Apply Number Concepts - HSPT Math

1. How many positive whole-number factors does $72$ have?
   Answer: 12 When you need to find the number of positive factors of a number, the most efficient approach is to use prime factorization. This method works because every factor of a number can be expressed as a product of its prime factors raised to various powers. First, let's find the prime factorization of 72. Start by dividing by the smallest prime: $72 = 8 \times 9 = 2^3 \times 3^2$. So $72 = 2^3 \times 3^2$. To find the total number of factors, use this formula: if a number has prime factorization $p_1^{a_1} \times p_2^{a_2} \times \ldots$, then the number of positive factors is $(a_1 + 1)(a_2 + 1)\ldots$ For $72 = 2^3 \times 3^2$, the number of factors is $(3 + 1)(2 + 1) = 4 \times 3 = 12$. This confirms answer choice C is correct. Let's examine why the other answers are wrong. Choice A (6) likely comes from incorrectly counting only some obvious factors like 1, 2, 3, 6, 8, 9, missing several others. Choice B (8) might result from counting the prime factorization elements incorrectly or forgetting the systematic approach. Choice D (15) could come from a calculation error in the formula or misidentifying the prime factorization. For factor problems on the HSPT, always use prime factorization rather than listing factors manually. The formula $(a_1 + 1)(a_2 + 1)\ldots$ is foolproof and prevents you from missing factors or making counting errors. Remember to add 1 to each exponent before multiplying.
2. If the least common multiple of $6$ and $n$ is $42$, which of the following is a possible value of $n$?
   Answer: 14 When you encounter LCM problems, you need to understand that the least common multiple of two numbers is the smallest positive number that both original numbers divide into evenly. To find which value of $n$ works, you can test each option by finding the LCM of 6 and that number. For choice C (14), let's find LCM(6, 14). First, find the prime factorizations: $6 = 2 \times 3$ and $14 = 2 \times 7$. The LCM uses the highest power of each prime factor: $2^1 \times 3^1 \times 7^1 = 42$. This matches our target, so C is correct. Let's check why the other options fail. For choice A (8): $8 = 2^3$, so LCM(6, 8) = $2^3 \times 3^1 = 24$, not 42. For choice B (12): $12 = 2^2 \times 3$, so LCM(6, 12) = $2^2 \times 3^1 = 12$, not 42. For choice D (30): $30 = 2 \times 3 \times 5$, so LCM(6, 30) = $2^1 \times 3^1 \times 5^1 = 30$, not 42. Notice that choices B and D actually give LCMs that are smaller than 42. This happens when one number is a multiple of the other (6 divides into both 12 and 30), making the larger number the LCM. Strategy tip: When working with LCM problems, always verify your answer by checking that both original numbers divide evenly into your result, and that no smaller number works.
3. Which of the following is not a multiple of $75$?
   Answer: 425 When you encounter a question asking which number is "not" a multiple of another number, you need to check whether each choice can be evenly divided by the given number. A multiple of 75 means $75 \times n$ for some whole number $n$, or equivalently, a number that gives no remainder when divided by 75. Let's check each option by dividing by 75: For choice A: $225 \div 75 = 3$ exactly, so 225 is a multiple of 75. For choice B: $300 \div 75 = 4$ exactly, so 300 is a multiple of 75. For choice D: $450 \div 75 = 6$ exactly, so 450 is a multiple of 75. For choice C: $425 \div 75 = 5.67...$ or $5\frac{2}{3}$. Since this doesn't give a whole number, 425 is not a multiple of 75. We can verify this: $75 \times 5 = 375$ and $75 \times 6 = 450$, so 425 falls between these multiples. Choice C is correct because 425 cannot be expressed as $75 \times n$ where $n$ is a whole number. The other choices (A, B, and D) are all actual multiples of 75, making them incorrect answers to this "not" question. Strategy tip: For "which is not" questions, systematically test each choice rather than trying to spot the answer. Also, remember that multiples of 75 must be multiples of both 25 and 3, which can sometimes provide a quicker check.
