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Prime factorization and exponent notation

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5100346
Which is the correct prime factorization of \(720\)? a) \(8\times9\times10\) b) \(2\times5\times7\times11\) c) \(2^3\times3^2\times5\) d) \(2^4\times3^2\times5\)

Hints

- In a prime factorization, every factor must be prime. - Check whether every number in each choice is prime. - You can also evaluate each choice and compare it with \(720\).

Solution

1. Decompose \(720\) as \(72\times10\). 2. Factor each part into primes: \(72=8\times9=2\times2\times2\times3\times3\), and \(10=2\times5\). 3. Combine repeated prime factors: \(2\times2\times2\times2\times3\times3\times5=2^4\times3^2\times5\).

Answer

d) \(2^4\times3^2\times5\)
5223836
1) Find the prime factorization of each number and write it as a power of one prime number: \(81\), \(125\), and \(64\). 2) Find the prime factorizations of \(200\) and \(450\). Use exponents for repeated prime factors.

Hints

- Factor each number until every factor is prime. - How can repeated equal prime factors be written using an exponent? - Begin with the smallest prime numbers.

Solution

1. Factor each prime power: \(81=3\times3\times3\times3=3^4\), \(125=5\times5\times5=5^3\), and \(64=2\times2\times2\times2\times2\times2=2^6\). 2. Factor the remaining numbers: \(200=2\times2\times2\times5\times5=2^3\times5^2\), and \(450=2\times3\times3\times5\times5=2\times3^2\times5^2\).

Answer

1) \(81=3^4\); \(125=5^3\); \(64=2^6\) 2) \(200=2^3\times5^2\); \(450=2\times3^2\times5^2\)
5118726
Find the prime factorization of each number and write it using exponents, such as \(2^2\times3\): \(60,126,\) and \(270\). Which prime factors are common to all three numbers?

Hints

- Recall the definition of a prime number. - Decompose each number until every factor is prime. - Count how many times each prime factor appears. - Compare the three lists of prime factors.

Solution

1. Factor \(60\): \(60=2\times30=2\times2\times15=2^2\times3\times5\). 2. Factor \(126\): \(126=2\times63=2\times3\times21=2\times3^2\times7\). 3. Factor \(270\): \(270=2\times135=2\times3\times45=2\times3^2\times15=2\times3^3\times5\). 4. The prime factors \(2\) and \(3\) appear in all three factorizations.

Answer

\(60=2^2\times3\times5\) \(126=2\times3^2\times7\) \(270=2\times3^3\times5\) The common prime factors are \(2\) and \(3\).
5118736
A number \(n\) has prime factorization \(2^2\times3\times7\). a) What is the value of \(n\)? b) What is the prime factorization of the number that is exactly five times \(n\)? c) Explain without calculating why a number greater than \(5\) that ends in \(5\) can never be prime.

Hints

- What happens to a prime factorization when the number is multiplied by another prime number? - Which ones digits indicate divisibility by \(5\)? - How many positive factors does a prime number have?

Solution

1. Evaluate the factorization: \(n=2^2\times3\times7=4\times3\times7=84\). 2. Multiplying \(n\) by \(5\) adds one factor of \(5\). The new prime factorization is \(2^2\times3\times5\times7\). 3. Every number ending in \(5\) is divisible by \(5\). If the number is greater than \(5\), then it has at least the positive factors \(1\), \(5\), and itself, so it is not prime.

Answer

a) \(n=84\) b) \(2^2\times3\times5\times7\) c) A number greater than \(5\) that ends in \(5\) is divisible by \(5\), so it has more than two positive factors and is not prime.
5119016
Find a natural number \(x\) that satisfies all three conditions. 1. \(30<x<50\) 2. \(x\) is a multiple of \(3\). 3. \(x\) has exactly \(6\) positive factors. Use prime factorizations to justify your choice. Give the number and list all its factors.

Hints

- First list the multiples of \(3\) in the interval. - Write the prime factorization of each candidate. - Use the exponents in a prime factorization to determine the number of positive factors.

