View topic - Find the last digit of a sum • Art of Problem Solving

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Jutaro Posts: 314 Location: Tokyo, Japan	Posted: Apr 29, 2007, 9:24 pm • # 1 For $0 \leq d \leq 9$ , we define the numbers $S_{d}=1+d+d^{2}+\cdots+d^{2006}$ Find the last digit of the number $S_{0}+S_{1}+\cdots+S_{9}.$ _________________ $\mathfrak{May\ your\ fire\ burn\ in\ our\ hearts,\ and\ let\ the\ legend\ survive!}$

skimnc

Posts: 204

Posted: Apr 29, 2007, 11:20 pm • # 2

There is probably a way better way to do this but I felt like trying it.

Hidden Text

$S_{d}=\sum_{n=0}^{2006}d^{n}= \frac{1 \cdot (1-d^{2007})}{1-d}=\frac{d^{2007}-1}{d-1}$

$S_{0}=1 \Rightarrow$ Last digit is $1$
$S_{1}=1+2006 \cdot 1=2007 \Rightarrow$ Last digit is $7$
$S_{2}=2^{2007}-1 \Rightarrow$ Last digit is $3$

Hidden Text

Note that:
$2^{1}-1=1$ ; Last digit is $1$ .
$2^{2}-1=3$ ; Last digit is $3$ .
$2^{3}-1=7$ ; Last digit is $7$ .
$2^{4}-1=15$ ; Last digit is $5$ .
$2^{5}-1=31$ ; Last digit is $1$ .
$2^{6}-1=63$ ; Last digit is $3$ .
$2^{7}-1=127$ ; Last digit is $7$ .
$2^{8}-1=255$ ; Last digit is $5$ .
$2^{9}-1=511$ ; Last digit is $1$ .
$\ldots$
$2^{2004}-1 \Rightarrow$ ; Last digit is $5$ .
$2^{2005}-1 \Rightarrow$ ; Last digit is $1$ .
$2^{2006}-1 \Rightarrow$ ; Last digit is $3$ .
$2^{2007}-1 \Rightarrow$ ; Last digit is $7$ .

So the $2^{2006}-1$ will have a last term of $3$ .

$S_{3}=\frac{3^{2006}-1}{3-1}=\frac{3^{2006}-1}{2}\Rightarrow$ Last digit is $4$ .

Hidden Text

Note that:
$\frac{3^{1}-1}{2}=1$ ; Last digit is $1$ .
$\frac{3^{2}-1}{2}=4$ ; Last digit is $4$ .
$\frac{3^{3}-1}{2}=13$ ; Last digit is $3$ .
$\frac{3^{4}-1}{2}=40$ ; Last digit is $0$ .
$\frac{3^{5}-1}{2}=121$ ; Last digit is $1$ .
$\frac{3^{6}-1}{2}=364$ ; Last digit is $4$ .
$\frac{3^{7}-1}{2}=1093$ ; Last digit is $3$ .
$\frac{3^{8}-1}{2}=3280$ ; Last digit is $0$ .
$\frac{3^{9}-1}{2}=9841$ ; Last digit is $1$ .
$\ldots$
$\frac{3^{2004}-1}{2}\Rightarrow$ ; Last digit is $0$ .
$\frac{3^{2005}-1}{2}\Rightarrow$ ; Last digit is $1$ .
$\frac{3^{2006}-1}{2}\Rightarrow$ ; Last digit is $4$ .
$\frac{3^{2007}-1}{2}\Rightarrow$ ; Last digit is $3$ .

Continue to note patterns and end up with:

$S_{0}\Rightarrow 1$
$S_{1}\Rightarrow 7$
$S_{2}\Rightarrow 7$
$S_{3}\Rightarrow 3$
$S_{4}\Rightarrow 1$
$S_{5}\Rightarrow 1$
$S_{6}\Rightarrow 7$
$S_{7}\Rightarrow 7$
$S_{8}\Rightarrow 3$
$S_{9}\Rightarrow 1$

$1+7+7+3+1+1+7+7+3+1=38 \Rightarrow$ Last digit is $\boxed{8}$ .

Last edited by skimnc on Apr 30, 2007, 12:59 pm, edited 1 time in total.

HTA

Posts: 330
Location: Hanoi , Vietnam
Viet Nam

Posted: Apr 30, 2007, 4:13 am • # 3

skimnc wrote:

There is probably a way better way to do this but I felt like trying it.

