- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,716

What is the remainder of \(\displaystyle a_{2013}\) divided by \(\displaystyle 7\)?

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,716

What is the remainder of \(\displaystyle a_{2013}\) divided by \(\displaystyle 7\)?

- Feb 13, 2012

- 1,704

Operating modulo 7 we have...

What is the remainder of \(\displaystyle a_{2013}\) divided by \(\displaystyle 7\)?

$$a_{0}=1$$

$$a_{1}=1$$

$$a_{2}= 1$$

$$a_{3} = 1 + 3 + 1 = 5$$

$$a_{4} = 5 + 3 + 1 = 2\ \text{mod}\ 7$$

$$a_{5} = 2 + 1 + 1 = 4\ \text{mod}\ 7$$

$$a_{6} = 4 + 6 + 5 = 1\ \text{mod}\ 7$$

$$a_{7} = 1 + 12 + 2 = 1\ \text{mod}\ 7$$

$$a_{8} = 1 + 3 + 4 = 1\ \text{mod}\ 7$$

$$a_{9} = 1 + 2 + 1 =5\ \text{mod}\ 7$$

... and we can stop because the sequence is mod 7 periodic with period 6. Now is $2013\ \text{mod}\ 6 = 3$, so that the requested number is $a_{3}=5$...

Kind regards

$\chi$ $\sigma$

Last edited:

- Moderator
- #3

- Feb 7, 2012

- 2,715

$a_{n+3} = a_{n+2} + 3a_{n+1} + a_{n}$,

$a_{n+4} = a_{n+3} + 3a_{n+2} + a_{n+1} = (a_{n+2} + 3a_{n+1} + a_{n}) + 3a_{n+2} + a_{n+1} = 4a_{n+2} + 4a_{n+1} + a_{n}$,

$a_{n+5} = a_{n+4} + 3a_{n+3} + a_{n+2} = (4a_{n+2} + 4a_{n+1} + a_{n}) + 3(a_{n+2} + 3a_{n+1} + a_{n}) + a_{n+2} = a_{n+2} + 6a_{n+1} + 4a_{n}$,

$a_{n+6} = a_{n+5} + 3a_{n+4} + a_{n+3} = (a_{n+2} + 6a_{n+1} + 4a_{n}) + 3(4a_{n+2} + 4a_{n+1} + a_{n}) + (a_{n+2} + 3a_{n+1} + a_{n}) = a_{n}$

(for all $n\geqslant0$). So the sequence repeats with period $6$. It starts with $(a_0,a_1,a_2,a_3,a_4,a_5) = (1,1,1,5,2,4)\pmod7$, and since $2013=3\pmod6$ it follows that $a_{2013} = a_3 = 5\pmod7.$

- Thread starter
- Admin
- #4

- Feb 14, 2012

- 3,716