Factors and Multiples - SHSAT Math

1. Which of the following integers has prime factorization of the form $p^{2}q$, where $p$ and $q$ are distinct primes?
   Answer: 50 When you encounter prime factorization problems, you need to break down each number into its prime factors and check if the pattern matches the given form. Here, you're looking for a number that can be written as $p^2q$, where $p$ and $q$ are different prime numbers. Let's examine each option by finding their prime factorizations: For choice A, $50 = 2 \times 25 = 2 \times 5^2$. This matches the form $p^2q$ where $p = 5$ and $q = 2$ (both prime and distinct). This is our answer. Choice B gives us $54 = 2 \times 27 = 2 \times 3^3$. This has the form $p^3q$, not $p^2q$, so it doesn't match our required pattern. For choice C, $70 = 2 \times 5 \times 7$. This is the product of three distinct primes with no repeated factors, so it has the form $pqr$, which doesn't match $p^2q$. Choice D yields $90 = 2 \times 45 = 2 \times 9 \times 5 = 2 \times 3^2 \times 5$. This involves three different primes with the form $p^2qr$, which again doesn't match our target pattern. Only choice A has exactly one prime raised to the second power multiplied by exactly one other distinct prime raised to the first power. Study tip: When working with prime factorization patterns, always write out the complete factorization first, then check if the exponents and number of distinct primes match the given form exactly. Don't just look for partial matches.
2. What is the least positive multiple of $45$ that is also divisible by $12$?
   Answer: 180 When you encounter a problem asking for the least positive multiple of one number that's also divisible by another, you're looking for the least common multiple (LCM) of both numbers. To find the LCM of 45 and 12, start by finding their prime factorizations: - $45 = 3^2 \times 5$ - $12 = 2^2 \times 3$ The LCM uses the highest power of each prime factor that appears: $2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$. This means 180 is the smallest number that both 45 and 12 divide into evenly. Let's verify: $180 ÷ 45 = 4$ and $180 ÷ 12 = 15$, so 180 is indeed a multiple of 45 that's divisible by 12. Choice A (60) is divisible by 12 since $60 ÷ 12 = 5$, but it's not a multiple of 45. Since $60 ÷ 45 = 1.33...$, this doesn't work. Choice B (90) is a multiple of 45 since $90 ÷ 45 = 2$, but when you divide $90 ÷ 12 = 7.5$, it's not evenly divisible by 12. Choice D (540) works for both conditions—it's $540 ÷ 45 = 12$ and $540 ÷ 12 = 45$—but it's not the least such number since 180 is smaller. Therefore, C (180) is correct. Study tip: When finding the LCM of two numbers, use prime factorization and take the highest power of each prime. This method is more reliable than guessing and checking, especially under time pressure.
3. A positive integer $n$ leaves a remainder of 7 when divided by 12 and a remainder of 4 when divided by 9. If $n$ is between 100 and 200, how many possible values of $n$ are there?
   Answer: 3 We need to solve the system: n ≡ 7 (mod 12) and n ≡ 4 (mod 9). From the first congruence, n = 12k + 7 for some integer k. Substituting into the second: 12k + 7 ≡ 4 (mod 9), so 12k ≡ -3 ≡ 6 (mod 9). Since 12 ≡ 3 (mod 9), we have 3k ≡ 6 (mod 9), which gives k ≡ 2 (mod 3). So k = 3j + 2 for some integer j. Therefore n = 12(3j + 2) + 7 = 36j + 24 + 7 = 36j + 31. The general solution is n ≡ 31 (mod 36). For 100 ≤ n ≤ 200: 100 ≤ 36j + 31 ≤ 200, so 69 ≤ 36j ≤ 169, giving 69/36 ≤ j ≤ 169/36, or 1.92 ≤ j ≤ 4.69. Since j must be an integer, j ∈ {2, 3, 4}. This gives n ∈ {103, 139, 175}. Let's verify: 103 = 8×12 + 7 ✓ and 103 = 11×9 + 4 ✓. 139 = 11×12 + 7 ✓ and 139 = 15×9 + 4 ✓. 175 = 14×12 + 7 ✓ and 175 = 19×9 + 4 ✓. Therefore there are 3 possible values.
4. Stage lights flash red every 9 seconds and blue every 12 seconds. After they flash together, how many seconds will pass before they flash together again for the first time?
