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    Hi
    Not sure how to answer this question. Btw I know how to express something as a product of its prime factors but not sure what this question wants me to do, thanks.

    Given that {25x}2 — 1 factorises to give (5{x}-1)(5{x}+1), express 2499 as a product of its prime factors.
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    25x^2-1=(5x-1)(5x+1)

    Let x=10

    LHS: 25*10^2-1=2500-1=2499

    So 2499=(5*10-1)(5*10+1)=49*51=7*7*51=7*7*17*3
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    (Original post by _Xenon_)
    Hi
    Not sure how to answer this question. Btw I know how to express something as a product of its prime factors but not sure what this question wants me to do, thanks.

    Given that {25x}2 —1 factorises to give (5{x}-1)(5{x}+1), express 2499 as a product of its prime factors.
    Think about the information given and how it relates to 2499. The rest should follow.
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    (Original post by Math12345)
    25x^2-1=(5x-1)(5x+1)

    Let x=10

    LHS: 25*10^2-1=2500-1=2499

    So 2499=(5*10-1)(5*10+1)=49*51=7*7*51
    Ah can I let x = 20 if I want so can it be anything?
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    (Original post by _Xenon_)
    Ah can I let x = 20 if I want so can it be anything?
    If you let x=20 then you get 25*20^2-1=10000-1=9999 which is not what you wanted the prime factorisation for.
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    (Original post by Math12345)
    If you let x=20 then you get 25*20^2-1=10000-1 which is not what you wanted the prime factorisation for.
    Oh so you think of a number which equals 2500 when 25 is multiplied by it so in this case that would be 10 giving 2500 and when you subtract 1 you get 2499 as it says 25x^2-1 ....


    Is 51 a prime number?!
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    (Original post by _Xenon_)
    Oh so you think of a number which equals 2500 when 25 is multiplied by it so in this case that would be 10 giving 2500 and when you subtract 1 you get 2499 as it says 25x^2-1 ....


    Is 51 a prime number?!
    Well, you know that 25x^2 - 1= (5x-1)(5x+1) and you've already said x=10, so replace the x in the brackets with 10, what do you get? You will then need to further decompose these two numbers into their prime factors.

    (51 is not prime).
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    (Original post by _Xenon_)
    Oh so you think of a number which equals 2500 when 25 is multiplied by it so in this case that would be 10 giving 2500 and when you subtract 1 you get 2499 as it says 25x^2-1 ....


    Is 51 a prime number?!

    Yes sorry forgot to cancel down 51 into 17*3
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    (Original post by Zacken)
    Well, you know that 25x^2 - 1= (5x-1)(5x+1) and you've already said x=10, so replace the x in the brackets with 10, what do you get? You will then need to further decompose these two numbers into their prime factors.

    (51 is not prime).
    Oh yes I'll do that and I thought it wasn't, the dude above made a mistake I think.
    He wrote 49*51=7*7*51 but 51 isn't prime...
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    (Original post by _Xenon_)
    Oh yes I'll do that and I thought it wasn't, the dude above made a mistake I think.
    He wrote 49*51=7*7*51 but 51 isn't prime so he's wrong, right?
    Well, not wrong. But he hasn't answered the question properly, yes. He's edited his post to give the correct version now.
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    (Original post by Math12345)
    Yes sorry forgot to cancel down 51 into 17*3
    Ah it's OK, I think I understand now. What about 105^2 as a product of its prime factors?

    Do I have to square it to get 11025 then break that down by prime factor decomposition? That's a massive number is there a simpler way?
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    (Original post by Zacken)
    Well, not wrong. But he hasn't answered the question properly, yes. He's edited his post to give the correct version now.
    Oh yeah sorry Thanks Maths12345 and Zacken
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    (Original post by _Xenon_)
    Ah it's OK, I think I understand now. What about 105^2 as a product of its prime factors?

    Do I have to square it to get 11025 then break that down by prime factor decomposition? That's a massive number is there a simpler way?
    105^2=(105)*(105)

    105=5*21=3*5*7

    so 105^2=(3*5*7)*(3*5*7)=3^2*5^2*7^  2
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    (Original post by Math12345)
    105^2=(105)*(105)

    105=5*21=3*5*7

    so 105^2=(3*5*7)*(3*5*7)=3^2*5^2*7^  2
    Oh nice, I've never been taught the method which you're using but that looks like something that will be very useful for me to learn. Thanks for your help.
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    (Original post by Math12345)
    105^2=(105)*(105)

    105=5*21=3*5*7

    so 105^2=(3*5*7)*(3*5*7)=3^2*5^2*7^  2
    How many prime number do you recommend I memorise? I know 2,3,5,7,11,13,17 already should I learn more or id that enough
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    (Original post by _Xenon_)
    How many prime number do you recommend I memorise? I know 2,3,5,7,11,13,17 already should I learn more or id that enough
    Learn them up to 100. There are only 25 primes up to 100. I doubt you will need any more for GCSE.
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    (Original post by Math12345)
    Learn them up to 100. There are only 25 primes up to 100. I doubt you will need any more for GCSE.
    OK thanks...
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    (Original post by _Xenon_)

    Is 51 a prime number?!
    Quick way to check for small factors - the first two obvious ones are
    if it's an even number it's divisible by 2;
    if it ends in 5 or 0 it's a multiple of 5;

    But also a useful one to check for primality - if the sum of the three digits is divisible by 3, then the number is divisible by 3; in your case 51 gives 5+1 = 6, divisible by 3 so not prime
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    (Original post by _Xenon_)
    OK thanks...
    It's possible but unlikely that you will lose a mark in the exam for not knowing a large prime. And it would probably only be a single mark.

    I recommend that you don't spend any time learning larger prime numbers since you only have 3 weeks. Focus on other topics.
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    (Original post by Zacken)
    Well, not wrong. But he hasn't answered the question properly, yes. He's edited his post to give the correct version now.
    (Original post by Math12345)
    105^2=(105)*(105)

    105=5*21=3*5*7

    so 105^2=(3*5*7)*(3*5*7)=3^2*5^2*7^  2
    (Original post by Metrododo)
    Quick way to check for small factors - the first two obvious ones are
    if it's an even number it's divisible by 2;
    if it ends in 5 or 0 it's a multiple of 5;

    But also a useful one to check for primality - if the sum of the three digits is divisible by 3, then the number is divisible by 3; in your case 51 gives 5+1 = 6, divisible by 3 so not prime
    notnek
    I found a trick that gives you the ANSWER!
    On casio calculators if it asks for product of prime factors you just type in the number, for example 210 then press shift and fact and it gives the answer, in index form!! But they do sometimes want working out but still very good.
 
 
 
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