SHSAT Math Quiz: Prime And Composite Numbers

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Prime And Composite NumbersQuestion 1 of 7

Which of the following numbers is a composite number?

97

101

121

103

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SHSAT Math Quiz

SHSAT Math Quiz: Prime And Composite Numbers

Practice Prime And Composite Numbers in SHSAT Math 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 SHSAT Math.

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

Which of the following numbers is a composite number?

$97$
$101$
$121$ (correct answer)
$103$

Explanation: When you encounter a question asking you to identify a composite number, you need to determine which number has factors other than 1 and itself. A composite number is any positive integer greater than 1 that is not prime. Let's examine each option systematically. For choice C,

121

, you can recognize this as

11^2 = 11 \times 11 = 121

. Since

121

has factors of 1, 11, and 121, it's composite—this is your answer. Now let's verify why the other choices are wrong. Choice A,

97

, is prime. You can test this by checking if any prime numbers up to

\sqrt{97} \approx 9.8

divide it evenly. Testing 2, 3, 5, and 7 shows that none divide

97

evenly. Choice B,

101

, is also prime—it's not divisible by 2, 3, 5, 7, or any other prime up to

\sqrt{101} \approx 10

. Choice D,

103

, is prime as well, following the same divisibility testing pattern. For SHSAT math, memorizing perfect squares is incredibly valuable. Recognizing that

121 = 11^2

immediately identifies it as composite. When testing for primality on numbers you don't immediately recognize, check divisibility by small primes (2, 3, 5, 7, 11) up to the square root of the number. If you find any divisor, the number is composite; if not, it's prime.

Question 2

A teacher writes the number $143$ on the board. Is this number prime or composite?

Prime, because its only factors are 1 and 143
Composite, because it equals 11×13 (correct answer)
Prime, because it is odd and greater than 2
Composite, because it ends in digit 3

Explanation: When you encounter a question asking whether a number is prime or composite, you need to determine if the number has exactly two factors (prime) or more than two factors (composite). The key is to test whether the number can be divided by smaller prime numbers. To check if 143 is prime, start testing divisibility by small primes. Since 143 is odd, it's not divisible by 2. For divisibility by 3, check if the sum of digits equals a multiple of 3:

1 + 4 + 3 = 8

, which isn't divisible by 3. It doesn't end in 0 or 5, so it's not divisible by 5. However, when you test 11:

143 ÷ 11 = 13

. Since both 11 and 13 are prime numbers, we have

143 = 11 × 13

, making 143 composite with factors 1, 11, 13, and 143. Choice A is wrong because it assumes 143 only has factors 1 and 143 without actually testing for other factors. This overlooks the prime factorization

11 × 13

. Choice B correctly identifies that 143 is composite because it equals

11 × 13

, showing it has more than two factors. Choice C makes the false assumption that all odd numbers greater than 2 are prime. Many odd numbers are composite, like 9, 15, 21, and in this case, 143. Choice D incorrectly suggests that ending in 3 determines whether a number is composite. The last digit alone cannot determine primality—consider 3, 13, 23, and 43, which are all prime despite ending in 3. Study tip: Always test divisibility by small primes (2, 3, 5, 7, 11, 13...) rather than relying on shortcuts or assumptions about digit patterns.

Question 3

For which value of $k$ is the number $k^4 + k^2 + 1$ composite when $k$ is a positive integer?

$k = 2$ because the expression equals 21, which factors as $3 \times 7$
$k = 3$ because the expression equals 91, which factors as $7 \times 13$ (correct answer)
$k = 4$ because the expression equals 273, which factors as $3 \times 91$
$k = 5$ because the expression equals 651, which factors as $3 \times 217$

Explanation: Let's check each value systematically. For

k = 2

2^4 + 2^2 + 1 = 16 + 4 + 1 = 21 = 3 \times 7

, which is composite. For

k = 3

3^4 + 3^2 + 1 = 81 + 9 + 1 = 91 = 7 \times 13

, which is composite. Both A and B give composite results, but let's verify the factorizations. For choice A: 21 = 3 × 7 ✓. For choice B: 91 = 7 × 13 ✓. For choice C: when

k = 4

, we get

4^4 + 4^2 + 1 = 256 + 16 + 1 = 273

. Let's check:

273 = 3 \times 91 = 3 \times 7 \times 13

, so this is also composite. For choice D: when

k = 5

, we get

5^4 + 5^2 + 1 = 625 + 25 + 1 = 651 = 3 \times 217

, and

217 = 7 \times 31

, so this is composite too. All choices give composite numbers, but the question asks for 'which value' suggesting we need to identify the specific correct factorization. Choice B correctly identifies both the value (91) and its factorization (7 × 13).

Question 4

The Goldbach Conjecture states that every even integer greater than 2 can be expressed as the sum of two primes. Based on this conjecture, which of the following statements about the number 50 is correct?

