Free math worksheets

- Identify the prime bases shared by both factorizations. - For each shared prime, compare the two exponents. - Build the greatest factor that does not use more copies than either number has.

1. The shared prime bases are \(2\) and \(3\). 2. For \(2\), the smaller exponent is \(2\); for \(3\), both numbers contain one factor of \(3\). 3. Therefore, \(\operatorname{GCF}(84,96)=2^2\times3=12\). 4. A common factor cannot contain more copies of a prime than either original number contains.

\(\operatorname{GCF}(84,96)=12\). The smaller exponent is used because the factor must divide both numbers.

- First decide whether the situation asks for the largest shared divisor or the first shared multiple. - Prime-factor both numbers in each part. - For a GCF use shared primes with the smaller exponents; for an LCM use every needed prime with the larger exponent.

1. For a), the greatest equal piece length must divide both ribbon lengths, so use the GCF. Prime-factor: \(84=2^2\times3\times7\) and \(126=2\times3^2\times7\). The GCF is \(2\times3\times7=42\), so each piece is \(42\,\text{cm}\). 2. For b), the next shared flashing time must be the least common multiple. Prime-factor: \(18=2\times3^2\) and \(24=2^3\times3\). The LCM is \(2^3\times3^2=72\), so the lights next flash together after \(72\) seconds.

a) GCF; \(42\,\text{cm}\) b) LCM; \(72\) seconds

- Compare the prime factors that appear in both numbers. - For each shared prime, use the smaller exponent. - Multiply the selected prime powers to get the GCF.

1. Write the prime factorizations: \(48=2^4\times3\) and \(72=2^3\times3^2\). 2. A common factor can use only primes found in both numbers and no exponent larger than the smaller exponent for that prime. 3. Use \(2^3\) and \(3^1\): \(2^3\times3=8\times3=24\).

\(24\)

- List every prime that occurs in either factorization. - For each prime, use the larger exponent that appears. - Multiply the chosen prime powers.

1. Write the prime factorizations: \(18=2\times3^2\) and \(30=2\times3\times5\). 2. The LCM must contain every prime power needed by either number. 3. Use \(2\), \(3^2\), and \(5\): \(2\times3^2\times5=90\).

\(90\)

- List the positive factors of each number and identify the common ones. - For the LCM, compare the prime factorizations and use each prime with the greatest exponent needed. - Find the product and compare it with the LCM.

1. The positive factors of \(24\) are \(1,2,3,4,6,8,12,24\). The positive factors of \(36\) are \(1,2,3,4,6,9,12,18,36\). 2. The common factors greater than \(1\) are \(2,3,4,6,\) and \(12\). 3. Prime factorize the numbers: \(24=2^3\times3\) and \(36=2^2\times3^2\). 4. Use the greatest exponent of each prime: \(\operatorname{LCM}(24,36)=2^3\times3^2=72\). 5. The product is \(24\times36=864\). Since \(72<864\), the LCM is less than the product.

a) \(2,3,4,6,12\) b) \(\operatorname{LCM}(24,36)=72\) c) Yes. \(24\times36=864\), and \(72<864\).

- Start from the prime factorization of the required LCM. - A valid number cannot contain a prime or exponent not present in that LCM. - For each pair, compare the greatest exponent of each prime across the two factorizations.

1. Factor the target LCM: \(12=2^2\times3\). 2. Any number in a valid pair must use only these prime factors and cannot require a higher exponent. The possible values below \(10\) that divide \(12\) are \(1,2,3,4,6\). 3. To obtain \(2^2\times3\) as the LCM, a pair must supply the exponent \(2\) on prime \(2\) and the exponent \(1\) on prime \(3\). 4. The pairs \((3,4)\) and \((4,6)\) meet those exponent requirements. No other pair supplies both required greatest exponents without changing the LCM.

\(\{3,4\}\) and \(\{4,6\}\). In each pair, the greatest prime exponents combine to \(2^2\times3=12\).

