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    Can anyone help with the following:
    Q) look at each of the sequences below:
    a) 198,205,212, 219, 226....
    b) 196, 205, 214, 223, 232.....
    What is the first number after 205 that will appear in both sequences. Work this out without writing every term in the sequence.
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    How much about sequences do you know?

    Here's one way to start...
    The nth term of the first sequence is 7n + 191, and the nth term of the second sequence is 9n + 187. Therefore, if the n_1th term in the first sequence is equal to the n_2th term in the second sequence, then
    7n_1 + 191 = 9n_2 + 187
    \implies 7n_1 + 4 = 9n_2.

    Can you work out where to go from here?
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    7n1 + 4 must be a multiple of 9
    Trial and error for n1?
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    (Original post by zanyzoya)
    Can anyone help with the following:
    Q) look at each of the sequences below:
    a) 198,205,212, 219, 226....
    b) 196, 205, 214, 223, 232.....
    What is the first number after 205 that will appear in both sequences. Work this out without writing every term in the sequence.
    i'm not sure i can't find a way to do this but i did write out about 6 extra terms for each of the sequences and got the answer which was 268, now the question is, is there a way to do this mathematically?
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    Zacken Student403 TeeEm
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    (Original post by zanyzoya)
    7n1 + 4 must be a multiple of 9Trial and error for n1?
    You're on the right lines - you could use trial and error, but you should be able to spot a pattern if you try successive values of n_1, which will in fact enable you to work out all values of n_1 that work!
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    (Original post by zanyzoya)
    Can anyone help with the following:
    Q) look at each of the sequences below:
    a) 198,205,212, 219, 226....
    b) 196, 205, 214, 223, 232.....
    What is the first number after 205 that will appear in both sequences. Work this out without writing every term in the sequence.

    There isn't any point tagging TeeEm, by the way.

    We have our first sequence as 191 + 7n for n \in \mathbb{N} and our second as 187 + 9n.

    Then we want to find non-negative integers m and n such that 191 + 7n = 187 + 9m

    This reduces to 9m - 7n = 4 which is a standard linear diophantine equation since 9 and 7 are relatively prime.

    We can use Bezout's identity to find solutions to this since we already know that m=n=2 is a solution so that the other (infinitely many) solutions are given by:

    \displaystyle 

\begin{equation*}\begin{cases} m =  2+ 7k \\ n = 2 + 9k \end{cases}\end{equation*}

    For natural k - so the next solution is given by taking m = 9, n = 11.
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    (Original post by Zacken)
    There isn't any point tagging TeeEm, by the way.

    We have our first sequence as 191 + 7n for n \in \mathbb{N} and our second as 187 + 9n.

    Then we want to find non-negative integers m and n such that 191 + 7n = 187 + 9m

    This reduces to 9m - 7n = 4 which is a standard linear diophantine equation since 9 and 7 are relatively prime.

    We can use Bezout's identity to find solutions to this since we already know that m=n=2 is a solution so that the other (infinitely many) solutions are given by:

    \displaystyle 

\begin{equation*}\begin{cases} m =  2+ 7k \\ n = 2 + 9k \end{cases}\end{equation*}

    For natural k - so the next solution is given by taking m = 9, n = 11.
    how come he left? ;(

    also this stuff i swear when i was doing some research on this stuff i saw all that ^^^^^^^ and i died from complicatedness xD
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    (Original post by Zacken)
    There isn't any point tagging TeeEm, by the way.

    We have our first sequence as 191 + 7n for n \in \mathbb{N} and our second as 187 + 9n.

    Then we want to find non-negative integers m and n such that 191 + 7n = 187 + 9m

    This reduces to 9m - 7n = 4 which is a standard linear diophantine equation since 9 and 7 are relatively prime.

    We can use Bezout's identity to find solutions to this since we already know that m=n=2 is a solution so that the other (infinitely many) solutions are given by:

    \displaystyle 

\begin{equation*}\begin{cases} m =  2+ 7k \\ n = 2 + 9k \end{cases}\end{equation*}

    For natural k - so the next solution is given by taking m = 9, n = 11.
    Fair enough - I did it in a more wordy way...

    Focus on the right-hand side of 7n_1 + 4 = 9n_2. You're essentially looking for a multiple of 9 which is 4 more than a multiple of 7. We already know that the first time this happens is when n_2 = 2. If you work it out (e.g. by noticing that 9 is 2 more than 7, and so each successive multiple of 9 will leave a remainder that is 2 more than the previous multiple of 9 when divided by 7), you'll find that the next time it happens is when n_2 = 9, so that9n_2 = 81 = 7(11) + 4\implies n_1 = 11. If you substitute in the values of n_1 and n_2, then you get that the next number which appears in both sequences is 268.
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    (Original post by MrLatinNerd)
    Fair enough - I did it in a more wordy way...
    That's a nice method - especially since we're only looking for the next common term instead of a collection of the infinitely many common terms, where my method becomes more tractable.

    Wordy is never a bad thing as long as you don't overdo it.
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    (Original post by thefatone)
    how come he left? ;(

    also this stuff i swear when i was doing some research on this stuff i saw all that ^^^^^^^ and i died from complicatedness xD
    It's not very complicated, you can wrap your head around it long enough if you start at it.
 
 
 
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