   Answer: 36 When you see a problem about events happening at regular intervals and need to find when they occur together, you're dealing with a least common multiple (LCM) problem. The question is asking when both lights will flash simultaneously again. To find when the red and blue lights flash together, you need the LCM of 9 and 12. Start by finding the prime factorization of each number: $9 = 3^2$ and $12 = 2^2 \times 3$. The LCM uses the highest power of each prime factor, so $\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = 36$. This means the lights flash together every 36 seconds. Looking at the wrong answers: Choice A (9) represents only the red light's cycle, ignoring the blue light entirely. Choice B (12) makes the same error but for the blue light's cycle. Choice C (18) is a common trap—it's half of 36, which you might get if you mistakenly found the average of 9 and 12, or made an error in your LCM calculation. The correct answer is D (36 seconds). Study tip: LCM problems often appear on the SHSAT when dealing with repeating cycles, whether it's lights, bells, or rotating schedules. Always remember that finding when events coincide requires the LCM, not addition, subtraction, or averaging of the individual periods. Practice finding LCMs quickly using prime factorization—it's faster and more reliable than listing multiples for larger numbers.
5. What is the least positive integer greater than 20 that is a multiple of every factor of $6$?
   Answer: 24 When you encounter a question asking for the least number that's a multiple of every factor of a given number, you're looking for the least common multiple (LCM) of all those factors. First, find all factors of 6. The factors are the numbers that divide evenly into 6: 1, 2, 3, and 6. Now you need the smallest positive integer greater than 20 that is divisible by all of these numbers. Since any number divisible by 6 is automatically divisible by both 2 and 3 (and certainly by 1), you're really looking for the smallest multiple of 6 that's greater than 20. The multiples of 6 are: 6, 12, 18, 24, 30, 36... The first one greater than 20 is 24. Let's verify: 24 ÷ 1 = 24, 24 ÷ 2 = 12, 24 ÷ 3 = 8, and 24 ÷ 6 = 4. All divisions result in whole numbers, confirming that 24 is divisible by every factor of 6. Choice (A) 22 is not divisible by 6, 3, or 2 (22 ÷ 2 = 11, but 22 ÷ 3 = 7.33...). Choice (C) 26 is divisible by 1 and 2, but not by 3 or 6 (26 ÷ 3 = 8.67...). Choice (D) 30 is divisible by all factors of 6, but it's not the least such number greater than 20. The answer is (B) 24. Strategy tip: When finding numbers divisible by all factors of n, focus on finding multiples of n itself, since n contains all its own factors as divisors.
6. If $n$ is a multiple of $18$ and $24 < n < 100$, how many possible values can $n$ have?
   Answer: 4 When you encounter a question asking for the number of values satisfying multiple conditions, you need to systematically find all multiples within the given range. Since $n$ is a multiple of 18, you can write $n = 18k$ where $k$ is a positive integer. The constraint $24 < n < 100$ becomes $24 < 18k < 100$. Dividing by 18: $\frac{24}{18} < k < \frac{100}{18}$, which simplifies to $1.33... < k < 5.55...$ Since $k$ must be an integer, the possible values are $k = 2, 3, 4, 5$. This gives you: - $k = 2$: $n = 36$ - $k = 3$: $n = 54$ - $k = 4$: $n = 72$ - $k = 5$: $n = 90$ All four values satisfy $24 < n < 100$, so there are 4 possible values. Choice A (3) likely results from missing one of the multiples, perhaps by incorrectly excluding 36 or 90 from the range. Choice C (5) might come from including $k = 1$ (giving $n = 18$, which violates $n > 24$) or $k = 6$ (giving $n = 108$, which violates $n < 100$). Choice D (6) probably stems from including both boundary violations or miscounting the multiples entirely. Strategy tip: For "how many values" questions involving multiples, convert to the form $n = \text{multiple} \times k$, solve the inequality for $k$, then count the integer values of $k$. Always double-check that your boundary values actually satisfy the original constraints.
