle-level SSAT skills: determining whether a number is prime or composite, focusing on number properties and integers. Prime numbers have only two distinct positive divisors: 1 and the number itself, while composite numbers have more than two. In this question, students must use the definitions and identify factors, for example, noting that 18, 20, and 21 have more than two divisors while 19 does not. Choice A is correct because it accurately reflects the composites in the set, showing understanding of composite identification. Choice D is incorrect because it demonstrates a common misconception, such as assuming only the prime is composite. To help students: Encourage practice with factorization and using divisibility rules. Teach them to verify by checking divisibility by known smaller primes (2, 3, 5, 7) and practice spotting common mistakes like assuming number patterns.

Question 2

How many prime numbers are there between $40$ and $50$ , inclusive?

$1$
$2$
$3$ (correct answer)
$4$
$5$

Explanation: When you encounter a question asking for prime numbers in a specific range, you need to systematically check each number for primality. A prime number has exactly two factors: 1 and itself. Let's examine each number from 40 to 50:

40: Even number, divisible by 2, 4, 5, 8, 10, 20 → not prime
41: Check divisibility by primes up to $\sqrt{41} \approx 6.4$ . Test 2, 3, 5: 41 is odd, $4+1=5$ (not divisible by 3), doesn't end in 0 or 5 → prime
42: Even number, obviously composite → not prime
43: Check divisibility by 2, 3, 5: 43 is odd, $4+3=7$ (not divisible by 3), doesn't end in 0 or 5 → prime
44: Even number → not prime
45: Ends in 5, divisible by 5 → not prime
46: Even number → not prime
47: Check divisibility by 2, 3, 5, 7: 47 is odd, $4+7=11$ (not divisible by 3), doesn't end in 0 or 5, and $47 ÷ 7 = 6.7...$ → prime
48: Even number → not prime
49: $7 \times 7 = 49$ → not prime
50: Even number → not prime

The prime numbers are 41, 43, and 47, giving us 3 primes total. Answer choice (A) 1 might come from only finding one obvious prime. Answer choice (B) 2 suggests missing one of the three primes. Answer choice (D) 4 likely includes 49, forgetting that

49 = 7^2

. Study tip: When checking for primes, you only need to test divisibility by primes up to the square root of your number. This saves time and reduces errors.

Question 3

If $p$ is a prime number greater than $2$ , which of the following expressions must represent a composite number?

$p + 2$
$p^2 - 1$ (correct answer)
$2p + 1$
$p + 4$
$3p - 2$

Explanation: When you encounter questions about prime numbers and composite numbers, remember that a composite number has factors other than 1 and itself, while a prime has exactly two factors. Since

p

is a prime greater than 2, it must be odd. Let's examine

p^2 - 1

, which we can factor as

(p-1)(p+1)

. Since

p

is an odd prime greater than 2, both

p-1

and

p+1

are even numbers greater than 1. This means

p^2 - 1

is the product of two integers both greater than 1, making it composite by definition. This factorization works for any odd prime: if

p = 3

, then

p^2 - 1 = 8 = 2 \times 4

; if

p = 5

, then

p^2 - 1 = 24 = 4 \times 6

. Choice A (

p + 2

) could be prime—when

p = 3

, we get

3 + 2 = 5

, which is prime. Choice C (

2p + 1

) also could be prime—when

p = 3

, we get

2(3) + 1 = 7

, which is prime. Choice D (

p + 4

) might be prime too—when

p = 3

, we get

3 + 4 = 7

, which is prime. The key insight is that only choice B must always be composite, regardless of which prime

p

you choose, because it always factors as

(p-1)(p+1)

. Study tip: When a question asks what "must" be true, look for algebraic expressions you can factor. Factoring often reveals why an expression will always have the required property.

Question 4

The largest prime factor of $210$ is how much greater than the smallest prime factor of $210$ ?

