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    how do u find the lowest common multiple of standard form numbers. I'm doing edexcel higher tier mathematics and am in my first year and this worksheet has caused me and my friend some issues!! please help
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    (Original post by neharajput)
    how do u find the lowest common multiple of standard form numbers. I'm doing edexcel higher tier mathematics and am in my first year and this worksheet has caused me and my friend some issues!! please help
    It would help if you posted the worksheet or a few of the questions.
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    a=2^10x5^9
    b=9,000,000
    c=2.4x10^9
    work out the lowest common multiple of a,b and c
    give your answer in standard form
    no calculator
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    (Original post by neharajput)
    a=2^10x5^9
    b=9,000,000
    c=2.4x10^9
    work out the lowest common multiple of a,b and c
    give your answer in standard form
    no calculator
    The easiest option is probably to express them all as products of prime factors.

    a is already in that form.

    b=2^6 \times 3^2 \times 5^6

    Can you express c as a product of prime factors?

    c=2.4 \times 10^9 = 24 \times 10^8 = ...

    Once you have them all as products of prime factors do you know how to continue?
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    so c would be 2^3 x 3...
    how would you express 10^8 as a product in its prime factors...
    thank you so much for your help, it has been much appreciated and could u let me know what you get as your final answer after you find the hcf and multiply it by the remaining prime factors, as I would be grateful to compare answers. These are grade 9 questions by the way in the new GCSE spec.
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    If you have a power of a composite number such as 10^8 then this is
    (2 x 5)^8 = 2^8 x 5^8.

    LCM is 36 000 000 000 if I remember correctly.
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    I got 2^6x5^6 as my HCF of all the numbers, however, multiplying them together did not give me the answer you got Do you mind writing the steps to get to that, as it would be much appreciated
    (Original post by BuryMathsTutor)
    If you have a power of a composite number such as 10^8 then this is
    (2 x 5)^8 = 2^8 x 5^8.

    LCM is 36 000 000 000 if I remember correctly.
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    Yes, that is the correct HCF. You said you wanted to find the LCM. I'm on my phone now so perhaps someone else will explain.
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    (Original post by BuryMathsTutor)
    Yes, that is the correct HCF. You said you wanted to find the LCM. I'm on my phone now so perhaps someone else will explain.
    Ok thankyou anyways
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    (Original post by neharajput)
    Ok thankyou anyways
    I'll do an example, and then I'd like you to have a go and post your working if you have problems.

    First off factorise each number into its prime factors and their exponents.

    E.g. 24=2^3\times 3 and 90=2\times 3^2\times 5


    To find the HCF:

    For each prime, find the minimum exponent across all numbers.

    Here the primes are 2,3, and 5
    Exponents of 2 are 3 and 1; minimum is 1
    Exponents of 3 are 1 and 2; minimum is 1
    Exponents of 5 are 0 (as it's not a factor of the first number) and 1; minimum 0

    We can ignore minimums of zero as anything to the power 0 is 1.

    So, HCF is 2^13^1=6

    To find the LCM:

    Again for each prime, we find the maximum exponent across all numbers.

    Here the primes are 2,3, and 5
    Exponents of 2 are 3 and 1; maximum is 3
    Exponents of 3 are 1 and 2; maximum is 2
    Exponents of 5 are 0 and 1; maximum is 1

    So, LCM is 2^33^25^1=360
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    So when you the HCF do you times it by the remaining prime factors to get the lowest common multiple?I tried this and i do not get the right answr
    (Original post by ghostwalker)
    I'll do an example, and then I'd like you to have a go and post your working if you have problems.

    First off factorise each number into its prime factors and their exponents.

    E.g. 24=2^3\times 3 and 90=2\times 3^2\times 5


    To find the HCF:

    For each prime, find the minimum exponent across all numbers.

    Here the primes are 2,3, and 5
    Exponents of 2 are 3 and 1; minimum is 1
    Exponents of 3 are 1 and 2; minimum is 1
    Exponents of 5 are 0 (as it's not a factor of the first number) and 1; minimum 0

    We can ignore minimums of zero as anything to the power 0 is 1.

    So, HCF is 2^13^1=6

    To find the LCM:

    Again for each prime, we find the maximum exponent across all numbers.

    Here the primes are 2,3, and 5
    Exponents of 2 are 3 and 1; maximum is 3
    Exponents of 3 are 1 and 2; maximum is 2
    Exponents of 5 are 0 and 1; maximum is 1

    So, LCM is 2^33^25^1=360
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    (Original post by neharajput)
    So when you the HCF do you times it by the remaining prime factors to get the lowest common multiple?I tried this and i do not get the right answr
    No. Read my previous post, and follow it.
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    The HCF for my equation is too big, which is where I'm getting confused..
    (Original post by ghostwalker)
    No. Read my previous post, and follow it.
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    (Original post by neharajput)
    The HCF for my equation is too big, which is where I'm getting confused..
    Post some working then, and say precisely where/what the problem is. What do you mean "it's too big"?
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    (Original post by ghostwalker)
    Post some working then, and say precisely where/what the problem is. What do you mean "it's too big"?
    It's okay now but thank you anyway
 
 
 
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