Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Do you have any ideas of your own about this? – Mark Bennet Jun 21 '14 at 17:26
I think I should try to find out inverse of each no. to obtain the missing no. – amitbadoni001 Jun 21 '14 at 17:31
What do you mean by group? – Sagnik Saha Jun 21 '14 at 17:37
A binary set that satisfies associativity property, closure property, have identity, and inverse. – amitbadoni001 Jun 21 '14 at 17:41

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.

share|cite|improve this answer
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 '14 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 '14 at 21:09
Right. The two methods are closely related since $\, 9^{-1}\equiv \color{blue}{-10}\pmod{91}.\ \ $ – Bill Dubuque Jun 21 '14 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.

share|cite|improve this answer

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 $29.$


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.

share|cite|improve this answer
But how did you know to choose $22$ first? – Bill Dubuque Jun 21 '14 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 '14 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 '14 at 22:21
I have just done it. Thank you for you remark. I agree that it makes the answer clearer. – mfl Jun 21 '14 at 22:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.