$3$
$5$ (correct answer)
$6$
$7$
$9$

Explanation: When you encounter questions about prime factors, you need to find the prime factorization of the given number first. Prime factors are the prime numbers that multiply together to give the original number. To find the prime factorization of

210

, start by dividing by the smallest prime numbers:

$210 ÷ 2 = 105$
$105 ÷ 3 = 35$
$35 ÷ 5 = 7$
$7 ÷ 7 = 1$

210 = 2 × 3 × 5 × 7

. The prime factors are

2, 3, 5,

and

7

. The smallest prime factor is

2

and the largest is

7

. The difference is

7 - 2 = 5

, which is choice B. Let's examine why the other answers are incorrect. Choice A (

3

) might come from finding

5 - 2 = 3

, confusing the second-largest prime factor

5

with the largest. Choice C (

6

) could result from incorrectly identifying

6

as a prime factor, but

6 = 2 × 3

is composite, not prime. Choice D (

7

) represents just the largest prime factor itself, not the difference between largest and smallest. Remember that prime factorization questions require you to break numbers down completely into prime components. Always double-check that your factors are actually prime numbers (only divisible by

1

and themselves). When finding differences between prime factors, carefully identify which is truly the largest and smallest before calculating.

Question 5

A number $N$ is the product of three consecutive positive integers. If the middle integer is prime, what can be concluded about $N$ ?

$N$ must be prime
$N$ must be composite (correct answer)
$N$ could be either prime or composite
$N$ must be even but not divisible by $4$
$N$ must be divisible by $6$

Explanation: When you encounter questions about products of consecutive integers, think about the fundamental properties these numbers must have based on their factors. Let's call the three consecutive positive integers

n-1

n

, and

n+1

, where

n

is the middle prime number. So

N = (n-1) \cdot n \cdot (n+1)

. Since

n

is prime and greater than 2 (the only even prime),

n

must be odd. This means both

n-1

and

n+1

are even integers. Therefore,

N

contains at least two factors of 2, plus the prime factor

n

, plus any other factors from

n-1

and

n+1

. Since

N

has multiple distinct prime factors, it must be composite. Let's verify with an example: if

n = 5

, then

N = 4 \cdot 5 \cdot 6 = 120

, which factors as

2^3 \cdot 3 \cdot 5

– definitely composite. Choice A is wrong because

N

has multiple factors beyond just itself and 1. Choice C is incorrect because we've proven

N

must always be composite, never prime. Choice D is wrong because while

N

is even, it's actually divisible by 4 (since it's the product of two even numbers times an odd number, giving us at least

2^2

as factors). Study tip: When dealing with consecutive integers, remember that among any set of consecutive numbers, you'll always find specific divisibility patterns. Products of consecutive integers are rarely prime because they accumulate multiple factors from their components.

Question 6

Which of the following statements about the number $1$ is correct?

$1$ is a prime number because it has exactly one positive divisor
$1$ is a composite number because it can be written as $1 × 1$
$1$ is neither prime nor composite (correct answer)
$1$ is both prime and composite
$1$ is prime when considered alone, composite in multiplication

Explanation: When you encounter questions about classifying numbers, you need to understand the precise definitions of prime and composite numbers. A prime number has exactly two positive divisors:

1

and itself. A composite number has more than two positive divisors. The number

1

is a special case that doesn't fit either category. Since

1

has only one positive divisor (itself), it fails to meet the "exactly two divisors" requirement for prime numbers. Similarly, since

1

doesn't have more than two divisors, it's not composite either. By mathematical convention,

1

is classified as neither prime nor composite, making answer choice C correct. Let's examine why the other options are incorrect. Choice A states that

1

is prime because it has exactly one divisor, but this misunderstands the definition—prime numbers must have exactly two positive divisors, not one. Choice B claims

1

is composite because it can be written as

1 × 1

, but being expressible as a product doesn't automatically make a number composite; the key is having more than two divisors. Choice D suggests

1

is both prime and composite, which is impossible since these categories are mutually exclusive by definition. Remember that

1

is the only positive integer that's neither prime nor composite—it's in a category by itself. On standardized tests, questions about

1

often try to trick you by testing whether you know the precise definitions rather than relying on intuition.

Question 7

The number $165$ can be expressed as $3 × 5 × 11$ . Based on this information, how many positive divisors does $165$ have?