Solution

1. The multiples of \(3\) between \(30\) and \(50\) are \(33,36,39,42,45,48\). 2. Use prime factorizations: \(33=3\cdot11\) and \(39=3\cdot13\) each have \(4\) factors; \(36=2^2\cdot3^2\) has \(9\) factors; \(42=2\cdot3\cdot7\) has \(8\) factors; \(45=3^2\cdot5\) has \(6\) factors; and \(48=2^4\cdot3\) has \(10\) factors. 3. The factors of \(45\) are \(1,3,5,9,15,45\), so \(x=45\).

Answer

\(x=45\); its factors are \(1,3,5,9,15,45\).
5197976
Consider the number \(130\). a) Find the prime factorization of \(130\). b) Use the prime factorization to list all positive factors of \(130\). c) How many positive factors does \(130\) have?

Hints

- Continue factoring until every factor is prime. - Each positive factor can be made by multiplying some of the prime factors. - Remember to include \(1\) and the number itself. - Organize the products systematically so that none are missed.

Solution

1. Factor the number: \(130=2\times65=2\times5\times13\). 2. Form every possible product using each prime factor either once or not at all: \(1,2,5,13,2\times5,2\times13,5\times13,\) and \(2\times5\times13\). 3. In numerical order, the positive factors are \(1,2,5,10,13,26,65,\) and \(130\). Therefore, \(130\) has \(8\) positive factors.

Answer

a) \(130=2\times5\times13\) b) \(1,2,5,10,13,26,65,130\) c) \(8\) positive factors
5197996
Compare the numbers of positive factors of two products of primes. a) How many positive factors does \(6=2\times3\) have? b) How many positive factors does \(9=3\times3\) have? c) Explain why the numbers of factors are different even though both numbers are products of two prime factors.

Hints

- List all positive factors of each number. - Compare the prime factors in the two factorizations. - How is multiplying two equal primes different from multiplying two distinct primes?

Solution

1. The positive factors of \(6\) are \(1,2,3,\) and \(6\), so \(6\) has \(4\) positive factors. 2. The positive factors of \(9\) are \(1,3,\) and \(9\), so \(9\) has \(3\) positive factors. 3. The prime factors of \(6\) are distinct, so each creates a different factor. In \(9=3^2\), the two prime factors are the same, so they do not create two different one-prime factors.

Answer

a) \(4\) positive factors: \(1,2,3,6\) b) \(3\) positive factors: \(1,3,9\) c) The number \(6\) has two distinct prime factors, while \(9\) repeats the same prime factor. Repetition produces fewer distinct products.
5203446
a) Find the prime factorization of \(72\). b) Use the prime factorization to find all positive factors of \(72\) that are between \(10\) and \(30\).

Hints

- Continue factoring until every factor is prime. - Positive factors are made by multiplying allowable numbers of the prime factors. - Try different combinations of the prime factors. - Check systematically which products are between \(10\) and \(30\).

Solution

1. Factor repeatedly: \(72=2\times36=2\times2\times18=2\times2\times2\times9=2^3\times3^2\). 2. Form factor products in the requested interval: \(2^2\times3=12\), \(2\times3^2=18\), and \(2^3\times3=24\). 3. Other nearby products, such as \(2^3=8\), \(3^2=9\), and \(2^2\times3^2=36\), are outside the interval. Therefore, the requested factors are \(12,18,\) and \(24\).

Answer

a) \(72=2^3\times3^2\) b) \(12,18,24\)
5203456
The prime factorization of \(150\) is \(2\times3\times5^2\). Decide whether each statement is true or false. Briefly justify each answer using the prime factorization. a) \(15\) is a factor of \(150\). b) \(20\) is a factor of \(150\). c) \(25\) is a factor of \(150\).

Hints

- What must be true about the prime factors of a factor of a number? - Find the prime factorizations of \(15\), \(20\), and \(25\). - Check whether \(150\) contains every required prime factor enough times.