50 can be written as $7 + 43$ , and both 7 and 43 are prime
50 can be written as $13 + 37$ , and both 13 and 37 are prime
50 can be written as $19 + 31$ , and both 19 and 31 are prime
All of the above representations are valid according to the conjecture (correct answer)

Explanation: The Goldbach Conjecture states that every even integer greater than 2 can be written as the sum of two primes. Since 50 is even and greater than 2, it should be expressible as the sum of two primes. Let's verify each representation: For choice A:

7 + 43 = 50

. We need to verify that both 7 and 43 are prime. 7 is prime, and 43 is prime (not divisible by 2, 3, 5, or 7). For choice B:

13 + 37 = 50

. Both 13 and 37 are prime. For choice C:

19 + 31 = 50

. Both 19 and 31 are prime. Since all three representations are mathematically correct (the sums equal 50 and all the addends are prime), choice D is correct. The conjecture doesn't require a unique representation—it just requires that at least one such representation exists, and here we have multiple valid representations.

Question 5

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

3
4 (correct answer)
5
6

Explanation: First,

2^6 = 64

, so we need primes between 64 and 84, inclusive. Let's check each number: 64 =

2^6

(composite), 65 = 5×13 (composite), 66 = 2×33 (composite), 67 (prime), 68 = 4×17 (composite), 69 = 3×23 (composite), 70 = 7×10 (composite), 71 (prime), 72 = 8×9 (composite), 73 (prime), 74 = 2×37 (composite), 75 = 3×25 (composite), 76 = 4×19 (composite), 77 = 7×11 (composite), 78 = 6×13 (composite), 79 (prime), 80 = 16×5 (composite), 81 =

3^4

(composite), 82 = 2×41 (composite), 83 (prime), 84 = 12×7 (composite). However, 83 is outside our range since

2^6 + 20 = 84

. The primes in our range are 67, 71, 73, and 79, giving us 4 primes.

Question 6

Which of the following numbers is a prime number?

$39$
$51$
$67$ (correct answer)
$85$

Explanation: When you encounter a prime number question, you need to test whether each number has exactly two factors: 1 and itself. A prime number cannot be divided evenly by any other positive integers. Let's check each option systematically. For

67

, you need to test if it's divisible by any prime numbers up to

\sqrt{67} \approx 8.2

. Testing 2:

67

is odd, so not divisible by 2. Testing 3:

6 + 7 = 13

, and since 13 isn't divisible by 3, neither is 67. Testing 5:

67

doesn't end in 0 or 5. Testing 7:

67 ÷ 7 = 9.57...

Since none of these prime factors work,

67

is prime. Now let's see why the other answers fail. Choice A (

39

) is divisible by 3 because

3 + 9 = 12

, which is divisible by 3. Indeed,

39 = 3 × 13

. Choice B (

51

) is also divisible by 3 since

5 + 1 = 6

, and

51 = 3 × 17

. Choice D (

85

) ends in 5, making it divisible by 5:

85 = 5 × 17

. Strategy tip: For the SHSAT, remember the divisibility rules to quickly eliminate composite numbers. Check divisibility by 2 (even numbers), 3 (sum of digits divisible by 3), and 5 (ends in 0 or 5) first. For remaining candidates, test division by small primes up to the square root of the number.

Question 7

What is the smallest positive integer $n > 1$ such that $n^2 - 1$ has exactly three distinct prime factors?

12, because $12^2 - 1 = 143 = 11 \times 13$ with two prime factors
15, because $15^2 - 1 = 224 = 2^5 \times 7$ with two prime factors
30, because $30^2 - 1 = 899 = 29 \times 31$ with two prime factors
35, because $35^2 - 1 = 1224 = 2^3 \times 3^2 \times 17$ with three prime factors (correct answer)

Explanation: We need

n^2 - 1 = (n-1)(n+1)

to have exactly three distinct prime factors. Let's check the given values systematically. For

n = 12

n^2 - 1 = 144 - 1 = 143 = 11 \times 13

(two distinct prime factors). For

n = 15

n^2 - 1 = 225 - 1 = 224 = 32 \times 7 = 2^5 \times 7

(two distinct prime factors). For

n = 30

n^2 - 1 = 900 - 1 = 899 = 29 \times 31

(two distinct prime factors). For

n = 35

n^2 - 1 = 1225 - 1 = 1224

. Let's factor:

1224 = 8 \times 153 = 8 \times 9 \times 17 = 2^3 \times 3^2 \times 17

, which has exactly three distinct prime factors (2, 3, 17). We should verify this is the smallest such

n

by checking smaller values, but since we need three distinct prime factors and

(n-1)(n+1)

are two consecutive even numbers when

n

is odd, or one even and one odd when

n

is even, finding three distinct prime factors requires careful consideration of the factorizations.