7. How many positive whole-number factors does $60$ have?
   Answer: 12 When you encounter a question about finding the number of factors of a number, you need to use the prime factorization method. This approach breaks down the number into its prime components and applies a systematic formula. First, find the prime factorization of 60. Start by dividing by the smallest primes: $60 = 2^2 \times 3^1 \times 5^1$. You can verify this: $4 \times 3 \times 5 = 60$. To find the total number of factors, add 1 to each exponent in the prime factorization, then multiply these results together. The formula is: if $n = p_1^{a_1} \times p_2^{a_2} \times p_3^{a_3}$, then the number of factors is $(a_1 + 1)(a_2 + 1)(a_3 + 1)$. For 60: $(2 + 1)(1 + 1)(1 + 1) = 3 \times 2 \times 2 = 12$. The answer is C. Choice A (8) likely comes from miscounting or using an incomplete method. Choice B (10) might result from adding the exponents instead of using the multiplication formula: $2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10$, but this is incorrect. Choice D (16) could come from incorrectly calculating $(2 + 2)(1 + 2)(1 + 2) = 4 \times 3 \times 3$ by adding 2 to each exponent instead of 1. Remember this key strategy: always use prime factorization for factor problems, then apply the formula $(exponent + 1)$ for each prime. This systematic approach prevents counting errors and works for any number.
8. Let $n = 2^{3}\cdot3^{2}\cdot5$. What is the sum of the greatest and least positive factors of $n$?
   Answer: 361 When you encounter a question about factors of a number given in prime factorization form, you need to identify the greatest and least positive factors, then find their sum. Given $n = 2^3 \cdot 3^2 \cdot 5$, let's first calculate the actual value: $n = 8 \cdot 9 \cdot 5 = 360$. The least positive factor of any positive integer is always 1, since 1 divides every integer. The greatest positive factor of any positive integer is the number itself, since every number divides itself. Therefore, the greatest factor of $n$ is 360. The sum of the greatest and least factors is $360 + 1 = 361$. Looking at the answer choices: Choice A gives us 361, which matches our calculation. Choice B (362) might tempt students who make an arithmetic error when adding 360 + 1. Choice C (365) could trap students who incorrectly think the greatest factor is $2^3 + 3^2 + 5 = 8 + 9 + 5 = 22$, then add the exponents to get $3 + 2 + 1 = 6$, leading to confused calculations. Choice D (721) is double 360 plus 1, which might result from mistakenly thinking you need to count both 1 and the number twice. Remember this key principle: for any positive integer, the smallest positive factor is always 1, and the largest positive factor is always the number itself. When you see prime factorization, calculate the actual number first—it often makes the problem much clearer.
9. Find the sum of the greatest common factor of $42$ and $56$ and the least common multiple of $4$ and $6$.
   Answer: 26 When you encounter problems involving both greatest common factor (GCF) and least common multiple (LCM), you need to find each value separately, then combine them as the problem requests. To find the GCF of 42 and 56, list the factors or use prime factorization. Since $42 = 2 \times 3 \times 7$ and $56 = 2^3 \times 7$, the GCF includes the lowest power of each common prime factor: $2^1 \times 7^1 = 14$. For the LCM of 4 and 6, you need the smallest number that both divide into evenly. Since $4 = 2^2$ and $6 = 2 \times 3$, the LCM includes the highest power of each prime factor: $2^2 \times 3^1 = 12$. The sum is $14 + 12 = 26$. Looking at the wrong answers: Choice A (20) might result from incorrectly finding the GCF as 8 instead of 14, perhaps by missing the factor of 7. Choice B (24) could come from reversing the operations—finding the LCM of 42 and 56 (which is 168) and making calculation errors, or finding $12 + 12$. Choice D (28) might result from doubling the GCF ($14 \times 2$) and forgetting about the LCM entirely. The correct answer is C. Remember this strategy: when working with GCF and LCM problems, always double-check your prime factorizations and make sure you're taking the minimum powers for GCF and maximum powers for LCM. These concepts appear frequently on the SHSAT, so practice identifying factor patterns quickly.
10. Which of the following numbers is a factor of $84$ but is NOT a multiple of $6$?
    Answer: 7 This question tests your understanding of factors and multiples, two fundamental concepts that often appear together on the SHSAT. You need to find a number that satisfies both conditions: it divides evenly into 84, AND it's not a multiple of 6. Let's start by finding the factors of 84. Since $84 = 2^2 \times 3 \times 7$, the factors are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. Now check which answer choices are factors: A) 7 ✓, B) 12 ✓, C) 18 ✗, D) 30 ✗. Next, determine which factors are NOT multiples of 6. A multiple of 6 must be divisible by both 2 and 3. Looking at our remaining candidates: 7 is odd (not divisible by 2), so it's not a multiple of 6. However, $12 = 6 \times 2$, so 12 IS a multiple of 6. Therefore, 7 is the only number that's a factor of 84 but NOT a multiple of 6. Here's why the other choices fail: B) 12 is indeed a factor of 84, but it's also $6 \times 2$, making it a multiple of 6. C) 18 isn't even a factor of 84 since $84 ÷ 18 = 4.67...$ D) 30 also isn't a factor of 84 since $84 ÷ 30 = 2.8$. Strategy tip: When dealing with factors and multiples, always check both conditions separately. First verify which numbers actually divide into the given number, then apply the second constraint to eliminate choices.
