HSPT Math Quiz: Apply Number Concepts

What this quiz covers

This quiz focuses on Apply Number Concepts, giving you a quick way to practice the rules, question types, and explanations that matter most for HSPT 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.

Question 1

1. 15
2. 16
3. 17 (correct answer)
4. 18

Explanation: When you encounter questions asking for digits in decimal representations of fractions, you need to perform long division to convert the fraction to its decimal form. To find the decimal representation of $\frac{5}{12}$, divide 5 by 12 using long division. Since 5 is smaller than 12, you'll get 0 before the decimal point and need to work with 50 ÷ 12. 50 ÷ 12 = 4 remainder 2, so the first digit after the decimal is 4. Bring down a 0 to get 20 ÷ 12 = 1 remainder 8, so the second digit is 1. Bring down a 0 to get 80 ÷ 12 = 6 remainder 8, so the third digit is 6. Bring down a 0 to get 80 ÷ 12 = 6 remainder 8, so the fourth digit is 6. Therefore, $\frac{5}{12} = 0.4166...$ and the first four digits after the decimal point are 4, 1, 6, and 6. The sum is 4 + 1 + 6 + 6 = 17, making C correct. Choice A (15) might result from miscounting digits or making arithmetic errors during division. Choice B (16) could come from adding incorrectly: 4 + 1 + 6 + 6 = 17, but getting 16 suggests dropping one from the sum. Choice D (18) might occur if you mistakenly included an extra digit or made an error in the long division process. Remember that converting fractions to decimals through long division requires careful tracking of remainders at each step. Double-check your division and addition to avoid simple computational errors.

Question 2

1. 14
2. 15 (correct answer)
3. 16
4. 17

Explanation: When you see expressions of the form $10^n - 1$, you're dealing with a pattern that creates numbers made entirely of 9's. Understanding this pattern is crucial for place value and powers of 10 problems. Let's explore what $10^n - 1$ produces. When $n = 1$: $10^1 - 1 = 10 - 1 = 9$ (one digit that's 9). When $n = 2$: $10^2 - 1 = 100 - 1 = 99$ (two digits that are 9's). When $n = 3$: $10^3 - 1 = 1000 - 1 = 999$ (three digits that are 9's). The pattern is clear: $10^n - 1$ always produces exactly $n$ digits, and every digit is 9. Since the problem states that $10^n - 1$ has exactly 15 digits that are 9's, we need $n = 15$. This gives us $10^{15} - 1 = 999,999,999,999,999$ (fifteen 9's). Looking at the wrong answers: (A) 14 would give us $10^{14} - 1$, which has only 14 digits of 9's, not 15. (C) 16 would produce $10^{16} - 1$, which has 16 digits of 9's, exceeding what we need. (D) 17 would create $10^{17} - 1$ with 17 digits of 9's, also too many. The answer is (B) 15. Study tip: Remember that $10^n - 1$ always creates exactly $n$ consecutive 9's. This direct relationship between the exponent and the number of 9's is a reliable pattern you can use on similar problems.

Question 3

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

Explanation: When you see a question asking how many zeros a number ends with, you're being asked to find trailing zeros - which come from factors of 10. Since $10 = 2 \times 5$, each trailing zero requires one factor of 2 and one factor of 5 in the prime factorization. Let's examine $2^4 \cdot 3^2 \cdot 5^3$. To find trailing zeros, count how many complete pairs of 2 and 5 you can make. You have four factors of 2 and three factors of 5. Since you need both a 2 and a 5 to make each factor of 10, you're limited by whichever you have fewer of. Here, you have only three factors of 5, so you can make exactly three factors of 10, giving you three trailing zeros. To verify: $2^4 \cdot 3^2 \cdot 5^3 = 16 \cdot 9 \cdot 125 = 18,000$, which indeed ends in three zeros. Looking at the wrong answers: Choice B (4) incorrectly counts the total number of factors of 2, ignoring that you need matching factors of 5. Choice C (5) might come from adding the exponents 2 and 3, which has no mathematical basis for this problem. Choice D (2) could result from miscounting the factors or making an arithmetic error. The correct answer is A. Key strategy: For trailing zeros, always take the minimum of the powers of 2 and 5 in the prime factorization. The factor that appears less frequently determines how many complete factors of 10 you can form.

Question 4

1. 25
2. 33
3. 50 (correct answer)
4. 100

Explanation: 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.

Question 5

1. 4
2. 6 (correct answer)
3. 7
4. 9

Explanation: 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.

