How to find the missing number? 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? 
 A: 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.
A: 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.
A: 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.
A: 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.
A: It is given that the list of integers form a group under multiplication modulo 91. A group is a closed set  meaning that any two elements under the operation must produce an element within the group.
1 is the identity element so we can skip it. I just start checking each in succession.
9 x 16 = 144 modulo 91 = 53. 53 is a member. 
9 x 22 = 198 modulo 91 = 16. 16 is a member.
9 x 53 = 477 modulo 91 = 22. 22 is a member.
9 x 74 = 666 modulo 91 = 29. 29 should be a member. 29 is missing.
Continuing to check will show this is the case.