Hidden Text

$S_{d}=\sum_{n=0}^{2006}d^{n}= \frac{1 \cdot (1-d^{2006})}{1-d}=\frac{d^{2006}-1}{d-1}$

$S_{0}=1 \Rightarrow$ Last digit is $1$
$S_{1}=1+2006 \cdot 1=2007 \Rightarrow$ Last digit is $7$
$S_{2}=2^{2006}-1 \Rightarrow$ Last digit is $3$

Hidden Text

Note that:
$2^{1}-1=1$ ; Last digit is $1$ .
$2^{2}-1=3$ ; Last digit is $3$ .
$2^{3}-1=7$ ; Last digit is $7$ .
$2^{4}-1=15$ ; Last digit is $5$ .
$2^{5}-1=31$ ; Last digit is $1$ .
$2^{6}-1=63$ ; Last digit is $3$ .
$2^{7}-1=127$ ; Last digit is $7$ .
$2^{8}-1=255$ ; Last digit is $5$ .
$2^{9}-1=511$ ; Last digit is $1$ .
$\ldots$
$2^{2004}-1 \Rightarrow$ ; Last digit is $5$ .
$2^{2005}-1 \Rightarrow$ ; Last digit is $1$ .
$2^{2006}-1 \Rightarrow$ ; Last digit is $3$ .

So the $2^{2006}-1$ will have a last term of $3$ .

$S_{3}=\frac{3^{2006}-1}{3-1}=\frac{3^{2006}-1}{2}\Rightarrow$ Last digit is $4$ .

Hidden Text

We can continue to note that the last digit of $S_{d}$ will always be the same last digit of $\frac{d^{2}-1}{d-1}$ aka it will be $d+1$ with the exception of $d=1,6$ in which case it is $\frac{d^{1}-1}{d-1}=1$

$S_{0}\Rightarrow 1$
$S_{1}\Rightarrow 7$
$S_{2}\Rightarrow 3$
$S_{3}\Rightarrow 4$
$S_{4}\Rightarrow 5$
$S_{5}\Rightarrow 6$
$S_{6}\Rightarrow 1$
$S_{7}\Rightarrow 8$
$S_{8}\Rightarrow 9$
$S_{9}\Rightarrow 0$

$1+7+3+4+5+6+1+8+9+0=44 \Rightarrow$ Last digit is $\boxed{4}$ .

my solution is the same as you but i dont think it's the best way

could anybody have another idea ?

skimnc Posts: 204	Posted: Apr 30, 2007, 4:41 am • # 4 I see why $1$ doesn't work out to be $d+1$ but why doesn't $6$ ? I noticed the pattern was it repeated every $5$ numbers but I have no clue as to why $6$ randomly decides to not work.

Kouichi Nakagawa

Posts: 1340
Location: Japan (日本)
Japan

Posted: Apr 30, 2007, 7:23 am • # 5

Hidden Text

$S_{0}\equiv 1 \pmod{10}$
$S_{1}\equiv 7 \pmod{10}$
$S_{2}\equiv 7 \pmod{10}$
$S_{3}\equiv 3 \pmod{10}$
$S_{4}\equiv 1 \pmod{10}$
$S_{5}\equiv 1 \pmod{10}$
$S_{6}\equiv 7 \pmod{10}$
$S_{7}\equiv 7 \pmod{10}$
$S_{8}\equiv 3 \pmod{10}$
$S_{9}\equiv 1 \pmod{10}$
Therefore, $\sum_{d=0}^{9}S_{d}\equiv 1+7+7+3+1+1+7+7+3+1=38\equiv \boxed{8}\pmod{10}$

Farenhajt Posts: 3482 Location: Belgrade	Posted: Apr 30, 2007, 11:32 am • # 6 Or we can sum the thing by the columns instead of rows: Ten ones give $10$ , ten first powers give $45$ , ten squares give $0+1+4+9+6+5+6+9+4+1=45$ , ten cubes give $45$ , ten fourth powers give $0+1+6+1+6+5+6+1+6+1=33$ . From fifth power on, the cycle repeats. Hence we have $10+501\cdot(45+45+45+33)+45+45\equiv 8\pmod{10}$

skimnc Posts: 204	Posted: Apr 30, 2007, 1:00 pm • # 7 I originally only carried the sum formula out to $r^{n}$ when it should of been $r^{n+1}$ .. I edited it.

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Find the last digit of a sum

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