Solution

1. Since \(15=3\times5\), and \(150\) contains both required prime factors, \(15\) is a factor of \(150\). 2. Since \(20=2^2\times5\), it requires two factors of \(2\). The factorization of \(150\) contains only one factor of \(2\), so \(20\) is not a factor. 3. Since \(25=5^2\), and \(150\) contains two factors of \(5\), \(25\) is a factor.

Answer

a) True, because \(15=3\times5\), and both prime factors occur in \(150\). b) False, because \(20=2^2\times5\), but \(150\) contains only one factor of \(2\). c) True, because \(25=5^2\), and \(150\) contains two factors of \(5\).
5203466
A number \(n\) has prime factorization \(2^2\times3\times11\). a) Find the value of \(n\). b) Give one factor of \(n\) that is between \(10\) and \(20\). c) Explain using the prime factorization why \(9\) is not a factor of \(n\).

Hints

- Multiply the factors in stages to find \(n\). - Which individual prime factors or products of factors are between \(10\) and \(20\)? - Find the prime factorization of \(9\). - Compare how many factors of \(3\) are required with how many occur in \(n\).

Solution

1. Evaluate the factorization: \(n=2^2\times3\times11=4\times3\times11=132\). 2. The prime factor \(11\) is between \(10\) and \(20\). Another possible factor is \(2^2\times3=12\). 3. Since \(9=3^2\), a number divisible by \(9\) must contain at least two factors of \(3\). The factorization of \(n\) contains only one factor of \(3\), so \(9\) is not a factor of \(n\).

Answer

a) \(n=132\) b) Possible answers include \(11\) and \(12\). c) Since \(9=3^2\) but \(n\) contains only one factor of \(3\), \(9\) is not a factor of \(n\).
5102186
A fraction in simplest form can be rewritten with a power of \(10\) as its denominator only when the denominator has no prime factors other than \(2\) and \(5\). a) Explain why \(\frac{1}{8}\) can be rewritten with denominator \(1000\), but not with denominator \(100\). b) Find the smallest power of \(10\) that can be used as a denominator for an equivalent fraction to \(\frac{7}{40}\). c) Give one denominator from \(10\) through \(15\) for which a fraction in simplest form cannot be rewritten with a power of \(10\) as its denominator. Explain.

Hints

- Write each denominator as a product of prime factors. - A power of \(10\) has equal numbers of factors of \(2\) and \(5\). - Look for a denominator containing a prime factor other than \(2\) or \(5\).

Solution

1. For a), \(8=2^3\). Since \(100=2^2\times5^2\), \(8\) is not a factor of \(100\). Since \(1000=2^3\times5^3\), \(8\) is a factor of \(1000\). 2. For b), \(40=2^3\times5\). The smallest power of \(10\) containing at least three factors of \(2\) and one factor of \(5\) is \(10^3=1000\). 3. For c), one answer is \(12\), because \(12=2^2\times3\) contains a factor of \(3\), and powers of \(10\) contain only factors of \(2\) and \(5\).

Answer

a) \(8\) is a factor of \(1000\), but not of \(100\). b) \(1000\) c) For example, \(12\), because it contains the prime factor \(3\).
5102196
Consider fractions that can be rewritten with denominator \(1000\). a) Find all possible denominators \(n\le100\) that divide \(1000\). b) A fraction with denominator \(m\) was multiplied by \(\frac{8}{8}\) to produce denominator \(1000\). Find \(m\). Give three different proper fractions with denominator \(m\) that are already in simplest form.

Hints

- Use the prime factorization \(1000=2^3\times5^3\). - Combine powers of \(2\) and \(5\) without exceeding \(100\). - For part b), divide \(1000\) by \(8\), then choose numerators relatively prime to the denominator.

