Welcome to our community

Be a part of something great, join today!

What is the remainder when a_{2013} is divided by 7?

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,706
Consider a sequence given by \(\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}\), where \(\displaystyle a_0=a_1=a_2=1\).

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

chisigma

Well-known member
Feb 13, 2012
1,704
Consider a sequence given by \(\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}\), where \(\displaystyle a_0=a_1=a_2=1\).

What is the remainder of \(\displaystyle a_{2013}\) divided by \(\displaystyle 7\)?
Operating modulo 7 we have...

$$a_{0}=1$$
$$a_{1}=1$$
$$a_{2}= 1$$
$$a_{3} = 1 + 3 + 1 = 5$$
$$a_{4} = 5 + 3 + 1 = 2\ \text{mod}\ 7$$
$$a_{5} = 2 + 1 + 1 = 4\ \text{mod}\ 7$$
$$a_{6} = 4 + 6 + 5 = 1\ \text{mod}\ 7$$
$$a_{7} = 1 + 12 + 2 = 1\ \text{mod}\ 7$$
$$a_{8} = 1 + 3 + 4 = 1\ \text{mod}\ 7$$
$$a_{9} = 1 + 2 + 1 =5\ \text{mod}\ 7$$

... 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:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,712
I agree with chisigma on the level of algebra, but not on the level of arithmetic. In fact, reducing all the coefficients mod 7 as we go along,

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

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

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

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

(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

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,706
Thanks to both chisigma and Opalg for the submission to this problem and I've been really impressed with the creativity that gone into the two approaches above and please allow me to thank you all again for the time that the two of you have invested to participate in this problem. :)