Free Math Worksheets

- 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 \(72<864\).

- Begin with the factors of \(12\) that are less than \(10\). - Check pairs by listing multiples or using prime factors. - Because the pairs are unordered, \(\{a, b\}\) and \(\{b, a\}\) count as the same pair.

1. Any number in a valid pair must be a factor of \(12\). The factors less than \(10\) are \(1, 2, 3, 4,\) and \(6\). 2. Check unordered pairs of these factors. 3. \(\operatorname{lcm}(3, 4)=12\), and \(\operatorname{lcm}(4, 6)=12\). 4. Every other pair has a least common multiple less than \(12\).

\(\{3, 4\}\) and \(\{4, 6\}\)

- List multiples of \(6\) and \(10\). - Compare the prime factorizations of the two denominators. - Consider what happens when the denominators share no factor greater than \(1\).

1. The least common multiple of \(6\) and \(10\) is \(30\). 2. The numbers \(6=2\times3\) and \(10=2\times5\) share a factor of \(2\). Their product includes that common factor twice, but the least common multiple needs it only once: \(2\times3\times5=30\). 3. The product of two denominators equals their least common multiple when the denominators are relatively prime.

a) \(30\) b) The denominators share the factor \(2\), so their product repeats a common factor. c) The product is the least common multiple when the denominators are relatively prime.

- 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: \(12\)

- 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 \(180\).

- Compare each least common multiple with the product of the two denominators. - Suppose a number divides both consecutive numbers. What else must it divide? - The difference between consecutive numbers is \(1\).

1. \(\operatorname{lcm}(3, 4)=12\), \(\operatorname{lcm}(4, 5)=20\), and \(\operatorname{lcm}(7, 8)=56\). 2. In each case, the least common multiple is the product of the two numbers. 3. If a number \(d\) divides both \(n\) and \(n+1\), then it divides their difference, \((n+1)-n=1\). Therefore, \(d=1\), so consecutive positive whole numbers are relatively prime. 4. The least common multiple of relatively prime numbers is their product, so the least common denominator is \(n(n+1)\).

a) \(12,20,56\) b) The least common denominator is the product of the consecutive numbers. c) Consecutive numbers have no common factor greater than \(1\), so their least common multiple is their product.

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

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

- What does it mean for two numbers to be relatively prime? - Try two small prime numbers as denominators and compare their product with their LCM. - Then try two denominators that share a factor. What changes in their prime factorizations?

1. If two denominators are relatively prime, their prime factorizations share no prime factor. The LCM therefore needs every prime factor from both denominators, so the LCM equals their product. 2. For example, \(3\) and \(5\) are relatively prime. Their product is \(3 \times 5 = 15\), and \(\operatorname{LCM}(3,5)=15\). 3. If the denominators share a prime factor, their product counts that shared factor from both numbers, while the LCM includes only the greatest needed power of each prime. Therefore the LCM is less than the product. 4. For example, \(4=2^2\) and \(6=2 \times 3\). Their product is \(24\), but \(\operatorname{LCM}(4,6)=2^2 \times 3=12\). Ben's claim is true.

Ben is correct. For relatively prime denominators, the LCM equals their product; for example, \(\operatorname{LCM}(3,5)=15=3 \times 5\). When the denominators share a factor, the LCM is smaller than their product; for example, \(\operatorname{LCM}(4,6)=12<24=4 \times 6\).

- 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: \(60\)