4. Which pair of numbers are consecutive multiples of $9$?
   Answer: 45 and 54 When you encounter questions about consecutive multiples, you're looking for two numbers that are both multiples of the given number and follow each other in sequence with no other multiples between them. To find consecutive multiples of 9, start by identifying what multiples of 9 look like. Since $9 \times 1 = 9$, $9 \times 2 = 18$, $9 \times 3 = 27$, and so on, consecutive multiples of 9 will differ by exactly 9. For example, 27 and 36 are consecutive multiples because $27 = 9 \times 3$ and $36 = 9 \times 4$. Let's check each option by verifying both numbers are multiples of 9 and consecutive: Choice A gives us 42 and 51. While $42 = 9 \times 4.67$ (not a whole number), 42 isn't even a multiple of 9, so this pair fails immediately. Choice B presents 45 and 54. Here, $45 = 9 \times 5$ and $54 = 9 \times 6$. Both are multiples of 9, and since they represent the 5th and 6th multiples, they are consecutive. Choice C offers 48 and 57. Since $48 = 9 \times 5.33$ (not whole), 48 isn't a multiple of 9, eliminating this option. Choice D shows 54 and 72. While $54 = 9 \times 6$ and $72 = 9 \times 8$, these represent the 6th and 8th multiples of 9. The 7th multiple (63) falls between them, so they're not consecutive. Therefore, B is correct. Study tip: When checking consecutive multiples, divide each number by the base to ensure you get consecutive whole numbers. This quick division check will save you time and prevent errors.
5. A six-digit number is formed using the digits 1, 2, 3, 4, 5, and 6 exactly once each. If this number is divisible by 4, what is the maximum possible value of the tens digit?
   Answer: 6 A number is divisible by 4 when the number formed by its last two digits is divisible by 4. The tens digit is the first of those two digits, so look for the largest possible tens digit. The largest digit available is 6. With 6 in the tens place, the ones digit must make 6_ divisible by 4. The number 64 works, since 64 ÷ 4 = 16, and 4 is one of the allowed digits. The other four digits fill the front, for example 123564, and 123564 ÷ 4 = 30891. A tens digit of 5 also works (endings 52 or 56), but 6 is larger, so stopping at 5 misses the 64 ending. A tens digit of 4 is not possible at all: the endings 40, 44, and 48 need a 0, a repeated 4, or an 8. A tens digit of 3 works with 32 or 36, but it is smaller. The maximum possible tens digit is 6.
6. If $12n$ is a multiple of $90$, which of the following could be the value of $n$?
   Answer: 15 When you see a question about one expression being a multiple of another, you need to work with prime factorization and divisibility rules. Here, $12n$ must be divisible by $90$, which means $90$ divides evenly into $12n$. First, find the prime factorizations: $12 = 2^2 \times 3$ and $90 = 2 \times 3^2 \times 5$. For $12n$ to be a multiple of $90$, the expression $12n = 2^2 \times 3 \times n$ must contain at least the same prime factors as $90$ with equal or greater powers. Comparing the factorizations, $12n$ already has $2^2$ (more than the $2^1$ needed) and $3^1$, but it needs $3^2$ and $5^1$. Therefore, $n$ must provide at least one more factor of $3$ and one factor of $5$. The smallest such $n$ would be $3 \times 5 = 15$. Testing choice C: $12 \times 15 = 180$, and $180 \div 90 = 2$. This works perfectly. Choice A gives $12 \times 10 = 120$, but $120 \div 90 = 1.33...$ (not a whole number). Choice B gives $12 \times 12 = 144$, but $144 \div 90 = 1.6$ (missing the factor of $5$). Choice D gives $12 \times 24 = 288$, and $288 \div 90 = 3.2$ (still missing the factor of $5$). Strategy tip: When checking if one number is a multiple of another, compare their prime factorizations. The multiple must contain all prime factors of the divisor with at least the same powers.
7. Which of the following numbers has an $8$ in the hundredths place and a $2$ in the ten-thousandths place?
   Answer: 7.6832 When you encounter decimal place value questions, you need to identify the specific position of each digit after the decimal point. The places go: tenths, hundredths, thousandths, ten-thousandths, and so on. Let's map out the decimal places for a number like $7.6832$: - $6$ is in the tenths place (first digit after decimal) - $8$ is in the hundredths place (second digit after decimal) - $3$ is in the thousandths place (third digit after decimal) - $2$ is in the ten-thousandths place (fourth digit after decimal) We need a number with $8$ in the hundredths place and $2$ in the ten-thousandths place. Choice A ($7.6832$) has $8$ in the hundredths place and $2$ in the ten-thousandths place—this matches perfectly. Choice B ($7.8632$) has $8$ in the tenths place (not hundredths) and $2$ in the ten-thousandths place. The $8$ is in the wrong position. Choice C ($7.6823$) has $8$ in the hundredths place correctly, but $3$ in the ten-thousandths place instead of $2$. Choice D ($7.6283$) has $2$ in the hundredths place (not $8$) and $3$ in the ten-thousandths place (not $2$). Both required digits are in wrong positions. Remember this pattern: after the decimal point, count positions carefully—tenths (1st), hundredths (2nd), thousandths (3rd), ten-thousandths (4th). Many students mix up these place values, so always count from the decimal point when identifying positions.