$6$
$8$ (correct answer)
$12$
$15$
$16$

Explanation: When you see a number expressed as a product of prime factors, you can use a systematic method to count all its positive divisors. The prime factorization

165 = 3^1 × 5^1 × 11^1

tells you exactly how many divisors exist. To find the number of positive divisors, add 1 to each prime's exponent, then multiply these results together. Since each prime appears once (exponent of 1), you calculate:

(1+1) × (1+1) × (1+1) = 2 × 2 × 2 = 8

divisors. You can verify this by listing them systematically. The divisors are formed by taking each prime factor either 0 times or 1 time:

1, 3, 5, 11, 15, 33, 55, 165

. That's exactly 8 divisors, confirming answer B. Answer A (6) likely comes from incorrectly counting only some combinations or forgetting that 1 and the number itself are always divisors. Answer C (12) might result from mistakenly using exponents of 2 instead of 1, giving

(2+1)(2+1)(2+1) = 27

, or some other calculation error. Answer D (15) could come from adding the exponents instead of using the proper formula, or from confusing this with a different type of counting problem. Remember this formula: for any number with prime factorization

p_1^{a_1} × p_2^{a_2} × ... × p_k^{a_k}

, the number of positive divisors is

(a_1+1)(a_2+1)...(a_k+1)

. This method works for any prime factorization and is much faster than listing divisors manually.

Question 8

Two prime numbers $p$ and $q$ satisfy $p + q = 30$ . How many different pairs $(p, q)$ are possible?

$2$
$3$ (correct answer)
$4$
$5$
$6$

Explanation: When you encounter problems about prime numbers that must sum to a given value, start by listing the relevant primes and checking each possibility systematically. Since

p + q = 30

and both numbers are prime, let's consider the prime numbers less than 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. For each prime

p

, we need

q = 30 - p

to also be prime. Testing systematically:

If $p = 2$ , then $q = 28$ , but 28 is not prime ( $28 = 4 \times 7$ )
If $p = 3$ , then $q = 27$ , but 27 is not prime ( $27 = 3^3$ )
If $p = 5$ , then $q = 25$ , but 25 is not prime ( $25 = 5^2$ )
If $p = 7$ , then $q = 23$ , and 23 is prime ✓
If $p = 11$ , then $q = 19$ , and 19 is prime ✓
If $p = 13$ , then $q = 17$ , and 17 is prime ✓

Notice that

(7,23)

(11,19)

, and

(13,17)

are our valid pairs. We don't need to check further because we'd just get the same pairs in reverse order. The answer is B) 3. Choice A) 2 undercounts the solutions. Choice C) 4 might result from counting reversed pairs as different or including non-prime solutions. Choice D) 5 likely comes from systematic errors in checking which numbers are prime. Strategy tip: For prime sum problems, always verify that both numbers in each pair are actually prime by checking for factors. Also remember that order typically doesn't matter—

(p,q)

and

(q,p)

represent the same pair.

Question 9

What is the product of all the different prime factors of 60?

30 (correct answer)
60
120
180
240

Explanation: When you see a question asking for the product of prime factors, you need to first find the prime factorization of the given number, then multiply each different prime factor exactly once. Let's find the prime factorization of 60. Start by dividing by the smallest prime numbers:

60 = 2^2 \times 3 \times 5 = 4 \times 3 \times 5

The prime factorization shows us that 60 has three different prime factors: 2, 3, and 5. Notice that 2 appears twice (as

2^2

), but we only count each different prime factor once when finding their product. The product of all different prime factors is:

2 \times 3 \times 5 = 30

Now let's examine why the other answers are incorrect. Answer (B) 60 is the original number itself, not the product of its prime factors—this would be a trap for students who confuse the question. Answer (C) 120 equals

2^3 \times 3 \times 5

, which might result from incorrectly including an extra factor of 2. Answer (D) 180 equals

2^2 \times 3^2 \times 5

, which could come from squaring some factors or making calculation errors. The correct answer is (A) 30. Strategy tip: For prime factorization problems, always break the number down completely into prime factors first, identify each unique prime (ignoring how many times each appears), then multiply those unique primes together. Don't let repeated factors confuse you—count each prime only once.

Question 10

Which of the following numbers has the most prime factors (counting repetitions)?

