Prime and Composite Numbers - SHSAT Math

1. Which of the following numbers is a composite number?
   Answer: $121$ 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.
2. What is the smallest positive integer $n > 1$ such that $n^2 - 1$ has exactly three distinct prime factors?
   Answer: 11, because $11^2 - 1 = 120 = 2^3 \times 3 \times 5$ with three prime factors Factor $n^2 - 1 = (n - 1)(n + 1)$ and test values of $n$ in order. $n = 2$ to $10$ give 3, 8, 15, 24, 35, 48, 63, 80 and 99. Each has at most two distinct prime factors (for example $24 = 2^3 \times 3$, $63 = 3^2 \times 7$, $99 = 3^2 \times 11$). $n = 11$ gives $120 = 2^3 \times 3 \times 5$, which has three distinct prime factors, so 11 is the smallest such $n$. The value 35 also works, since $1224 = 2^3 \times 3^2 \times 17$, but it is not the smallest. The values 12 and 30 give $143 = 11 \times 13$ and $899 = 29 \times 31$, each with only two prime factors.
3. How many prime numbers are there between $2^6$ and $2^6 + 20$, inclusive?
   Answer: 5 First, $2^6 = 64$ and $2^6 + 20 = 84$, so check the whole numbers from 64 to 84, including both ends. Every even number is composite, so only the odd numbers need checking: 65 = 5 × 13, 67 is prime, 69 = 3 × 23, 71 is prime, 73 is prime, 75 = 3 × 25, 77 = 7 × 11, 79 is prime, 81 = $3^4$, and 83 is prime. The primes are 67, 71, 73, 79 and 83, which is 5 primes. A count of 4 comes from leaving out 83, even though 83 is less than 84 and the range is inclusive. A count of 3 misses two of the primes, and a count of 6 treats a composite such as 77 or 81 as prime.
4. A teacher writes the number $143$ on the board. Is this number prime or composite?
   Answer: Composite, because it equals 11×13 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.
5. Which of the following numbers is a prime number?
   Answer: $67$ 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.
6. 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?
   Answer: All of the above representations are valid according to the conjecture 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.

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