8. Which of the following is the prime factorization of $198$?
   Answer: $2 \times 3^{2} \times 11$ Prime factorization questions require you to break down a number into its smallest prime factors. When you see a question like this, your goal is to systematically divide the given number by prime numbers until you can't divide anymore. To find the prime factorization of 198, start by dividing by the smallest prime numbers. First, 198 is even, so divide by 2: $198 ÷ 2 = 99$. Now factor 99. Since the digits of 99 sum to 18 (which is divisible by 3), divide by 3: $99 ÷ 3 = 33$. Continue with 33: $33 ÷ 3 = 11$. Since 11 is prime, you're done. Therefore: $198 = 2 × 3 × 3 × 11 = 2 × 3^2 × 11$. Looking at the wrong answers: Choice A gives $2 × 3 × 33$, but 33 isn't prime since $33 = 3 × 11$. This represents incomplete factorization. Choice C shows $2^2 × 3 × 11 = 4 × 3 × 11 = 132$, which equals the wrong number entirely. Choice D gives $2 × 9 × 11$, but 9 isn't prime since $9 = 3^2$, making this another case of incomplete factorization. Choice B, $2 × 3^2 × 11$, correctly shows all prime factors with proper exponents. Strategy tip: Always verify your prime factorization by multiplying back: $2 × 3^2 × 11 = 2 × 9 × 11 = 198$. Also remember that prime factorization must use only prime numbers (2, 3, 5, 7, 11, etc.) — if you see composite numbers like 9, 15, or 33 in an answer, it's incomplete.
9. The number $M = 3^{100} \cdot 7^{50}$ is divisible by $21^k$ for some positive integer $k$. What is the largest possible value of $k$?
   Answer: 50 When you encounter problems about the highest power of one number that divides another, you need to think about prime factorization and how exponents combine. First, let's factor $21 = 3 \cdot 7$, so $21^k = 3^k \cdot 7^k$. For $21^k$ to divide $M = 3^{100} \cdot 7^{50}$, we need $3^k$ to divide $3^{100}$ and $7^k$ to divide $7^{50}$. This means we need $k \leq 100$ and $k \leq 50$ simultaneously. The limiting factor is the smaller exponent. Since $k$ must satisfy both conditions, the largest possible value is $k = 50$. We can verify: $21^{50} = 3^{50} \cdot 7^{50}$ divides $3^{100} \cdot 7^{50}$ because $50 \leq 100$ and $50 \leq 50$. However, $21^{51} = 3^{51} \cdot 7^{51}$ would not divide $M$ because $7^{51}$ cannot divide $7^{50}$. Looking at the wrong answers: (A) 25 is too conservative—we can go higher since both prime powers allow it. (B) 33 falls into the same trap of not pushing to the true maximum. (D) 100 ignores the constraint from the $7^{50}$ factor; while $3^{100}$ could support $k = 100$, the $7^{50}$ term limits us to $k = 50$. The answer is C) 50. Strategy tip: In divisibility problems involving composite numbers, always break down to prime factors first, then find the limiting exponent—the smallest one determines your maximum.
10. If $K = 15^4 \times 8^2 \times 5^3$, how many trailing zeros does the integer K have when written in standard form?
    Answer: 6 The number of trailing zeros in an integer is determined by the number of times 10 is a factor in its prime factorization. Since $10 = 2 \times 5$, we need to find the number of pairs of 2 and 5 in the prime factorization of K. First, express K in terms of its prime factors: $K = (3 \times 5)^4 \times (2^3)^2 \times 5^3$. Using the rules of exponents, this simplifies to: $K = (3^4 \times 5^4) \times (2^6) \times 5^3$. Now, combine the powers of the same bases: $K = 2^6 \times 3^4 \times 5^{(4+3)} = 2^6 \times 3^4 \times 5^7$. The number of factors of 2 is 6, and the number of factors of 5 is 7. The number of pairs of (2, 5) is limited by the smaller of these two exponents, which is 6. Therefore, K has 6 trailing zeros.
11. The product $P = 1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot 99$ is the product of all odd integers from 1 to 99. How many factors of 10 are in $P$?
