Could you help me out with a piece of homework, I really do not know, how to solve this. Even how to begin.

$8$ natural numbers are written in a row. Each number, beginning from the third is a sum of last two numbers before it. Which maximum value has the first number if the last is $2018$?

Hope that my knowledge is solid enough for this task.

  • $\begingroup$ Do you mean the sum of any two of the numbers before it (which makes the problem rather trivial), or the sum of the last two before it? $\endgroup$ – Robert Israel Jan 24 '12 at 20:49
  • $\begingroup$ No, I mean two consecutive numbers $\endgroup$ – user1131662 Jan 24 '12 at 21:00

so far we know there are 8 numbers, so we can name them $$a_1,a_2,\ldots,a_8$$, and every number starting from 3rd the sum of two previous ones. so we know $$\begin{eqnarray} a_3 &=& a_2+a_1 &=& 1*&a_2&+1*&a_1& \\ a_4 &=& a_3+a_2 &=& 2*&a_2&+1*&a_1& \\ a_5 &=& a_4+a_3 &=& 3*&a_2&+2*&a_1& \\ a_6 &=& a_5+a_4 &=& 5*&a_2&+3*&a_1& \\ a_7 &=& a_6+a_5 &=& 8*&a_2&+5*&a_1& \\ a_8 &=& a_7+a_6 &=& 13*&a_2&+8*&a_1& \\ \end{eqnarray}$$

so we know whatever values we choose for $a_1$ and $a_2$, we will always get $a_8=13*a_2+8*a_1$ and from the problem we know $a_8= 2018$. the rest is easy, you just have to minimize value for $a_2$ (eg. $a_2 = 2$) and calculate the value for $a_1$ (which would be 249)

  • $\begingroup$ I think you've written $s_1$ where you meant to write $a_1$. $\endgroup$ – Gerry Myerson Jan 24 '12 at 23:51
  • $\begingroup$ @GerryMyerson you were right, fixed. $\endgroup$ – Ali1S232 Jan 25 '12 at 3:19
  • $\begingroup$ @Gajet Sorry, but I have problems understanding the last thing, I mean how do we get $2$ for $a_2$? Everything else's clear, thanks. $\endgroup$ – user1131662 Jan 25 '12 at 9:14
  • $\begingroup$ I had to minimize the value for $a_2$, and I knew it's a positive integer. assuming $a_2=1$ will results in $a_1=\frac{2005}{8}$ which is not an integer. My next try was $a_2=2$! $\endgroup$ – Ali1S232 Jan 25 '12 at 11:25

Assuming that you mean "... the sum of the last two numbers before it":

Every number is the sequence is a linear function of the two first ones. So write down the sequences starting from $1,0$ and $0,1$. Then you only need to puzzle out how to get $2018$ as a non-negative linear combination of the two final element.

Edit: The OP has clarified that he means "two consecutive numbers", but apparently not necessarily the previous two ones. In that case the answer is 2018, as evidenced by the sequence $$2018,0,2018,2018,2018,2018,2018,2018$$

  • 1
    $\begingroup$ ... or a positive linear combination if your "natural numbers" don't include $0$. $\endgroup$ – Robert Israel Jan 24 '12 at 20:52

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