$\sum _{i=1}^{10}i{d}_{i}\equiv 0(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(11))$

Prove that if you swap two adjacnt digits in an ISBN#, it is no longer a valid ISBN#.

Talon Mcbride
2022-07-17
Answered

All books are assigned a 10 digit ISBN (${d}_{10}{d}_{9}{d}_{8}...{d}_{2}{d}_{1}$) which has the following property:

$\sum _{i=1}^{10}i{d}_{i}\equiv 0(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(11))$

Prove that if you swap two adjacnt digits in an ISBN#, it is no longer a valid ISBN#.

$\sum _{i=1}^{10}i{d}_{i}\equiv 0(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(11))$

Prove that if you swap two adjacnt digits in an ISBN#, it is no longer a valid ISBN#.

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Damarion Pierce

Answered 2022-07-18
Author has **11** answers

Step 1

Let ${S}_{1}$ be the "right" sum and ${S}_{2}$ be the sum with ${d}_{i}$ and ${d}_{i+1}$ transposed.

Then ${S}_{1}-{S}_{2}=i{d}_{i}+(i+1){d}_{i+1}-(i{d}_{i+1}+(i+1){d}_{i})={d}_{i+1}-{d}_{i}$

Step 2

Since ${S}_{1}\equiv 0\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$ you have that $-{S}_{2}\equiv {d}_{i+1}-{d}_{i}\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$, or better ${S}_{2}\equiv {d}_{i}-{d}_{i+1}\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$

If ${d}_{i}\not\equiv {d}_{i+1}\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$ then the ${S}_{2}$ is obviously not the sum of a valid ISBN since ${S}_{2}\not\equiv 0\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$.

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.

Let ${S}_{1}$ be the "right" sum and ${S}_{2}$ be the sum with ${d}_{i}$ and ${d}_{i+1}$ transposed.

Then ${S}_{1}-{S}_{2}=i{d}_{i}+(i+1){d}_{i+1}-(i{d}_{i+1}+(i+1){d}_{i})={d}_{i+1}-{d}_{i}$

Step 2

Since ${S}_{1}\equiv 0\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$ you have that $-{S}_{2}\equiv {d}_{i+1}-{d}_{i}\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$, or better ${S}_{2}\equiv {d}_{i}-{d}_{i+1}\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$

If ${d}_{i}\not\equiv {d}_{i+1}\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$ then the ${S}_{2}$ is obviously not the sum of a valid ISBN since ${S}_{2}\not\equiv 0\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$.

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.

Paxton Hoffman

Answered 2022-07-19
Author has **6** answers

Step 1

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 $\sum _{i=1}^{10}{d}_{i}\equiv 0\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$

Step 2

For the new sum to be divisible by 11,

${d}_{n}-{d}_{n+1}\equiv 0\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$

${d}_{n}-{d}_{n+1}\equiv k\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$ 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.

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 $\sum _{i=1}^{10}{d}_{i}\equiv 0\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$

Step 2

For the new sum to be divisible by 11,

${d}_{n}-{d}_{n+1}\equiv 0\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$

${d}_{n}-{d}_{n+1}\equiv k\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}11)$ 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.

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What classes of graphs result from $\overline{T}$?

I need help in characterizing the classes of graphs that results from taking the complementary of a tree, i.e., the graph that results from removing the edges of a tree from a complete graph. More formally, let $T=(V,E)$ be an n-vertex tree with vertex set V and edge set E. Are there known results on the classes of graphs defined by $\overline{T}$?

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If there are no (or few) results about general T, can we say something about $\overline{T}$ if T belongs to a class of trees, for example caterpillar trees (trees in which the removal of all leaves produces a path graph), lobster trees (trees in which the removal of all leaves produces a caterpillar tree), ...?

Any help will be appreciated. Thank you.

I need help in characterizing the classes of graphs that results from taking the complementary of a tree, i.e., the graph that results from removing the edges of a tree from a complete graph. More formally, let $T=(V,E)$ be an n-vertex tree with vertex set V and edge set E. Are there known results on the classes of graphs defined by $\overline{T}$?

There are two trivial cases. If $T={S}_{n}$, i.e., is a star tree (one single vertex has degree $n-1$ and the other vertices have degree 1), we have that $\overline{{S}_{n}}=\{v\}\cup {K}_{n-1}$, i.e., $\overline{{S}_{n}}$ is a graph with an isolate vertex (degree 0) and a $(n-1)$ vertex clique. I've got the feeling that $\overline{{P}_{n}}$ (where ${P}_{n}$ is an n-vertex path graph) has a precise characterization but I can't put a name to it.

If there are no (or few) results about general T, can we say something about $\overline{T}$ if T belongs to a class of trees, for example caterpillar trees (trees in which the removal of all leaves produces a path graph), lobster trees (trees in which the removal of all leaves produces a caterpillar tree), ...?

Any help will be appreciated. Thank you.

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