11. What is the greatest integer less than $100$ that has exactly 9 positive factors?
    Answer: 36 When you encounter questions about the number of factors a number has, you need to use the prime factorization formula. If a number has prime factorization $p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}$, then the number of factors is $(a_1 + 1)(a_2 + 1)...(a_k + 1)$. To find numbers with exactly 9 factors, we need $(a_1 + 1)(a_2 + 1)... = 9$. Since $9 = 9 \times 1 = 3 \times 3$, there are two possible forms: - $p^8$ (where $8 + 1 = 9$) - $p^2 \times q^2$ (where $(2 + 1)(2 + 1) = 9$) For the first form, $2^8 = 256 > 100$, so this won't work. For the second form $p^2 \times q^2$, we need the largest such number under 100. Using the smallest primes: $2^2 \times 3^2 = 4 \times 9 = 36$. We could try $2^2 \times 5^2 = 100$, but that equals 100, not less than 100. Other combinations like $3^2 \times 5^2 = 225$ are too large. Let's verify: $36 = 2^2 \times 3^2$ has $(2+1)(2+1) = 9$ factors: 1, 2, 3, 4, 6, 9, 12, 18, 36. Looking at the wrong answers: (A) $30 = 2 \times 3 \times 5$ has $(1+1)(1+1)(1+1) = 8$ factors. (B) $32 = 2^5$ has $5+1 = 6$ factors. (D) $40 = 2^3 \times 5$ has $(3+1)(1+1) = 8$ factors. The answer is C. Strategy tip: For factor-counting problems, always start by finding the possible ways to factor the target number of factors, then work backwards to find numbers with those prime factorization patterns.
12. If $k$ is a factor of $72$ and $k>10$, which of the following could be $k$?
    Answer: 24 When you encounter a question about factors, you need to find all numbers that divide evenly into the given number. Since $k$ is a factor of $72$ with the condition $k > 10$, you must first identify which factors of $72$ are greater than $10$. To find the factors of $72$, break it down: $72 = 8 \times 9 = 2^3 \times 3^2$. The complete list of factors is: $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$. The factors greater than $10$ are: $12, 18, 24, 36, 72$. Choice C ($24$) is correct because $72 ÷ 24 = 3$, confirming that $24$ divides evenly into $72$ with no remainder. Choice A ($14$) is wrong because $72 ÷ 14 = 5.14...$ This doesn't result in a whole number, so $14$ is not a factor of $72$. Choice B ($20$) is wrong because $72 ÷ 20 = 3.6$. Again, this isn't a whole number, so $20$ cannot be a factor of $72$. Choice D ($30$) is wrong because $72 ÷ 30 = 2.4$. Since this doesn't yield a whole number, $30$ is not a factor of $72$. Strategy tip: When testing if a number is a factor, perform the division and check if the result is a whole number. If you get a decimal or remainder, it's not a factor. For efficiency on the SHSAT, you can also check if the prime factorization of your potential factor can be "built" using only the prime factors of the original number.
13. Which integer between 50 and 70 has exactly four positive factors?
    Answer: 55 When you see a question about the number of positive factors, you need to understand how factors relate to prime factorization. A number's factors come from all possible combinations of its prime factors. To find which number has exactly four positive factors, let's check each option by finding their prime factorizations: For choice A) 54: $54 = 2 \times 3^3$. The factors are 1, 2, 3, 6, 9, 18, 27, and 54 — that's 8 factors, not 4. For choice B) 55: $55 = 5 \times 11$. Since both 5 and 11 are prime, the factors are 1, 5, 11, and 55 — exactly 4 factors. For choice C) 56: $56 = 2^3 \times 7$. Using the factor formula $(3+1)(1+1) = 8$, this gives 8 factors. For choice D) 59: This is a prime number, so it only has 2 factors: 1 and 59. Only 55 has exactly four factors, making B correct. Here's the key insight: numbers with exactly four positive factors fall into two categories. The first is the cube of a prime (like $2^3 = 8$), and the second is the product of two distinct primes (like $5 \times 11 = 55$). Remember this pattern: when looking for numbers with exactly four factors, check if the number is either a prime cubed or the product of two different primes. This will save you time on factor-counting problems.