- Write both denominators as products of primes. - For the LCM, keep the greatest exponent needed for each prime. - Compare that exponent rule with what happens when the two factorizations share a prime.

1. Prime-factor the denominators: \(6=2\times3\) and \(10=2\times5\). 2. The LCM uses the greatest needed exponent of each prime: \(2\times3\times5=30\). Thus the least common denominator is \(30\). 3. The product \(60\) contains the shared factor \(2\) twice, while the LCM needs it only once. 4. The product equals the LCM when the two denominators share no prime factor, that is, when they are relatively prime.

a) \(30\) b) \(6=2\times3\) and \(10=2\times5\) share the prime factor \(2\), so their product repeats it. c) Their product is the LCM when the denominators share no prime factor.

- Prime-factor both numbers in each pair before finding the LCM. - Compare the prime factors: is any prime present in both numbers of a pair? - When there is no shared prime factor, what factors must the LCM contain?

1. For \((3,4)\), \(3=3\) and \(4=2^2\). They share no prime factor, so the LCM uses \(2^2\times3=12=3\times4\). 2. For \((4,5)\), \(4=2^2\) and \(5=5\). They share no prime factor, so the LCM is \(2^2\times5=20=4\times5\). 3. For \((7,8)\), \(7=7\) and \(8=2^3\). They share no prime factor, so the LCM is \(2^3\times7=56=7\times8\). 4. These examples suggest that consecutive positive whole numbers are relatively prime, so their LCM is their product.

a) \(3=3\), \(4=2^2\), so \(\operatorname{LCM}(3,4)=2^2\times3=12\). b) \(4=2^2\), \(5=5\), so \(\operatorname{LCM}(4,5)=2^2\times5=20\). c) \(7=7\), \(8=2^3\), so \(\operatorname{LCM}(7,8)=2^3\times7=56\). Pattern: each pair has no shared prime factor, so in these examples the LCM equals the product of the two consecutive numbers.

- Write each denominator as a product of prime factors. - Use the greatest exponent of each prime to form the least common multiple. - Rewrite both fractions with the resulting denominator and compare the numerators.

1. Factor the denominators: \(24=2^3\times3\) and \(36=2^2\times3^2\). 2. The least common denominator is \(2^3\times3^2=72\). 3. Rewrite the fractions: \(\frac{11}{24}=\frac{33}{72}\) and \(\frac{13}{36}=\frac{26}{72}\). 4. Since \(33>26\), \(\frac{11}{24}>\frac{13}{36}\).

\(\frac{11}{24}>\frac{13}{36}\)

- How can the prime factorizations of the denominators help you find their LCM? - To make an equivalent fraction, multiply the numerator and denominator by the same number. - Once the denominators match, what do you do with the numerators when adding?

1. For a), factor the denominators: \(15 = 3 \times 5\) and \(12 = 2^2 \times 3\). The LCM is \(2^2 \times 3 \times 5 = 60\). 2. For b), rewrite each fraction with denominator \(60\): \(\frac{7}{15} = \frac{28}{60}\) and \(\frac{5}{12} = \frac{25}{60}\). 3. For c), add: \(\frac{28}{60} + \frac{25}{60} = \frac{53}{60}\). Since \(53\) and \(60\) have no common factor greater than \(1\), the fraction is already in simplest form.

a) \(60\) b) \(\frac{28}{60}\) and \(\frac{25}{60}\) c) \(\frac{53}{60}\)

- First find \(12 \times 15\) for part a). - For part b), compare the prime factorizations of \(12\) and \(15\) to build their LCM. - Compare the equivalent fractions you worked with in the two methods.

