Determine whether each of the ISBNs below is correct

[Math100]

Homework Statement

The International Standard Book Number (ISBN) used in many libraries consists of nine digits $a_{1}a_{2}...a_{9}$ followed by a tenth check digit $a_{10}$, which satisfies
$a_{10}\equiv \sum^{9}_{k=1} ka_{k}\pmod {11}$.
Determine whether each of the ISBNs below is correct:
(a) 0-07-232569-0 (United States).
(b) 91-7643-497-5 (Sweden).
(c) 1-56947-303-10 (England).

Relevant Equations

None.

(a)
Consider the ISBN $0-07-232569-0$ from the United States.
Observe that
\begin{align*}
&\sum^{9}_{k=1} ka_{k}\pmod {11}\\
&\equiv (a_{1}+2a_{2}+\dotsb +9a_{9})\pmod {11}\\
&\equiv (0+2\cdot 0+3\cdot 7+4\cdot 2+5\cdot 3+6\cdot 2+7\cdot 5+8\cdot 6+9\cdot 9)\pmod {11}\\
&\equiv (21+8+15+12+35+48+81)\pmod {11}\\
&\equiv 220\pmod {11}\\
&\equiv 0\pmod {11}.\\
\end{align*}
Thus $\sum^{9}_{k=1} ka_{k}\pmod {11}\equiv a_{10}$.
Therefore, the ISBN $0-07-232569-0$ is correct.

(b)
Consider the ISBN $91-7643-497-5$ from Sweden.
Observe that
\begin{align*}
&\sum^{9}_{k=1} ka_{k}\pmod {11}\\
&\equiv (a_{1}+2a_{2}+\dotsb +9a_{9})\pmod {11}\\
&\equiv (9+2\cdot 1+3\cdot 7+4\cdot 6+5\cdot 4+6\cdot 3+7\cdot 4+8\cdot 9+9\cdot 7)\pmod {11}\\
&\equiv (9+2+21+24+20+18+28+72+63)\pmod {11}\\
&\equiv 257\pmod {11}\\
&\equiv 4\pmod {11}.\\
\end{align*}
Thus $\sum^{9}_{k=1} ka_{k}\pmod {11}\not\equiv a_{10}$ because $a_{10}=5$.
Therefore, the ISBN $91-7643-497-5$ is not correct.

(c)
Consider the ISBN $1-56947-303-10$ from England.
Observe that
\begin{align*}
&\sum^{9}_{k=1} ka_{k}\pmod {11}\\
&\equiv (a_{1}+2a_{2}+\dotsb +9a_{9})\pmod {11}\\
&\equiv (1+2\cdot 5+3\cdot 6+4\cdot 9+5\cdot 4+6\cdot 7+7\cdot 3+8\cdot 0+9\cdot 3)\pmod {11}\\
&\equiv (1+10+18+36+20+42+21+27)\pmod {11}\\
&\equiv 175\pmod {11}\\
&\equiv 10\pmod {11}.\\
\end{align*}
Thus $\sum^{9}_{k=1} ka_{k}\pmod {11}\equiv a_{10}$.
Therefore, the ISBN $1-56947-303-10$ is correct.

 

[fresh_42]

Homework Statement:: The International Standard Book Number (ISBN) used in many libraries consists of nine digits $a_{1}a_{2}...a_{9}$ followed by a tenth check digit $a_{10}$, which satisfies
$a_{10}\equiv \sum^{9}_{k=1} ka_{k}\pmod {11}$.
Determine whether each of the ISBNs below is correct:
(a) 0-07-232569-0 (United States).
(b) 91-7643-497-5 (Sweden).
(c) 1-56947-303-10 (England).
Relevant Equations:: None.

(a)
Consider the ISBN $0-07-232569-0$ from the United States.
Observe that
\begin{align*}
&\sum^{9}_{k=1} ka_{k}\pmod {11}\\
&\equiv (a_{1}+2a_{2}+\dotsb +9a_{9})\pmod {11}\\
&\equiv (0+2\cdot 0+3\cdot 7+4\cdot 2+5\cdot 3+6\cdot 2+7\cdot 5+8\cdot 6+9\cdot 9)\pmod {11}\\
&\equiv (21+8+15+12+35+48+81)\pmod {11}\\
&\equiv 220\pmod {11}\\
&\equiv 0\pmod {11}.\\
\end{align*}
Thus $\sum^{9}_{k=1} ka_{k}\pmod {11}\equiv a_{10}$.
Therefore, the ISBN $0-07-232569-0$ is correct.

(b)
Consider the ISBN $91-7643-497-5$ from Sweden.
Observe that
\begin{align*}
&\sum^{9}_{k=1} ka_{k}\pmod {11}\\
&\equiv (a_{1}+2a_{2}+\dotsb +9a_{9})\pmod {11}\\
&\equiv (9+2\cdot 1+3\cdot 7+4\cdot 6+5\cdot 4+6\cdot 3+7\cdot 4+8\cdot 9+9\cdot 7)\pmod {11}\\
&\equiv (9+2+21+24+20+18+28+72+63)\pmod {11}\\
&\equiv 257\pmod {11}\\
&\equiv 4\pmod {11}.\\
\end{align*}
Thus $\sum^{9}_{k=1} ka_{k}\pmod {11}\not\equiv a_{10}$ because $a_{10}=5$.
Therefore, the ISBN $91-7643-497-5$ is not correct.

(c)
Consider the ISBN $1-56947-303-10$ from England.
Observe that
\begin{align*}
&\sum^{9}_{k=1} ka_{k}\pmod {11}\\
&\equiv (a_{1}+2a_{2}+\dotsb +9a_{9})\pmod {11}\\
&\equiv (1+2\cdot 5+3\cdot 6+4\cdot 9+5\cdot 4+6\cdot 7+7\cdot 3+8\cdot 0+9\cdot 3)\pmod {11}\\
&\equiv (1+10+18+36+20+42+21+27)\pmod {11}\\
&\equiv 175\pmod {11}\\
&\equiv 10\pmod {11}.\\
\end{align*}
Thus $\sum^{9}_{k=1} ka_{k}\pmod {11}\equiv a_{10}$.
Therefore, the ISBN $1-56947-303-10$ is correct.

I got the same result. But that only reduces the chances. We still could have made both the same error, at least modulo 11.

 

[Math100]

I got the same result. But that only reduces the chances. We still could have made both the same error, at least modulo 11.

What's the error then?

 

[fresh_42]

What's the error then?

That was a joke. I don't think you made one. However, I checked the calculations only in mind.

The first book I checked here, however, has a ten-digit number plus a check digit.

 

[fresh_42]

Let's test $3-411-014420-2.$

$3+4\cdot 2+ 3+4+6+4\cdot 7+4\cdot 8+2\cdot 9 \equiv 3 \not\equiv 2\pmod{11}.$ Hmm ...

 

[Math100]

Let's test $3-411-014420-2.$

$3+4\cdot 2+ 3+4+6+4\cdot 7+4\cdot 8+2\cdot 9 \equiv 3 \not\equiv 2\pmod{11}.$ Hmm ...

I know, it's a bit weird.

 

[anuttarasammyak]

They say a new 13 digits code has been applied since 2007.

 

[fresh_42]

They say a new 13 digits code has been applied since 2007.

My book example is older.