Question 6

1. 45
2. 60
3. 90 (correct answer)
4. 180

Explanation: 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 $18$:

• $15 = 3 \times 5$
• $18 = 2 \times 3^2$

Question 7

1. 2,000,000
2. 200,000 (correct answer)
3. 20,000
4. 2,000

Explanation: When you encounter a question about place value, you need to identify which position a digit occupies in a number and understand what that position represents in our base-ten number system. In the number $4,205,687$, let's identify the place value of each digit by working from right to left. The rightmost digit (7) is in the ones place, then 8 is in the tens place, 6 is in the hundreds place, 5 is in the thousands place, 0 is in the ten-thousands place, 2 is in the hundred-thousands place, and 4 is in the millions place. The digit 2 sits in the hundred-thousands place, which means its value is $2 \times 100,000 = 200,000$. This makes choice B correct. Let's examine why the other answers are wrong. Choice A (2,000,000) represents the millions place value - this would be correct if the 2 were in the leftmost position where the 4 currently sits. Choice C (20,000) represents the ten-thousands place value, which is one position to the right of where our 2 actually appears. Choice D (2,000) represents the thousands place value, which is two positions to the right of the 2's actual location. For place value questions, always count positions carefully from right to left, and remember the pattern: ones, tens, hundreds, thousands, ten-thousands, hundred-thousands, millions. Using commas as guides helps too - the 2 appears right after the first comma from the left, confirming it's in the hundred-thousands place.

Question 8

1. 30 (correct answer)
2. 18
3. 28
4. 45

Explanation: When you encounter questions about prime factors, you need to break each number down into its prime factorization—expressing it as a product of prime numbers only. The question asks for exactly three different prime factors, so you're looking for three distinct primes that multiply together (possibly with repetition) to form the number. Let's examine each option systematically: Choice A (30): $30 = 2 \times 3 \times 5$. This gives us exactly three different prime factors: 2, 3, and 5. This matches what we're looking for. Choice B (18): $18 = 2 \times 3^2 = 2 \times 3 \times 3$. While 3 appears twice, there are only two different prime factors: 2 and 3. This doesn't meet our requirement. Choice C (28): $28 = 4 \times 7 = 2^2 \times 7 = 2 \times 2 \times 7$. This has only two different prime factors: 2 and 7. The factor 2 appears twice, but we only count distinct primes. Choice D (45): $45 = 9 \times 5 = 3^2 \times 5 = 3 \times 3 \times 5$. This has only two different prime factors: 3 and 5. Only choice A has exactly three different prime factors, making it the correct answer. Study tip: When counting prime factors, always distinguish between "different prime factors" and "total prime factors." Write out the complete prime factorization first, then count the unique primes. Don't be fooled by repeated factors—focus on how many distinct primes appear in the factorization.

Question 9

1. $2 \times 3 \times 33$
2. $2 \times 3^{2} \times 11$ (correct answer)
3. $2^{2} \times 3 \times 11$
4. $2 \times 9 \times 11$

Explanation: 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.

Question 10

1. 56 (correct answer)
2. 84
3. 108
4. 126

Explanation: 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.

Question 11

1. 42 and 51
2. 45 and 54 (correct answer)
3. 48 and 57
4. 54 and 72

Explanation: 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.

Question 12

1. 992 (correct answer)
2. 996
3. 988
4. 984

Explanation: 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.

Question 13

1. 4
2. 10
3. 20 (correct answer)
4. 40

Explanation: 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.

Question 14

1. 9:36 AM
2. 9:42 AM
3. 10:12 AM (correct answer)
4. 10:24 AM

Explanation: When you see a problem asking when two events that occur at regular intervals will happen together again, you're dealing with a least common multiple (LCM) problem. The key insight is that both bells will ring together at time intervals equal to the LCM of their individual periods. To find when the bells ring together again, you need the LCM of 18 and 24 minutes. Start by finding the prime factorization of each number: $18 = 2 \times 3^2$ and $24 = 2^3 \times 3$. The LCM uses the highest power of each prime factor: $LCM = 2^3 \times 3^2 = 8 \times 9 = 72$ minutes. Since both bells rang together at 9:00 AM, they'll ring together again 72 minutes later. Converting 72 minutes to hours and minutes: $72 = 60 + 12 = 1 \text{ hour } 12 \text{ minutes}$. Adding this to 9:00 AM gives 10:12 AM, which is answer choice C. Choice A (9:36 AM) represents 36 minutes after 9:00 AM, which is double the school bell's period but not a multiple of 24. Choice B (9:42 AM) is 42 minutes later, which isn't a multiple of either 18 or 24. Choice D (10:24 AM) represents 84 minutes, which while close to 72, doesn't work since neither bell rings at 84-minute intervals from the start time. For LCM problems on the HSPT, always find the least common multiple of the given periods, then add that time interval to your starting point. Don't get trapped by answer choices that are simple multiples of just one of the periods.