14. Which number is both a factor of $96$ and a multiple of $12$?
    Answer: 24 When you encounter a problem asking for a number that satisfies two conditions simultaneously, you need to check both requirements carefully. Here, you're looking for a number that is both a factor of 96 (divides evenly into 96) and a multiple of 12 (can be expressed as 12 times some integer). Let's test each option systematically. For choice B) 24: First, check if 24 is a factor of 96 by dividing: $96 ÷ 24 = 4$. Since this gives a whole number, 24 is indeed a factor of 96. Second, check if 24 is a multiple of 12: $24 ÷ 12 = 2$. Since 24 equals 12 times 2, it's a multiple of 12. Choice B satisfies both conditions. Now let's see why the other answers fail. Choice A) 18 is not a factor of 96 because $96 ÷ 18 = 5.33...$, which isn't a whole number. Choice C) 54 also fails the factor test: $96 ÷ 54 = 1.77...$, not a whole number. Choice D) 108 fails immediately because it's larger than 96, so it cannot possibly be a factor of 96. Study tip: When checking factors and multiples simultaneously, start by eliminating any numbers larger than the given number (they can't be factors), then systematically test the remaining choices for both conditions. Remember that factors divide evenly into the original number, while multiples are the result of multiplying by whole numbers.
15. What is the product of the distinct prime factors of $210$?
    Answer: 210 When you encounter questions about "distinct prime factors," you need to find the prime factorization first, then work with those unique prime numbers. Let's find the prime factorization of 210. Start by dividing by the smallest primes: - $210 ÷ 2 = 105$ - $105 ÷ 3 = 35$ - $35 ÷ 5 = 7$ - $7 ÷ 7 = 1$ So $210 = 2 × 3 × 5 × 7$ The distinct prime factors are 2, 3, 5, and 7. The question asks for their product: $2 × 3 × 5 × 7 = 210$. This confirms answer choice D is correct. Now let's examine why the other answers are wrong: Choice A (21) equals $3 × 7$, which is the product of only two of the prime factors. This represents a common error where students might miss some factors during prime factorization. Choice B (35) equals $5 × 7$, again using only two prime factors. Students might arrive at this by incorrectly factoring 210 or stopping their prime factorization too early. Choice C (70) equals $2 × 5 × 7$, missing the factor of 3. This could happen if you factor 210 as $3 × 70$ and mistakenly think 70 is prime, or if you overlook one prime during the factorization process. Study tip: When finding distinct prime factors, always complete the full prime factorization to ensure you don't miss any factors. Double-check by multiplying your prime factors back together—they should equal the original number.
16. If $k$ is the greatest common factor of 180 and 252, and $m$ is the least common multiple of 180 and 252, what is the value of $\frac{k+m}{k}$?
    Answer: 36 Factor both numbers: 180 = 2² × 3² × 5 and 252 = 2² × 3² × 7. The GCF uses only the shared prime powers: k = 2² × 3² = 36. The LCM uses every prime to its highest power: m = 2² × 3² × 5 × 7 = 1260. Then (k + m)/k = (36 + 1260)/36 = 1296/36 = 36. A quicker route: (k + m)/k = 1 + m/k = 1 + 1260/36 = 1 + 35 = 36. A common slip is to stop at m/k = 35 and forget to add 1. The answers 85, 127 and 169 would need m/k to be 84, 126 or 168, but here the LCM is only 35 times the GCF, so none of those values can be right.
17. A digital clock chimes every 20 minutes, and a cuckoo clock chimes every 28 minutes. If both chime together at 6:00 AM, at what time will they chime together next?
    Answer: 8:20 AM When you see a problem about events that repeat at different intervals and happen "together," you're dealing with a least common multiple (LCM) problem. The key is finding when both patterns will align again. The digital clock chimes every 20 minutes, and the cuckoo clock chimes every 28 minutes. To find when they'll chime together again, you need the LCM of 20 and 28. First, find the prime factorization of each number: - $20 = 2^2 \times 5$ - $28 = 2^2 \times 7$ The LCM uses the highest power of each prime factor: $2^2 \times 5 \times 7 = 4 \times 5 \times 7 = 140$ So they'll chime together again after 140 minutes. Converting: $140 ÷ 60 = 2$ hours and $20$ minutes. Starting from 6:00 AM, this gives us 8:20 AM. Let's examine why the other answers are wrong. Choice A (6:20 AM) represents just one cycle of the digital clock—it ignores the cuckoo clock entirely. Choice B (7:40 AM) is 100 minutes later, which isn't a multiple of 28, so the cuckoo clock won't chime then. Choice D (8:40 AM) is 160 minutes later, which works for the digital clock (8 cycles × 20 minutes) but not the cuckoo clock, since 160 ÷ 28 isn't a whole number. Strategy tip: In "simultaneous event" problems, always find the LCM of the time intervals. Don't get trapped by answers that work for only one of the patterns—both conditions must be satisfied simultaneously.
