# 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 equiv 0(mod (11))

All books are assigned a 10 digit ISBN (${d}_{10}{d}_{9}{d}_{8}...{d}_{2}{d}_{1}$) which has the following property:
$\sum _{i=1}^{10}i{d}_{i}\equiv 0\left(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(11\right)\right)$
Prove that if you swap two adjacnt digits in an ISBN#, it is no longer a valid ISBN#.
You can still ask an expert for help

## Want to know more about Discrete math?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Damarion Pierce
Step 1
Let ${S}_{1}$ be the "right" sum and ${S}_{2}$ be the sum with ${d}_{i}$ and ${d}_{i+1}$ transposed.
Then ${S}_{1}-{S}_{2}=i{d}_{i}+\left(i+1\right){d}_{i+1}-\left(i{d}_{i+1}+\left(i+1\right){d}_{i}\right)={d}_{i+1}-{d}_{i}$
Step 2
Since ${S}_{1}\equiv 0\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11\right)$ you have that $-{S}_{2}\equiv {d}_{i+1}-{d}_{i}\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11\right)$, or better ${S}_{2}\equiv {d}_{i}-{d}_{i+1}\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11\right)$
If ${d}_{i}\not\equiv {d}_{i+1}\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11\right)$ then the ${S}_{2}$ is obviously not the sum of a valid ISBN since ${S}_{2}\not\equiv 0\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11\right)$.
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.
###### Did you like this example?
Paxton Hoffman
Step 1
Suppose you swap ${d}_{n}$ and ${d}_{n+1}$
Then the sum becomes $\left(\sum _{i=1}^{10}{d}_{i}\right)+{d}_{n}-{d}_{n+1}$
Since $\sum _{i=1}^{10}{d}_{i}\equiv 0\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11\right)$
Step 2
For the new sum to be divisible by 11,
${d}_{n}-{d}_{n+1}\equiv 0\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11\right)$
${d}_{n}-{d}_{n+1}\equiv k\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11\right)$ where $0\le k\le 9$ 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.