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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?

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Do you have any ideas of your own about this? –  Mark Bennet Jun 21 at 17:26
    
I think I should try to find out inverse of each no. to obtain the missing no. –  amitbadoni001 Jun 21 at 17:31
    
What do you mean by group? –  Sagnik Saha Jun 21 at 17:37
    
A binary set that satisfies associativity property, closure property, have identity, and inverse. –  amitbadoni001 Jun 21 at 17:41

3 Answers 3

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.

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As is often true, this way is easier in residue system that is balanced (least magnitude), using $\, {-}10$ instead of $\,9,\,$ see my answer. –  Bill Dubuque Jun 21 at 20:44
    
@BillDubuque I see what you are doing, and how that works, and why it might be useful. But here I just need to use $9\times 16=-1+9\times 6=53$ and $9\times 74= -7 +9\times 4=29$, for example, which is what I meant by easy arithmetic. –  Mark Bennet Jun 21 at 21:09
    
Right. The two methods are closely related since $\, 9^{-1}\equiv \color{blue}{-10}\pmod{91}.\ \ $ –  Bill Dubuque Jun 21 at 22:40

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.

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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.

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But how did you know to choose $22$ first? –  Bill Dubuque Jun 21 at 20:48
    
I didn't choose it first. I have checked the powers of $9$ and $16$ before, but they appear in the list. –  mfl Jun 21 at 21:25
    
It would be helpful to say that in your answer, since otherwise it is not clear what method was used (and if it works generally). –  Bill Dubuque Jun 21 at 22:21
    
I have just done it. Thank you for you remark. I agree that it makes the answer clearer. –  mfl Jun 21 at 22:26

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