Solution

1. Since \(1000=2^3\times5^3\), its positive factors no greater than \(100\) are \(1,2,4,5,8,10,20,25,40,50,\) and \(100\). 2. For b), \(8m=1000\), so \(m=1000\div8=125\). 3. Since \(125=5^3\), a numerator not divisible by \(5\) gives a fraction in simplest form. Examples are \(\frac{1}{125},\frac{2}{125},\) and \(\frac{3}{125}\).

Answer

a) \(1,2,4,5,8,10,20,25,40,50,100\) b) \(m=125\); examples: \(\frac{1}{125},\frac{2}{125},\frac{3}{125}\)
5118746
Two numbers are given by their prime factorizations: \(A=2^2\times3\times5\) and \(B=2\times3^2\times5\). a) Find the values of \(A\) and \(B\). b) Find the prime factorization of \(A\times B\). c) What is the least prime number that is not a factor of either \(A\) or \(B\)? d) Is \(A+B\) divisible by \(5\)? Justify your answer using the prime factorizations without first finding the sum.

Hints

- How can you combine two products of prime factors into one product? - Check the prime numbers in order: \(2,3,5,7,11,\ldots\) - If two numbers share a factor, what does that tell you about their sum?

Solution

1. Evaluate each factorization: \(A=4\times3\times5=60\) and \(B=2\times9\times5=90\). 2. Combine like prime factors in the product: \(A\times B=(2^2\times3\times5)(2\times3^2\times5)=2^{2+1}\times3^{1+2}\times5^{1+1}=2^3\times3^3\times5^2\). 3. The primes \(2\), \(3\), and \(5\) occur in at least one of the factorizations. The next prime, \(7\), occurs in neither. 4. Both \(A\) and \(B\) contain a factor of \(5\), so both are multiples of \(5\). The sum of two multiples of \(5\) is also divisible by \(5\).

Answer

a) \(A=60\), \(B=90\) b) \(2^3\times3^3\times5^2\) c) \(7\) d) Yes. Both numbers contain a factor of \(5\), so their sum is divisible by \(5\).
5197696
Solve the number riddle. My number is less than \(100\). It is divisible by both \(4\) and \(9\). It has exactly nine positive factors. What is my number?

Hints

- Which numbers less than \(100\) are divisible by both \(4\) and \(9\)? - First find the least common multiple, then list its multiples in the range. - Use the prime factorization of each candidate to determine its number of positive factors. - A number with an odd number of positive factors is a perfect square.

Solution

1. Since \(4=2^2\) and \(9=3^2\), their least common multiple is \(2^2\times3^2=36\). 2. The positive multiples of \(36\) less than \(100\) are \(36\) and \(72\). 3. The prime factorization \(36=2^2\times3^2\) gives \((2+1)(2+1)=9\) positive factors. 4. The prime factorization \(72=2^3\times3^2\) gives \((3+1)(2+1)=12\) positive factors. 5. Therefore, the only number satisfying all conditions is \(36\).

Answer

\(36\)
5197986
Suppose \(n\) is the product of four distinct prime numbers \(a\), \(b\), \(c\), and \(d\), so \(n=abcd\). a) Use the letters to list all positive factors of \(n\). b) Find the total number of positive factors of \(n\). c) Check your result by listing the positive factors of \(210=2\times3\times5\times7\).

Hints

- Organize the factors by how many of the four primes are used. - How many ways can you choose none, one, two, three, or all four letters? - A tree diagram or table can help you find every combination.

Solution

1. Group the factors by the number of prime factors used. Using none gives \(1\). Using one gives \(a,b,c,d\). Using two gives \(ab,ac,ad,bc,bd,cd\). Using three gives \(abc,abd,acd,bcd\). Using all four gives \(abcd\). 2. The number of factors is \(1+4+6+4+1=16\). 3. For \(210=2\times3\times5\times7\), the positive factors are \(1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,\) and \(210\). This list has \(16\) factors, confirming the result.

Answer

a) \(1\); \(a,b,c,d\); \(ab,ac,ad,bc,bd,cd\); \(abc,abd,acd,bcd\); \(abcd\) b) \(16\) positive factors c) \(1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,210\)

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