Sequences question

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.

Reply 1

MrLatinNerd

7

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_1

th term in the first sequence is equal to the

n_2

th 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?

Reply 2

zanyzoya

OP

7n1 + 4 must be a multiple of 9
Trial and error for n1?

Reply 3

thefatone

6

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?

Reply 4

thefatone

6

@Zacken @Student403 @TeeEm

Reply 5

MrLatinNerd

7

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!

(edited 8 years ago)

Reply 6

Zacken

22

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.

Original post by thefatone

@Zacken @Student403 @TeeEm

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

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:

Unparseable latex formula:

\displaystyle [br]\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

.

Reply 7

thefatone

6

Original post by Zacken

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

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:

Unparseable latex formula:

\displaystyle [br]\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

Reply 8

MrLatinNerd

7

Original post by Zacken

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

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:

Unparseable latex formula:

\displaystyle [br]\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 that

9n_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.

(edited 8 years ago)

Reply 9

Zacken

22

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.