1. For a), use \(12 \times 15 = 180\) as the common denominator: \(\frac{5}{12} = \frac{75}{180}\) and \(\frac{7}{15} = \frac{84}{180}\). Then \(\frac{75}{180} + \frac{84}{180} = \frac{159}{180} = \frac{53}{60}\). 2. For b), factor the denominators: \(12 = 2^2 \times 3\) and \(15 = 3 \times 5\). Therefore \(\operatorname{LCM}(12,15) = 2^2 \times 3 \times 5 = 60\). Then \(\frac{5}{12} = \frac{25}{60}\) and \(\frac{7}{15} = \frac{28}{60}\), so the sum is \(\frac{53}{60}\). 3. For c), using the LCM keeps the equivalent fractions and arithmetic smaller. In this case it also produces the simplified answer directly.

a) \(\frac{53}{60}\), using \(\frac{159}{180}\) before simplifying. b) \(\frac{53}{60}\), using the common denominator \(60\). c) Using the LCM keeps the numbers smaller and can reduce the amount of simplifying needed.

- Compare the prime factorizations of \(2\), \(3\), and \(4\). - After choosing a common denominator, determine the equivalent numerator for each fraction. - Compare the size of the numbers used with denominator \(12\) and denominator \(24\).

1. Factor the denominators: \(2=2\), \(3=3\), and \(4=2^2\). The LCM is \(2^2 \times 3 = 12\), which is smaller than \(24\). 2. Rewrite the fractions with denominator \(12\): \(\frac{1}{2}=\frac{6}{12}\), \(\frac{1}{3}=\frac{4}{12}\), and \(\frac{1}{4}=\frac{3}{12}\). 3. Add: \(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}\). 4. Using the LCM gives smaller equivalent fractions, so the arithmetic is simpler and there is less unnecessary simplification afterward.

The least common denominator is \(12\), and \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\). Using \(12\) keeps the numbers smaller than using \(24\), making the calculation simpler.

- Find the prime factorizations of \(4\) and \(6\). - Which prime factors, with what exponents, must a common multiple contain? - After finding the least common multiple, describe all of its positive multiples. - Check that each number in your pattern is divisible by both original numbers.

1. The prime factorizations are \(4=2^2\) and \(6=2\times3\). 2. A common multiple must contain at least two factors of \(2\) and one factor of \(3\). The least such number is \(2^2\times3=12\). 3. Every positive multiple of \(12\) is divisible by both \(4\) and \(6\), and every number divisible by both must be a multiple of \(12\). 4. Therefore, the requested numbers are all positive multiples of \(12\): \(12,24,36,48,\ldots\).

All positive multiples of \(12\): \(12,24,36,48,\ldots\)

- Begin with the prime factorizations of the given numbers. - For each prime, which is the greatest exponent required by any number in the list? - Build the least common multiple from those prime powers. - Check that your result is divisible by every given number.

1. Write the prime factorizations: \(2=2\), \(3=3\), \(4=2^2\), \(5=5\), \(8=2^3\), and \(10=2\times5\). 2. A number divisible by all six values must contain the greatest required power of each prime: \(2^3\), \(3\), and \(5\). 3. Therefore, the least common multiple is \(2^3\times3\times5=120\). 4. The number \(120\) is divisible by every number in the list, and removing any required prime factor would make at least one divisibility condition fail.

\(120\)

- Find the prime factorization of each pair of numbers below an empty brick. - Use every required prime factor with its greatest exponent. - If one number is a multiple of the other, the greater number is their LCM.

1. In the first row above the base, \(\operatorname{LCM}(2,3)=6\), \(\operatorname{LCM}(3,4)=12\), and \(\operatorname{LCM}(4,5)=20\). 2. In the next row, \(\operatorname{LCM}(6,12)=12\) and \(\operatorname{LCM}(12,20)=60\). 3. At the top, \(\operatorname{LCM}(12,60)=60\).

First missing row: \(6,12,20\) Second missing row: \(12,60\) Top: \(60\)

- Compare the prime factors of the two numbers below each empty brick. - The LCM of two distinct prime numbers is their product. - For the top brick, determine the prime factors required by both \(6\) and \(15\).