    Answer: 0 When you see a question asking about factors of 10 in a product, remember that each factor of 10 requires one factor of 2 and one factor of 5. So you need to count how many pairs of 2s and 5s are available in the prime factorization. Let's examine what we're working with: $P = 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot \ldots \cdot 99$. This product contains only odd numbers. Since every even number is divisible by 2, and we're excluding all even numbers, our product $P$ contains no factors of 2 whatsoever. Even though $P$ contains odd multiples of 5 (like 5, 15, 25, 35, 45, 55, 65, 75, 85, 95), and some of these contribute multiple factors of 5 (for example, 25 = 5²), we cannot form even a single factor of 10 because we have zero factors of 2 to pair with them. Looking at the wrong answers: (B) 12 and (C) 24 likely come from students who counted only the factors of 5 in the product, forgetting that factors of 10 require both 2s and 5s. (D) 49 might result from counting all odd numbers from 1 to 99, which is indeed 50 numbers, but this completely misses the concept of what creates factors of 10. The answer is (A) 0. Key strategy: For factor-of-10 problems, always check for both prime factors (2 and 5). The number of factors of 10 equals the minimum of the count of 2s and count of 5s in the prime factorization. If either is zero, you get zero factors of 10.
12. A student counts by $45$s: $45,\;90,\;135,\;180,\ldots$. What is the 8th number in this pattern?
    Answer: 360 When you encounter a counting pattern question, you're working with an arithmetic sequence where each term increases by the same amount. Here, the student counts by 45s, meaning each number is 45 more than the previous one. To find the 8th term in this sequence, you can use the arithmetic sequence formula: $a_n = a_1 + (n-1)d$, where $a_1$ is the first term, $n$ is the position, and $d$ is the common difference. With $a_1 = 45$, $n = 8$, and $d = 45$: $a_8 = 45 + (8-1) \times 45 = 45 + 7 \times 45 = 45 + 315 = 360$ Alternatively, since we're counting by 45s, the 8th number is simply $8 \times 45 = 360$. This works because the sequence starts at 45 (which is $1 \times 45$). Choice A (315) represents a common error where students calculate $(n-1) \times d$ but forget to add the first term: $7 \times 45 = 315$. Choice C (405) is what you'd get if you incorrectly used $n = 9$ instead of $n = 8$: $9 \times 45 = 405$. Choice D (450) results from using $n = 10$: $10 \times 45 = 450$. For arithmetic sequences that start with the common difference (like counting by a number), remember that the $n$th term equals $n$ times that difference. This shortcut saves time and reduces calculation errors on sequence problems.
13. If $p$ is the least common multiple of $24$ and $30$, and $q$ is their greatest common factor, what is $\dfrac{p}{q}$?
    Answer: 20 When you encounter problems involving least common multiple (LCM) and greatest common factor (GCF), start by finding the prime factorizations of both numbers. For 24: $24 = 2^3 \times 3^1$ For 30: $30 = 2^1 \times 3^1 \times 5^1$ To find the LCM, take the highest power of each prime factor that appears: $p = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$ To find the GCF, take the lowest power of each common prime factor: $q = 2^1 \times 3^1 = 6$ Therefore, $\frac{p}{q} = \frac{120}{6} = 20$, which is answer choice C. Let's examine why the other answers are incorrect: A) 4 might result from confusing the relationship between factors or making calculation errors with the prime factorizations. B) 10 could come from incorrectly finding either the LCM or GCF. For instance, you might get this if you mistakenly calculated the GCF as 12 instead of 6. D) 40 is a common trap answer that appears when students mix up the LCM and GCF calculations or make arithmetic errors in the division. Remember this key relationship: for any two numbers, their product equals their LCM times their GCF. You can use this as a check: $24 \times 30 = 720$ and $120 \times 6 = 720$ ✓. This relationship can also provide an alternative solution method when you're short on time.
