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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2Do you have any ideas of your own about this? – Mark Bennet Jun 21 '14 at 17:26
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I think I should try to find out inverse of each no. to obtain the missing no. – Jun 21 '14 at 17:31
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What do you mean by group? – thesagniksaha Jun 21 '14 at 17:37
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A binary set that satisfies associativity property, closure property, have identity, and inverse. – Jun 21 '14 at 17:41
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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.
Bill Dubuque
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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.
Mark Bennet
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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 '14 at 20:44
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@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
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Right. The two methods are closely related since $, 9^{-1}\equiv \color{blue}{-10}\pmod{91}.\ \ $ – Bill Dubuque Jun 21 '14 at 22:40
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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.
Jack D'Aurizio
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+1.Your approach for this question is something different. Thank you. Can you suggest a book regarding this? – Avinash N Jul 03 '19 at 11:46
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@AvinashN: I guess that any good text on basic Algebra/Number Theory should do the job. I used Hernstein's Algebra in my youth, it should not be too difficult to find. – Jack D'Aurizio Jul 03 '19 at 11:51
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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.
mfl
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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
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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
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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
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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.
Oscar Lanzi
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Error the claim "this forces $S=0$" is unjustified and requires proof. Note that $x = ex$ does not imply $x =0$ e.g. $,13 = 8\cdot13$ in $\Bbb Z_{91}\ $ (cf. unwarranted cancellation error in this "proof") $\ \ $ – Bill Dubuque May 07 '23 at 16:48
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It can be fixed: hint: $,(e-1)S \equiv 0\Rightarrow S\equiv 0,$ if $e-1$ is a unit (invertible) – Bill Dubuque May 07 '23 at 18:16
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Yes, your edit fixes the error per my above hint, but I think it is worthwhile to leave my comments since this is an oft-recurring but much more subtle version of dividing by zero (even made by very experienced mathematicians as the link in my first comment shows), so it will help readers be aware of such pitfalls. – Bill Dubuque May 08 '23 at 00:12
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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.
