ISBN# Question Discrete Math I was having a lot of trouble solving this:
All books are assigned a 10 digit ISBN# ($d_{10}d_9d_8...d_2d_1$) which has the following property:
$\sum_{i = 1}^{10}id_i \equiv 0(\mod(11))$
Prove that if you swap two adjacnt digits in an ISBN#, it is no longer a valid ISBN#.
This is what I have so far:
Suppose you have an ISBN# $d_{10}d_9d_8...d_2d_1$ such that $\sum_{i = 1}^{10}id_i \equiv 0(\mod(11))$. We can write $11|1d_1 + 2d_2 + ... + 10d_{10}$
I'm stuck here though, hints please?
 A: Suppose you swap $d_n$ and $d_{n+1}$
Then the sum becomes  $(\sum_{i=1}^{10} d_i )+d_n-d_{n+1}$
Since $\sum_{i=1}^{10} d_i \equiv 0 \pmod {11}$
For the new sum to be divisible by $11$, 
$d_n-d_{n+1} \equiv 0\pmod {11}$
$d_n-d_{n+1} \equiv k\pmod {11}$ where $0\leq k \leq 9$ and to be divisble by $11$ the $2$ adjacent digits must be same which means that swapping, in this case, doesn't change the ISBN.
A: Well, you've done everything except consider the transposed number.
Let $S_1$ be the "right" sum and $S_2$ be the sum with $d_i$ and $d_{i+1}$ transposed.
Then $S_1-S_2=id_i+(i+1)d_{i+1}-(id_{i+1}+(i+1)d_i)=d_{i+1}-d_i$
Since $S_1\equiv0\pmod {11}$ you have that $-S_2\equiv d_{i+1}-d_i\pmod{11}$, or better 
$S_2\equiv d_{i}-d_{i+1}\pmod{11}$
If $d_i\not\equiv d_{i+1}\pmod{11}$ then the $S_2$ is obviously not the sum of a valid ISBN since $S_2\not\equiv 0\pmod {11}$.
On the other hand, if they are the same digit mod 11, then transposing them makes no difference, and you do get a valid word.  This is a inaccuracy in the problem statement.  For example, the all zeros ISBN number is still valid no matter how many transpositions you do.