$72$
$84$
$90$
$96$ (correct answer)
$108$

Explanation: When you encounter questions about prime factors, you need to find the prime factorization of each number and count all the prime factors, including repetitions. This means if a prime appears multiple times in the factorization, you count it each time it appears. Let's find the prime factorization of each option: For choice A:

72 = 2^3 \times 3^2 = 2 \times 2 \times 2 \times 3 \times 3

. This gives us 5 prime factors total. For choice B:

84 = 2^2 \times 3 \times 7 = 2 \times 2 \times 3 \times 7

. This gives us 4 prime factors total. For choice C:

90 = 2 \times 3^2 \times 5 = 2 \times 3 \times 3 \times 5

. This gives us 4 prime factors total. For choice D:

96 = 2^5 \times 3 = 2 \times 2 \times 2 \times 2 \times 2 \times 3

. This gives us 6 prime factors total. Choice A is incorrect because 72 has only 5 prime factors, not the maximum of 6. Choice B is wrong because 84 has only 4 prime factors. Choice C is also incorrect because 90 has only 4 prime factors. Choice D is correct because 96 has 6 prime factors, which is more than any other option. To tackle these problems efficiently, start by dividing each number by 2 repeatedly until you can't anymore, then try 3, then 5, and so on. Remember that the question asks for counting repetitions, so

2^5

contributes 5 factors to your total count, not just 1.

Question 11

Prime numbers have exactly two factors; composite numbers have more than two. Examples: 2, 3, 5, 7 are prime; 4 and 6 are composite. Is 1 prime or composite?

1 is prime because it is odd.
1 is composite because it has many factors.
1 is neither prime nor composite. (correct answer)
1 is prime because it has one factor.

Explanation: This question tests middle-level SSAT skills: determining whether a number is prime or composite, focusing on number properties and integers. Prime numbers have only two distinct positive divisors: 1 and the number itself, while composite numbers have more than two. In this question, students must use the definitions and identify factors, for example, recognizing that 1 has only one divisor, which is itself. Choice C is correct because it accurately reflects that 1 is neither prime nor composite, showing understanding of number classification. Choice A is incorrect because it demonstrates a common misconception, such as assuming 1 is prime due to its oddness. To help students: Encourage practice with factorization and using divisibility rules. Teach them to verify by checking divisibility by known smaller primes (2, 3, 5, 7) and practice spotting common mistakes like assuming number patterns.

Question 12

A student claims that every odd number greater than $1$ is prime. Which of the following is the smallest counterexample to this claim?

$9$ (correct answer)
$15$
$21$
$25$
$27$

Explanation: This question tests your understanding of prime numbers and asks you to find a counterexample—an instance that disproves a general claim. The student's claim is that "every odd number greater than 1 is prime," so you need to find the smallest odd number greater than 1 that is NOT prime. A prime number has exactly two factors: 1 and itself. A composite number has more than two factors. Let's check the odd numbers systematically: 3 is prime (factors: 1, 3), 5 is prime (factors: 1, 5), 7 is prime (factors: 1, 7), but 9 is composite because

9 = 3 \times 3

, giving it factors 1, 3, and 9. Since 9 is the smallest odd number greater than 1 that isn't prime, it's the smallest counterexample to the claim. This makes A)

9

correct. Looking at the other choices: B)

15

is indeed composite (

15 = 3 \times 5

), but it's larger than 9, so it's not the smallest counterexample. C)

21

is also composite (

21 = 3 \times 7

), but again, it's larger than 9. D)

25

is composite too (

25 = 5 \times 5

), but it's the largest option given. When looking for counterexamples on the SSAT, always work systematically from the smallest possibilities. Don't assume the first few cases represent the general pattern—mathematical claims often have exceptions, and finding the smallest one demonstrates the most efficient logical reasoning.

Question 13

If $n = 2^3 × 3^2 × 7$ , which of the following is NOT a prime factor of $n^2$ ?

$2$
$3$
$5$ (correct answer)
$7$
All of the above are prime factors of $n^2$

Explanation: When you encounter questions about prime factors of powers, remember that squaring a number doesn't create new prime factors—it only increases the exponents of existing ones. Given

n = 2^3 × 3^2 × 7

, let's find

n^2

. When you square a number written in prime factorization, you double each exponent:

n^2 = (2^3 × 3^2 × 7)^2 = 2^6 × 3^4 × 7^2

. The prime factors of

n^2

are exactly the same as those of

n

: the numbers 2, 3, and 7. Only their exponents have changed. Looking at each choice: Choice A) 2 appears in

n^2

with exponent 6, so it's definitely a prime factor. Choice B) 3 appears in

n^2

with exponent 4, so it's a prime factor. Choice D) 7 appears in

n^2

with exponent 2, so it's a prime factor. Choice C) 5 never appears in the original factorization of

n

, and since squaring only affects exponents (not which primes are present), 5 cannot be a prime factor of

n^2

. The answer is C. Remember this key principle: when you raise a number to any power, the prime factors remain the same—only their exponents change. If a prime doesn't divide the original number, it won't divide any power of that number either. This makes these questions much faster to solve without lengthy calculations.