14. What is the greatest common factor (GCF) of $84$ and $126$?
    Answer: 42 When you encounter a greatest common factor (GCF) problem, you're looking for the largest number that divides evenly into both given numbers. There are two main approaches: listing factors or using prime factorization. Let's use prime factorization, which is typically faster for larger numbers. First, break down each number into its prime factors: $84 = 4 \times 21 = 4 \times 3 \times 7 = 2^2 \times 3 \times 7$ $126 = 2 \times 63 = 2 \times 9 \times 7 = 2 \times 3^2 \times 7$ To find the GCF, take the lowest power of each common prime factor: - Both numbers contain $2$: take $2^1$ (the lower power) - Both numbers contain $3$: take $3^1$ (the lower power) - Both numbers contain $7$: take $7^1$ Therefore: $\text{GCF} = 2^1 \times 3^1 \times 7^1 = 2 \times 3 \times 7 = 42$ Looking at the wrong answers: (A) 14 represents $2 \times 7$, missing the factor of 3 that both numbers share. (B) 21 represents $3 \times 7$, missing the factor of 2. (D) 63 represents $3^2 \times 7$, but since 84 only has $3^1$, this cannot divide evenly into 84. Study tip: Always double-check your GCF by verifying it divides evenly into both original numbers. Here: $84 ÷ 42 = 2$ and $126 ÷ 42 = 3$, confirming that 42 is correct.
15. The least common multiple (LCM) of $15$ and $18$ is:
    Answer: 90 When you encounter LCM questions, you're finding the smallest positive number that both given numbers divide into evenly. There are two reliable methods: listing multiples or using prime factorization. Let's use prime factorization for $15$ and $18$: - $15 = 3 \times 5$ - $18 = 2 \times 3^2$ To find the LCM, take the highest power of each prime factor that appears: $2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90$. You can verify this by checking that $90 ÷ 15 = 6$ and $90 ÷ 18 = 5$, confirming both divide evenly. Looking at the wrong answers: Choice (A) $45$ only works for $15$ (since $45 ÷ 15 = 3$) but fails for $18$ because $45 ÷ 18 = 2.5$. Choice (B) $60$ works for $15$ ($60 ÷ 15 = 4$) but not for $18$ ($60 ÷ 18 = 3.33...$). Choice (D) $180$ is a common multiple of both numbers ($180 ÷ 15 = 12$ and $180 ÷ 18 = 10$), but it's not the least common multiple—it's actually $2 \times 90$. Remember that the LCM will always be greater than or equal to the larger of your two numbers, but you're looking for the smallest such number. When prime factorization seems complex, you can also list multiples systematically: $15, 30, 45, 60, 75, 90...$ and $18, 36, 54, 72, 90...$ until you find the first match.
16. The sum of the digits of a two-digit number is 12. If the digits are reversed, the new number is 18 less than the original number. What is the original number?
    Answer: 75 This type of problem combines digit manipulation with algebraic thinking. When you see questions about reversing digits and comparing the results, set up equations using the place values of the digits. Let's call the original two-digit number $10a + b$, where $a$ is the tens digit and $b$ is the units digit. The reversed number would be $10b + a$. From the given information, you can write two equations: First, the sum of digits equals 12: $a + b = 12$ Second, the reversed number is 18 less than the original: $10a + b - (10b + a) = 18$ Simplifying the second equation: $10a + b - 10b - a = 18$, which becomes $9a - 9b = 18$, or $a - b = 2$. Now solve the system: $a + b = 12$ and $a - b = 2$. Adding these equations gives $2a = 14$, so $a = 7$. Substituting back: $b = 5$. The original number is 75. Let's verify: 75 has digits summing to 12 ✓, and 75 - 57 = 18 ✓. Choice A (57) has the digits in reverse order - this would be the "new" number after reversing, not the original. Choice B (66) has digits summing to 12, but reversing gives the same number, so the difference would be 0, not 18. Choice D (84) has digits summing to 12, but 84 - 48 = 36, not 18. When solving digit reversal problems, always define your variables clearly using place value notation, set up both constraint equations, and verify your answer by checking both conditions.
17. $5! = 120$. How many factors of $120$ are even?
    Answer: 12 When you encounter a question about finding factors with specific properties, you need to use prime factorization to systematically count them. First, find the prime factorization of 120. Since $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$, we can write: $120 = 2^3 \times 3^1 \times 5^1$ To find all factors of 120, you combine the prime factors in every possible way. The total number of factors is $(3+1)(1+1)(1+1) = 16$. For a factor to be even, it must contain at least one factor of 2. The even factors are those that include $2^1$, $2^2$, or $2^3$ in their prime factorization. Count the even factors: For each even factor, you must choose at least one power of 2 (3 choices: $2^1$, $2^2$, or $2^3$), any power of 3 (2 choices: $3^0$ or $3^1$), and any power of 5 (2 choices: $5^0$ or $5^1$). This gives us $3 \times 2 \times 2 = 12$ even factors. Answer choice A) 8 likely comes from miscounting or using an incorrect method. Answer choice B) 10 might result from confusion about which combinations are valid. Answer choice D) 14 could come from incorrectly subtracting only 2 odd factors from the total 16, when there are actually 4 odd factors. Remember: To count factors with specific properties, use prime factorization and systematically consider all valid combinations of the prime powers.
