How to find the missing number?

Question

A teacher intended to give a typist a list of nine integers that form a group under multiplication modulo 91. But one of the nine integers was inadvertently left out, so that the list appeared as $1,9,16,22,53,74,79,81.$ Which integer was left out?

Answer 1

If it has to be a multiplicative group, then $22^2$ must be an element of the group. Since $22^2=29\mod 91$ the missing element is $29.$

Edit

The first attempt to solve the problem is to compute $9^0=1,9^1=9,9^2=3,9^4=1$ (mod $91$) which belongs to the list. Then $16^0=1,16^1=16,16^2=74,16^3=1$ (mod $91$) which belongs to the list. Next $22^2=29\mod 91$ which is not in the list.

Answer 2

Multiply everything by a non-unit element - $9$ looks easiest, because $9\times 10\equiv -1 \mod 91$ which makes the arithmetic particularly easy.

$9\times 1=9; 9\times 9 = 81; $

$9\times 16 = 53; 9\times 22 = 16; $

$9\times 53 = 22; 9\times 74 = 29; $

$9\times 79 =74; 9\times 81 = 1$

and check $9 \times 29 = 79$

So $29$ is the missing number.

Answer 3

The list $\,L \equiv 1,\color{#0a0}9,\color{blue}{-10},\color{#c00}{-12},\,\ldots\pmod{91}.\,$ The map $\,f(x) = \color{blue}{-10}x \,$ is a permutation on $\,G\,$ with action $\ f(\color{#0a0}9)\equiv -90\equiv 1\in L,\ \ f(\color{blue}{-10})\equiv 100\equiv 9\in L,\ \ f(\color{#c00}{-12})\equiv 120\equiv 29\not\in\! L,\,$ bingo!

Remark $\ $ This method always works. Indeed if we use the permutation $\,f(x) = ax\,$ for $\,a\not\equiv 1\,$ then the missing element $\,m\,$ will be discovered when we compute $\,f(a^{-1}m) \equiv m\,$ (note $\,a^{-1}m \in L\,$ else $\,a^{-1}m \equiv m\,$ so $\,a\equiv 1),\,$ which is clear when viewed as rotation of the cycles of the permutation $\,f.$

To simplify arithmetic, I ordered the elements in $\,L\,$ least-magnitude first, using balanced (least-magnitude) remainders/reps, and chose $\,a\equiv \color{blue}{-10},\,$ for easy multiplication.

Answer 4

Let $e$ be any element onthe group. Then multiplying by $e$ leaves the set of elements invariant and therefore the sum of the elements $S$ must satisfy

$S=eS$

for all $e$ in the group. This forces $S=0$ by choosing a multiplier $e$, such as $9$, for which $e-1$ is invertible ($8×57\equiv1\bmod91$). (Alternatively, the combination of $e=22$ with $e=53$ would give this result, since these would resoectvely force $S\equiv0\bmod13$ and $S\equiv0\bmod7$). So the missing element must give a sum of $0 \bmod 91$ when combined with the given ones.

You may then verify that the given elements sum to $62\bmod 91$, from which you need $29$ to complete the required zero sum.

Answer 5

Well, $\mathbb{Z}/(91\mathbb{Z})^*\simeq \mathbb{Z}/(7\mathbb{Z})^*\times\mathbb{Z}/(13\mathbb{Z})^*$ is an abelian group of order $6\cdot 12$ and there are not so many subgroups of order $9$. The given list reduced $\!\!\pmod{7}$ equals $$ 1,2,2,1,4,4,2,4 $$ and reduced $\!\!\pmod{13}$ equals $$ 1,9,3,9,1,9,1,3 $$ so the associated subgroup (isomorphic to $\mathbb{Z}/(3\mathbb{Z})\times\mathbb{Z}/(3\mathbb{Z})$) is the set of squares $\!\!\pmod{7}$ and fourth powers $\!\!\pmod{13}$, $\{1,2,4\}\times\{1,3,9\}$. The missing element is $\equiv 1\pmod{7}$ and $\equiv 3\pmod{13}$, so it is $\color{red}{29}$ by the CRT.