Question 14

If $n$ is the smallest positive integer that is divisible by both $12$ and $18$ , how many prime factors does $n$ have (counting repetitions)?

$3$
$4$ (correct answer)
$5$
$6$
$7$

Explanation: When you see a question asking for the smallest positive integer divisible by two numbers, you're looking for their least common multiple (LCM). To find the LCM and count prime factors with repetitions, you need to work with prime factorizations. First, let's find the prime factorization of each number:

$12 = 2^2 \times 3^1$
$18 = 2^1 \times 3^2$

To find the LCM, take the highest power of each prime that appears in either factorization:

Highest power of 2: $2^2$ (from 12)
Highest power of 3: $3^2$ (from 18)

Therefore,

n = \text{LCM}(12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36

Now count the prime factors with repetitions:

2^2 \times 3^2

gives us two 2's and two 3's, for a total of 4 prime factors. Looking at the wrong answers: Choice (A) gives 3, which might come from incorrectly counting only unique primes plus one repetition. Choice (C) gives 5, which could result from mistakenly adding the exponents (2 + 2 = 4) then adding 1. Choice (D) gives 6, which might come from multiplying the exponents (2 × 2 = 4) then adding 2, or from confusing this with the total number of divisors. The correct answer is (B) 4. Study tip: When counting prime factors with repetitions, always write out the complete prime factorization and count each prime the number of times its exponent indicates. Don't overthink it—just count every prime factor individually.

Question 15

The number $91$ can be written as the product of two prime numbers. What is the sum of these two prime numbers?

$20$
$18$
$16$ (correct answer)
$14$
$12$

Explanation: When you encounter a question about writing a number as the product of prime numbers, you're working with prime factorization. This means breaking down a composite number into its prime building blocks. To find the prime factorization of

91

, start by testing small prime numbers as divisors. Since

91

is odd, it's not divisible by

2

. Check if it's divisible by

3

by adding its digits:

9 + 1 = 10

, which isn't divisible by

3

, so

91

isn't either. Next, try

5

- since

91

doesn't end in

0

5

, it's not divisible by

5

. Now test

7

91 ÷ 7 = 13

. Since

13

is also prime, we have

91 = 7 × 13

. The sum of these two prime factors is

7 + 13 = 16

. Looking at the wrong answers: Choice A (

20

) would come from pairs like

11 + 9

, but

9

isn't prime. Choice B (

18

) might result from

5 + 13

, but

5 × 13 = 65

, not

91

. Choice D (

14

) could come from

7 + 7

, but

7 × 7 = 49

, not

91

. The correct answer is C (

16

). Strategy tip: When finding prime factorization, work systematically through small primes:

2, 3, 5, 7, 11, 13...

Use divisibility rules to save time - check if the sum of digits is divisible by

3

, or if the number ends in

0

5

for divisibility by

5

Question 16

A positive integer $n$ has exactly three positive divisors. What type of number must $n$ be?

The square of a prime number (correct answer)
The cube of a prime number
The product of two distinct primes
A prime number itself
The square of a composite number

Explanation: When you encounter questions about the number of divisors a positive integer has, you need to think systematically about how divisors work based on prime factorization. Let's work through what it means for a number to have exactly three positive divisors. If

n = p^a \cdot q^b \cdot r^c \cdots

where

p, q, r

are distinct primes, then the number of positive divisors is

(a+1)(b+1)(c+1)\cdots

For this product to equal 3, and since 3 is prime, we need exactly one factor in this expression, meaning

n

has only one prime in its factorization. So

n = p^a

for some prime

p

. Since we need

(a+1) = 3

, we get

a = 2

, making

n = p^2

. Let's verify: if

n = p^2

(like

4 = 2^2

9 = 3^2

), the divisors are

1, p,

and

p^2

— exactly three divisors. Looking at the wrong answers: Choice B (cube of a prime) gives us

p^3

, which has divisors

1, p, p^2, p^3

— that's four divisors. Choice C (product of two distinct primes) gives us

pq

, which has divisors

1, p, q, pq

— again four divisors. Choice D (prime number) has only two divisors:

1

and itself. Strategy tip: When counting divisors, always use the formula

(a+1)(b+1)(c+1)\cdots

from the prime factorization. For questions asking about a specific number of divisors, work backwards from this formula to determine what the prime factorization must look like.