18. The greatest common divisor of two positive integers is 15. Their least common multiple is 300. If one of the integers is 60, what is the other integer?
    Answer: 75 When you encounter problems involving both the greatest common divisor (GCD) and least common multiple (LCM) of two numbers, remember this fundamental relationship: For any two positive integers, $\text{GCD} \times \text{LCM} = \text{product of the two numbers}$. Let's call the unknown integer $x$. We know that $\text{GCD}(60, x) = 15$ and $\text{LCM}(60, x) = 300$. Using our formula: $15 \times 300 = 60 \times x$ $4500 = 60x$ $x = 75$ We can verify this works: The factors of 60 are related to $2^2 \times 3 \times 5$, and the factors of 75 are $3 \times 5^2$. Their GCD is indeed $3 \times 5 = 15$, and their LCM is $2^2 \times 3 \times 5^2 = 300$. Let's check why the other answers fail. Choice (A) 45: If we apply our formula, $15 \times 300 = 60 \times 45$ gives us $4500 = 2700$, which is false. Choice (C) 90: Similarly, $15 \times 300 = 60 \times 90$ gives us $4500 = 5400$, which doesn't work. Choice (D) 120: Here $15 \times 300 = 60 \times 120$ gives us $4500 = 7200$, also incorrect. The answer is (B) 75. Study tip: Memorize the GCD-LCM relationship formula. It's your fastest tool for these problems, and you can always verify your answer by checking that the GCD and LCM values actually work with your calculated number.
19. Which of the following numbers is a multiple of $14$ but not a multiple of $6$?
    Answer: 56 This question tests your understanding of multiples and how to check divisibility by composite numbers. When determining if a number is a multiple of another number, you need to check if the division results in a whole number with no remainder. Let's check each option systematically. For a number to be a multiple of 14, it must be divisible by 14. For it to NOT be a multiple of 6, division by 6 must leave a remainder. Choice A: $56 ÷ 14 = 4$ (exact), so 56 is a multiple of 14. $56 ÷ 6 = 9.33...$ (not exact), so 56 is NOT a multiple of 6. This satisfies both conditions. Choice B: $84 ÷ 14 = 6$ (exact), so 84 is a multiple of 14. However, $84 ÷ 6 = 14$ (exact), so 84 IS also a multiple of 6. This fails the "not a multiple of 6" requirement. Choice C: $108 ÷ 14 = 7.71...$ (not exact), so 108 is NOT a multiple of 14. This immediately eliminates this choice since it fails the first condition. Choice D: $126 ÷ 14 = 9$ (exact), so 126 is a multiple of 14. However, $126 ÷ 6 = 21$ (exact), so 126 IS also a multiple of 6. Like choice B, this fails the second condition. Only choice A meets both criteria: it's a multiple of 14 but not a multiple of 6. Strategy tip: For "multiple but not multiple" questions, always check both conditions systematically. Many wrong answers will satisfy one condition but not both, making this a good way to eliminate choices quickly.
20. What is the greatest three-digit multiple of $8$?
    Answer: 992 When you're looking for the greatest three-digit multiple of a number, you need to work backwards from the largest possible three-digit number, which is 999. To find the greatest three-digit multiple of 8, divide 999 by 8: $999 ÷ 8 = 124.875$. Since we need a whole number of groups of 8, we round down to 124. This means the greatest three-digit multiple of 8 is $124 × 8 = 992$. You can verify this by checking that 992 is divisible by 8: $992 ÷ 8 = 124$ exactly, with no remainder. Looking at the wrong answers: Choice B (996) might seem tempting since it's closer to 999, but $996 ÷ 8 = 124.5$, which isn't a whole number, so 996 isn't a multiple of 8. Choice C (988) is indeed a multiple of 8 ($988 ÷ 8 = 123.5$... wait, that's not right either — let me recalculate: $988 ÷ 8 = 123.5$, so this isn't a multiple of 8). Actually, $984 ÷ 8 = 123$, so choice D (984) is a multiple of 8, but it's smaller than 992. The correct answer is A (992). Study tip: When finding the greatest multiple of a number within a range, always divide the upper limit by that number, round down to the nearest whole number, then multiply back. This systematic approach prevents you from getting distracted by answer choices that look close but aren't actually multiples.

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