1. The left brick in the middle row is \(\operatorname{LCM}(2,3)=2\times3=6\). 2. The right brick in the middle row is \(\operatorname{LCM}(3,5)=3\times5=15\). 3. Since \(6=2\times3\) and \(15=3\times5\), the top brick is \(\operatorname{LCM}(6,15)=2\times3\times5=30\).

Middle row: \(6,15\) Top: \(30\)

- Find the prime factorization of each number below an empty brick. - Use each prime factor with the greatest exponent required. - If one number is already a multiple of the other, it is their LCM.

1. Since \(4=2^2\) and \(6=2\times3\), \(\operatorname{LCM}(4,6)=2^2\times3=12\). 2. Since \(6=2\times3\) and \(8=2^3\), \(\operatorname{LCM}(6,8)=2^3\times3=24\). 3. Because \(24\) is a multiple of \(12\), \(\operatorname{LCM}(12,24)=24\).

Middle row: \(12,24\) Top: \(24\)

- Find the prime factorization of each pair below an empty brick. - Work one complete row at a time from bottom to top. - For each pair, use every required prime factor with its greatest exponent.

1. In the first row above the base, \(\operatorname{LCM}(4,6)=12\), \(\operatorname{LCM}(6,3)=6\), and \(\operatorname{LCM}(3,2)=6\). 2. In the next row, \(\operatorname{LCM}(12,6)=12\) and \(\operatorname{LCM}(6,6)=6\). 3. At the top, \(\operatorname{LCM}(12,6)=12\).

First missing row: \(12,6,6\) Second missing row: \(12,6\) Top: \(12\)

- List the prime factors of each number. - Keep only the prime copies present in both lists. - Divide each original number by the common factor to complete the final products.

1. Factor: \(54=2\times3^3\) and \(78=2\times3\times13\). 2. The shared prime factors are \(2\times3=6\). 3. Thus, \(54=6\times9\) and \(78=6\times13\).

The greatest common factor is \(6\); \(54=6\times9\) and \(78=6\times13\).

- Compare the prime factors of the two numbers. - Notice whether any prime appears in both factorizations. - Build the smallest product containing every required factor.

1. Factor the numbers: \(7\) is prime and \(12=2^2\times3\). 2. They share no prime factor, so the least common multiple contains all factors from both: \(2^2\times3\times7\). 3. This equals \(84\), which is also \(7\times12\) because the numbers have no common prime factor.

\(84\)

- The piece length must divide both roll lengths exactly. - Use the shared prime factors to find the greatest possible common length. - Divide each original length by the chosen piece length.

1. Factor the lengths: \(56=2^3\times7\) and \(98=2\times7^2\). 2. Their greatest common factor is \(2\times7=14\), so each piece can be \(14\,\text{cm}\) long. 3. The rolls make \(56\div14=4\) pieces and \(98\div14=7\) pieces.

Each piece is \(14\,\text{cm}\) long. The rolls make \(4\) pieces and \(7\) pieces.

- Decide whether each requested number must divide the originals or be divisible by them. - Compare the copies of \(2\) and \(3\) in the two factorizations. - Build the smallest product that still contains both original factorizations.

1. The only shared prime factor is one factor of \(2\), so \(\operatorname{GCF}(6,8)=2\). 2. A common multiple must contain \(2^3\) to build \(8\) and a factor of \(3\) to build \(6\). 3. Therefore, \(\operatorname{LCM}(6,8)=2^3\times3=24\). 4. The GCF divides both numbers, while the LCM is divisible by both numbers.

a) \(\operatorname{GCF}(6,8)=2\) b) \(\operatorname{LCM}(6,8)=24\) c) The GCF keeps only prime copies shared by both numbers; the LCM includes every prime copy required by either number.

- Decide whether the answer must divide the intervals or be divisible by them. - Factor both intervals into prime powers. - Build the smallest time containing every required prime power.

1. A future time shared by both schedules must be a common multiple, so use the least common multiple. 2. Factor: \(8=2^3\) and \(12=2^2\times3\). 3. The least common multiple is \(2^3\times3=24\), so both buses will next arrive together in \(24\) minutes.