Question 17

Prime numbers have exactly two factors; composite numbers have more than two. Examples: 2, 3, 5, 7 are prime; 4 and 6 are composite. Select all the numbers that are prime from this list: 31, 33, 35, 37.

31 and 37 (correct answer)
33 and 35
31, 33, and 37
31, 35, and 37

Explanation: This question tests middle-level SSAT skills: determining whether a number is prime or composite, focusing on number properties and integers. Prime numbers have only two distinct positive divisors: 1 and the number itself, while composite numbers have more than two. In this question, students must use the definitions and identify factors, for example, checking that 31 and 37 have only two divisors each. Choice A is correct because it accurately reflects the numbers' properties, showing understanding of prime identification. Choice B is incorrect because it demonstrates a common misconception, such as identifying composites as primes. To help students: Encourage practice with factorization and using divisibility rules. Teach them to verify by checking divisibility by known smaller primes (2, 3, 5, 7) and practice spotting common mistakes like assuming number patterns.

Question 18

A prime has exactly two factors; a composite has more than two. Small examples: 2, 3, 5, 7 are prime; 4, 6 are composite. Select the prime numbers from the list: 22, 23, 24, 25.

Only 23 (correct answer)
22 and 23
23 and 25
22, 23, and 25

Explanation: This question tests middle-level SSAT skills: determining whether a number is prime or composite, focusing on number properties and integers. Prime numbers have only two distinct positive divisors: 1 and the number itself, while composite numbers have more than two. In this question, students must use the definitions and identify factors, for example, checking that only 23 has exactly two divisors in the list. Choice A is correct because it accurately reflects the prime in the list, showing understanding of prime identification. Choice B is incorrect because it demonstrates a common misconception, such as including a composite like 22 as prime. To help students: Encourage practice with factorization and using divisibility rules. Teach them to verify by checking divisibility by known smaller primes (2, 3, 5, 7) and practice spotting common mistakes like assuming number patterns.

Question 19

Prime numbers have exactly two factors; composite numbers have more than two. Examples: 2, 3, 5, 7 are prime; 4 and 6 are composite. How many prime numbers are there between 30 and 40?

1
2 (correct answer)
3
4

Explanation: This question tests middle-level SSAT skills: determining whether a number is prime or composite, focusing on number properties and integers. Prime numbers have only two distinct positive divisors: 1 and the number itself, while composite numbers have more than two. In this question, students must use the definitions and identify factors, for example, listing primes like 31 and 37 between 30 and 40. Choice B is correct because it accurately reflects the count of primes in the range, showing understanding of prime identification. Choice A is incorrect because it demonstrates a common misconception, such as undercounting the primes in the interval. To help students: Encourage practice with factorization and using divisibility rules. Teach them to verify by checking divisibility by known smaller primes (2, 3, 5, 7) and practice spotting common mistakes like assuming number patterns.

Question 20

Prime numbers have exactly two factors, 1 and itself; composite numbers have more than two. Using 2, 3, 4, 5, 6, and 7 as examples, decide: Is 29 prime or composite?

29 is composite because it is odd.
29 is prime because it has two factors. (correct answer)
29 is composite because it has four factors.
29 is prime because it is not even.

Explanation: This question tests middle-level SSAT skills: determining whether a number is prime or composite, focusing on number properties and integers. Prime numbers have only two distinct positive divisors: 1 and the number itself, while composite numbers have more than two. In this question, students must use the definitions and identify factors, for example, checking if 29 is prime because its only divisors are 1 and 29. Choice B is correct because it accurately reflects the number's properties, showing understanding of prime number identification. Choice A is incorrect because it demonstrates a common misconception, such as assuming all odd numbers are composite, but oddness does not determine compositeness. To help students: Encourage practice with factorization and using divisibility rules. Teach them to verify by checking divisibility by known smaller primes (2, 3, 5, 7) and practice spotting common mistakes like assuming number patterns.