18. Jessica has 18 red beads and 30 blue beads. She wants to create identical bracelets using all the beads with none left over. What is the greatest number of bracelets she can make?
    Answer: 6 When you see a problem about dividing items equally with "none left over," you're looking for the greatest common factor (GCF). Jessica needs to distribute all her beads evenly across identical bracelets, so each bracelet must contain the same number of red beads and the same number of blue beads. To find the GCF of 18 and 30, you can use prime factorization or the Euclidean algorithm. Let's factor: $18 = 2 \times 3^2$ and $30 = 2 \times 3 \times 5$. The GCF is the product of common prime factors: $2 \times 3 = 6$. This means Jessica can make 6 bracelets, each containing $18 ÷ 6 = 3$ red beads and $30 ÷ 6 = 5$ blue beads. Choice A (3) is incorrect because while 3 divides both 18 and 30, it's not the greatest common factor. You could make 3 bracelets, but that wouldn't maximize the number of bracelets possible. Choice C (12) is wrong because 12 doesn't divide evenly into 18. You'd have $18 ÷ 12 = 1.5$ red beads per bracelet, which is impossible. Choice D (15) fails for the same reason—15 doesn't divide evenly into 18. You'd need $18 ÷ 15 = 1.2$ red beads per bracelet. Remember: when you see "identical groups using all items with none left over," you're finding the GCF of the given quantities. The question asks for the greatest number of groups possible, which is always the GCF itself.
19. If $a$ and $b$ are positive integers such that $\gcd(a,b) = 18$ and $\text{lcm}(a,b) = 1080$, what is the number of possible ordered pairs $(a,b)$?
    Answer: 8 Use $\gcd(a,b) \times \text{lcm}(a,b) = ab$: $ab = 18 \times 1080 = 19440$. Because the GCD is 18, write $a = 18m$ and $b = 18n$, where $m$ and $n$ share no common factor. Then $324mn = 19440$, so $mn = 60 = 2^2 \times 3 \times 5$. Since $m$ and $n$ share no factor, each prime power (4, 3 and 5) must go entirely to $m$ or entirely to $n$. That gives $2^3 = 8$ choices: (1, 60), (4, 15), (3, 20), (5, 12) and their reverses. So there are 8 ordered pairs, such as (18, 1080) and (72, 270). 12 counts every ordered factor pair of 60, including pairs like (2, 30) and (6, 10) that share a factor of 2; those would make the GCD 36, not 18. 6 counts the factor pairs of 60 without order. 10 matches neither count.
20. If $p$ is the greatest prime such that $p^{2}$ is a factor of $180$, what is the value of $p$?
    Answer: 3 When you see a question asking for the greatest prime whose square divides a number, you need to find the prime factorization and identify which prime appears with the highest even exponent. Let's find the prime factorization of 180. Start by dividing by the smallest primes: $180 = 4 \times 45 = 4 \times 9 \times 5 = 2^2 \times 3^2 \times 5^1$ So $180 = 2^2 \times 3^2 \times 5^1$. For a prime $p$ to have $p^2$ as a factor of 180, the prime must appear with an exponent of at least 2 in the factorization. Looking at our factorization, both 2 and 3 appear with exponent 2, while 5 appears with exponent 1. Since $2^2 = 4$ and $3^2 = 9$ are both factors of 180, but $5^2 = 25$ is not (since 5 only appears once), the primes whose squares divide 180 are 2 and 3. The greatest of these is 3. Now for the wrong answers: (A) 2 is incorrect because while $2^2$ does divide 180, it's not the greatest such prime. (C) 5 is wrong because $5^2 = 25$ does not divide 180 — you'd need $5^2$ in the factorization, but we only have $5^1$. (D) 7 is incorrect because 7 doesn't even appear in the prime factorization of 180, so $7^2 = 49$ cannot be a factor. The answer is (B) 3. Strategy tip: Always find the complete prime factorization first, then check which primes have exponents ≥ 2. The greatest of these primes is your answer.

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