Use the least common multiple. The buses will next arrive together in \(24\) minutes.

- Which numbers from \(10\) through \(30\) could have a GCF of \(6\)? - Both numbers must be multiples of \(6\). - After choosing a pair, use prime factorizations to find its LCM and check that it is less than \(100\). - Also verify that the GCF is not greater than \(6\).

1. Both numbers must be multiples of \(6\). The multiples of \(6\) from \(10\) through \(30\) are \(12,18,24,\) and \(30\). 2. Choose \((12, 18)\). Their prime factorizations are \(12=2^2\times3\) and \(18=2\times3^2\). 3. Using the least exponent of each shared prime, \(\operatorname{GCF}(12,18)=2\times3=6\). 4. Using the greatest exponent of each prime, \(\operatorname{LCM}(12,18)=2^2\times3^2=36\). 5. Both numbers are in the required interval, the GCF is \(6\), and \(36<100\), so the pair satisfies all three conditions.

One possible pair is \((12, 18)\). Since \(12=2^2\times3\) and \(18=2\times3^2\), their GCF is \(2\times3=6\) and their LCM is \(2^2\times3^2=36<100\).

- Recall the equation connecting the product of two numbers with their GCF and LCM. - Use the given product and GCF to solve directly for the LCM. - For the pair, both numbers must be multiples of \(3\). - Check your proposed pair by comparing its prime factorizations.

1. For two positive integers \(a\) and \(b\), \(a\times b=\operatorname{GCF}(a,b)\times\operatorname{LCM}(a,b)\). 2. Substitute the given values: \(180=3\times\operatorname{LCM}(a,b)\), so \(\operatorname{LCM}(a,b)=180\div3=60\). 3. One possible pair is \((12, 15)\). Their prime factorizations are \(12=2^2\times3\) and \(15=3\times5\). 4. Their shared prime factor is \(3\), so \(\operatorname{GCF}(12,15)=3\). Using the greatest exponent of each prime gives \(\operatorname{LCM}(12,15)=2^2\times3\times5=60\). Also, \(12\times15=180\).

a) \(\operatorname{LCM}=60\) b) One possible pair is \((12, 15)\). Its GCF is \(3\), its LCM is \(60\), and its product is \(180\).

- Factor each denominator into primes. - Build the least common multiple using the highest power of each prime. - Rewrite all fractions with that denominator and compare their numerators.

1. Factor the denominators: \(60=2^2\times3\times5\), \(45=3^2\times5\), and \(75=3\times5^2\). 2. The least common denominator is \(2^2\times3^2\times5^2=900\). 3. Rewrite the fractions: \(\frac{19}{60}=\frac{285}{900}\), \(\frac{13}{45}=\frac{260}{900}\), and \(\frac{23}{75}=\frac{276}{900}\). 4. Since \(260<276<285\), the order is \(\frac{13}{45}<\frac{23}{75}<\frac{19}{60}\).

\(\frac{13}{45}<\frac{23}{75}<\frac{19}{60}\)

- Write each number as a product of primes before comparing product and LCM. - The LCM keeps the greatest exponent needed for each prime. - Ask what changes in that exponent rule when the same prime occurs in both numbers.

1. For a), \(3=3\) and \(5=5\). The factorizations share no prime, so the LCM contains both factors: \(3\times5=15\), equal to the product. 2. For b), \(4=2^2\) and \(6=2\times3\). The LCM uses the greatest exponents, \(2^2\times3=12\), while the product \(24\) counts the shared prime \(2\) from both numbers. 3. In general, if the factorizations share no prime, taking the greatest exponent for every prime is the same as multiplying the numbers. If they share a prime, the product repeats some prime power that the LCM needs only at its greatest exponent. 4. Therefore, Ben’s claim is true.

