# finite series

• May 21st 2011, 06:39 PM
sixstringartist
finite series
Ahoy all late nighters. looking for a some clarity on a problem I have. Please go easy. Im sure to screw something up when writing this.

Ok so heres the problem as I know it. It is an algorithm with the following behavior:
${t}_{0} = 0 \\ {t}_{1} = {t}_{0}*17 + {x}_{0} \\ {t}_{2} = {t}_{1}*17 + {x}_{1} \\ {t}_{3} = {t}_{2}*17 + {x}_{2} \\ \vdots \\ {t}_{n} = {t}_{n-1}*17 + {x}_{n-1} \\$

which I believe is the following series

${t}_{n} = 17^{n-1}{x}_{0} + 17^{n-2}{x}_{1} + \hdots + 17^{1}{x}_{n-2} + 17^{0}{x}_{n-1}$

Now all the ${x}_{n}$'s are unknown and are an integer in the range 1-26
Given a value for ${t}_{n}$ can a unique sum be found? I suspect there will be multiple solutions. Could someone help me simplify this?

If it helps, assume ${t}_{n} = 22640$
• May 21st 2011, 08:39 PM
chisigma
Quote:

Originally Posted by sixstringartist
Ahoy all late nighters. looking for a some clarity on a problem I have. Please go easy. Im sure to screw something up when writing this.

Ok so heres the problem as I know it. It is an algorithm with the following behavior:
${t}_{0} = 0 \\ {t}_{1} = {t}_{0}*17 + {x}_{0} \\ {t}_{2} = {t}_{1}*17 + {x}_{1} \\ {t}_{3} = {t}_{2}*17 + {x}_{2} \\ \vdots \\ {t}_{n} = {t}_{n-1}*17 + {x}_{n-1} \\$

which I believe is the following series

${t}_{n} = 17^{n-1}{x}_{0} + 17^{n-2}{x}_{1} + \hdots + 17^{1}{x}_{n-2} + 17^{0}{x}_{n-1}$

Now all the ${x}_{n}$'s are unknown and are an integer in the range 1-26
Given a value for ${t}_{n}$ can a unique sum be found? I suspect there will be multiple solutions. Could someone help me simplify this?

If it helps, assume ${t}_{n} = 22640$

Is...

$t_{n}= 17^{n-1}\ x_{0} + 17^{n-3}\ x_{1} + ... + 17\ x_{n-2} + x_{n-1} = 22640$ (1)

Now is...

$\text {int} (\frac{22640}{17})= 1331 \implies x_{n-1}= 22640-1331*17=13$

$\text {int} (\frac{1331}{17})= 78 \implies x_{n-2}= 1331-78*17=5$

$\text {int} (\frac{78}{17})= 4 \implies x_{n-3}= 78-4*17=10$

$\text {int} (\frac{4}{17})= 0 \implies x_{n-4}= 4-0*17=4$

... so that is n=4 and...

$t_{n}= 17^{3}*4 + 17^{2}*10 + 17*5 + 13 = 22640$ (2)

Kind regards

$\chi$ $\sigma$
• May 21st 2011, 09:43 PM
chisigma
The solution we found in the last post requires all the $x_{n}$ in the range 1-16... if the range is 1-26, then another possible value of $x_{2}$ is 22 and that produces the solution $x_{1}=9$ and $x_{0}=4$ so that is...

$t_{n}= 17^{3}*4 + 17^{2}*9+17*22+13= 22640$

... and from that it derives the other solution $x_{1}=26$ and $x_{0}=3$ so that is...

$t_{n}= 17^{3}*3 + 17^{2}*26+17*22+13= 22640$

No more solution exists...

Kind regards

$\chi$ $\sigma$
• May 22nd 2011, 06:29 AM
sixstringartist
you are doing a modulus operation no?

I understand now. Thanks for the help