Question 15

1. 2
2. 3 (correct answer)
3. 5
4. 9

Explanation: When you encounter a divisibility problem like this, you need to recall the divisibility rule for 9: a number is divisible by 9 if and only if the sum of its digits is divisible by 9. For the three-digit number $8\square7$, you need to find which digit makes the sum $8 + \square + 7$ divisible by 9. This means you need $15 + \square$ to be divisible by 9. Since $15 = 9 + 6$, we know that 15 leaves a remainder of 6 when divided by 9. To make the total sum divisible by 9, you need to add enough to eliminate this remainder of 6. The smallest positive number that does this is 3, because $15 + 3 = 18$, and 18 is divisible by 9 ($18 ÷ 9 = 2$). Let's check why each wrong answer fails: Choice A gives us $8 + 2 + 7 = 17$, and $17 ÷ 9 = 1$ remainder $8$, so 817 is not divisible by 9. Choice C gives us $8 + 5 + 7 = 20$, and $20 ÷ 9 = 2$ remainder $2$, so 857 is not divisible by 9. Choice D gives us $8 + 9 + 7 = 24$, and while 24 is divisible by 9, choice B (3) is the correct answer since $8 + 3 + 7 = 18$ is also divisible by 9, and B appears before D in the options. Remember: for divisibility by 9, always check if the sum of digits is a multiple of 9. This rule works for any size number and is much faster than doing long division.

Question 16

1. 6
2. 10
3. 15 (correct answer)
4. 20

Explanation: When you encounter questions about digit values in large numbers, you're working with place value concepts. Each digit's value depends on its position, and comparing these values often reveals important relationships. In the number 7,164,059, you need to identify where each digit appears. The digit 6 is in the ten-thousands place, giving it a value of $6 \times 10,000 = 60,000$. The digit 4 is in the thousands place, giving it a value of $4 \times 1,000 = 4,000$. To find how many times larger 60,000 is than 4,000, divide: $\frac{60,000}{4,000} = 15$. So the value of digit 6 is 15 times the value of digit 4. Choice (A) 6 incorrectly assumes you're comparing just the face values of the digits themselves (6 ÷ 4 = 1.5, which doesn't even match). Choice (B) 10 represents the common misconception that adjacent place values always differ by a factor of 10 – while place values themselves differ by 10, the actual digit values depend on what digits occupy those places. Choice (D) 20 might result from miscounting place values or calculation errors. Remember that place value questions require two steps: first identify each digit's actual value by considering its position, then perform the requested comparison. Don't get distracted by the digits' face values – a 2 in the hundreds place (200) has greater value than a 9 in the tens place (90). Always multiply each digit by its place value first.

Question 17

1. 40
2. 56
3. 64
4. 80 (correct answer)

Explanation: This problem combines two conditions: divisibility by 8 and digit sum equals 8. When you encounter multi-condition problems like this, work systematically through possibilities, starting with the smallest candidates. To find the answer, you need numbers whose digits add to 8, then check which ones are divisible by 8. Start small and work up: 17 (digits: 1+7=8, but 17÷8=2.125), 26 (digits: 2+6=8, but 26÷8=3.25), 35 (digits: 3+5=8, but 35÷8=4.375), 44 (digits: 4+4=8, but 44÷8=5.5), 53 (digits: 5+3=8, but 53÷8=6.625), 62 (digits: 6+2=8, but 62÷8=7.75), 71 (digits: 7+1=8, but 71÷8=8.875), 80 (digits: 8+0=8, and 80÷8=10 exactly). So 80 is the smallest positive integer satisfying both conditions. Looking at the choices: A) 40 has digits that sum to 4+0=4, not 8. B) 56 has digits that sum to 5+6=11, not 8. C) 64 has digits that sum to 6+4=10, not 8. While B and C are divisible by 8, they fail the digit sum requirement. D) 80 satisfies both conditions: 8+0=8 and 80÷8=10. Remember that "smallest" problems require checking all possibilities up to your answer. Don't just verify that answer choices work—make sure no smaller number works first. Always double-check both conditions when dealing with multi-requirement problems.

Question 18

1. 225
2. 300
3. 425 (correct answer)
4. 450

Explanation: 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.

Question 19

1. 7.6832 (correct answer)
2. 7.8632
3. 7.6823
4. 7.6283

Explanation: 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)

Question 20

1. 8
2. 10
3. 12 (correct answer)
4. 14

Explanation: 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.