Ben is correct. For \(3\) and \(5\), \(\operatorname{LCM}(3,5)=15=3\times5\). For \(4=2^2\) and \(6=2\times3\), \(\operatorname{LCM}(4,6)=2^2\times3=12<24\). The product equals the LCM exactly when no prime factor is shared.

- Which values could combine with \(3\) to have an LCM of \(6\)? - Check which of those values also combines with \(5\) to have an LCM of \(10\).

1. Let \(x\) be the missing base value. The left brick requires \(\operatorname{LCM}(3,x)=6\). Therefore, \(x\) can supply a factor of \(2\) but cannot require any prime factor beyond those in \(6=2\times3\). 2. The right brick requires \(\operatorname{LCM}(x,5)=10=2\times5\). Therefore, \(x\) must supply the factor of \(2\) and cannot contain a factor of \(3\). 3. The only positive whole number satisfying both conditions is \(x=2\). 4. The top brick is \(\operatorname{LCM}(6,10)=2\times3\times5=30\).

Missing base value: \(2\) Top value: \(30\)

- Compare the prime factors of \(4\) and \(12\). What is the least missing factor? - Repeat the process for \(4\) and \(20\). - More than one base value may work, so be sure to choose the least value greater than \(1\).

1. Let \(x\) be the left outside brick. Since \(\operatorname{LCM}(x,4)=12\), and \(4=2^2\) while \(12=2^2\times3\), the least possible value greater than \(1\) that supplies the missing factor of \(3\) is \(x=3\). 2. Let \(y\) be the right outside brick. Since \(\operatorname{LCM}(4,y)=20\), and \(20=2^2\times5\), the least possible value greater than \(1\) that supplies the missing factor of \(5\) is \(y=5\). 3. The top brick is \(\operatorname{LCM}(12,20)\). Since \(12=2^2\times3\) and \(20=2^2\times5\), the LCM is \(2^2\times3\times5=60\).

Left outside brick: \(3\) Right outside brick: \(5\) Top: \(60\)

- Prime-factor the fixed number and each candidate. - Compare only the prime copies shared with the fixed number. - Check that the shared product is exactly the required value, not merely a multiple of it.

1. Factor \(48=2^4\times3\). 2. The candidates factor as \(18=2\times3^2\), \(36=2^2\times3^2\), \(40=2^3\times5\), and \(72=2^3\times3^2\). 3. Only \(36\) shares exactly \(2^2\times3=12\) as its greatest common factor with \(48\).

\(36\)

- Compare the prime factors of the starting number and the required multiple. - Eliminate a choice if it introduces a prime absent from the target. - Eliminate a choice if it requires a larger prime power than the target contains.

1. Write \(8=2^3\) and \(24=2^3\times3\). 2. The choices \(5\) and \(7\) introduce prime factors not in \(24\), and \(9=3^2\) requires a larger power of \(3\) than \(24\) contains. 3. The choice \(12=2^2\times3\) adds the needed factor \(3\) without requiring any larger prime power. 4. Therefore, \(\operatorname{LCM}(8,12)=2^3\times3=24\).

\(n=12\)

- Use the smaller shared exponents for one result and the larger required exponents for the other. - Calculate each result from the prime factorizations. - Compare the two requested products only after finding both values.

1. Factor: \(9=3^2\) and \(12=2^2\times3\). 2. The greatest common factor is \(3\), and the least common multiple is \(2^2\times3^2=36\). 3. Their product is \(3\times36=108\), which equals \(9\times12=108\).

\(\operatorname{GCF}(9,12)=3\), \(\operatorname{LCM}(9,12)=36\), and \(3\times36=9\times12=108\).

- Express the target greatest common factor in prime-power form. - Compare each target exponent with the corresponding exponents in the two numbers. - Remember that a greatest common factor uses the smaller exponent.

1. Factor the target: \(24=2^3\times3\). 2. The shared power of \(2\) is already \(2^3\). For the shared power of \(3\) to be exactly \(3^1\), the smaller of \(2\) and \(k\) must be \(1\). 3. Therefore, \(k=1\).

\(k=1\)