lahuerasvv
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- Nov 30, 2008
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Can anybody help me. I need somebody to explain to me how do I fin the LCD of 24, 40, 48?
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Finding LCD of 24, 40, 48
- Thread starter lahuerasvv
- Start date
Re: Finding LCD
Assuming 24, 40 and 48 are denominators of fractions, simply find the smallest number that those three numbers divide into exactly. One way to do this is to list all the multiple of each number, then find the smallest multiple that is common to all your lists.
For instance, if you were looking for the least common multiples of 4, 6 and 8 you could do it this way.
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, etc.
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, etc.
8, 16, 24, 32, 40, 48, 56, 64, 72, etc.
lahuerasvv said:Can anybody help me. I need somebody to explain to me how do I fin the LCD of 24, 40, 48?
Assuming 24, 40 and 48 are denominators of fractions, simply find the smallest number that those three numbers divide into exactly. One way to do this is to list all the multiple of each number, then find the smallest multiple that is common to all your lists.
For instance, if you were looking for the least common multiples of 4, 6 and 8 you could do it this way.
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, etc.
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, etc.
8, 16, 24, 32, 40, 48, 56, 64, 72, etc.
Re: Finding LCD
Hello, lahuerasvv!
Find the prime factorization of each number.
. . \(\displaystyle \begin{array}{ccc} 24 &=& 2^3\cdot3 \\ 40 &=& 2^3\cdot5 \\ 48 &=& 2^4\cdot2 \end{array}\)
List one of each different prime factor: .\(\displaystyle 2,3,5\)
Apply the largest exponent for each factor.
. . The largest exponent on 2 is "4": .\(\displaystyle 2^4\)
. . The largest exponent on 3 is "1": .\(\displaystyle 3^1\)
. . The largest exponent on 5 is "1": .\(\displaystyle 5^1\)
Then multiply these factors.
\(\displaystyle \text{Therefore: }\;\text{LCM}(24,40,48) \;=\;2^4\cdot3^1\cdot5^1 \;=\;240\)
Hello, lahuerasvv!
Find the LCD of 24, 40, 48
Find the prime factorization of each number.
. . \(\displaystyle \begin{array}{ccc} 24 &=& 2^3\cdot3 \\ 40 &=& 2^3\cdot5 \\ 48 &=& 2^4\cdot2 \end{array}\)
List one of each different prime factor: .\(\displaystyle 2,3,5\)
Apply the largest exponent for each factor.
. . The largest exponent on 2 is "4": .\(\displaystyle 2^4\)
. . The largest exponent on 3 is "1": .\(\displaystyle 3^1\)
. . The largest exponent on 5 is "1": .\(\displaystyle 5^1\)
Then multiply these factors.
\(\displaystyle \text{Therefore: }\;\text{LCM}(24,40,48) \;=\;2^4\cdot3^1\cdot5^1 \;=\;240\)