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I want to know whether there is a mental way to solve modulus equations viz. $9^{-1} \bmod 11$ or any equation where the inverse is less than the mod.

I can solve $11^{-1} \bmod 9$, but the reverse way confuses me.

Zubin Kadva
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    $9$ is $-2$ and $2\cdot 6\equiv 1$, so $9^{-1}\equiv -6$. – Hagen von Eitzen Oct 02 '16 at 20:40
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    Why is $11^{-1}\mod{9}$ easier for you? But if it is somehow easier if the modulus is smaller than the number to be inverted, note: $9^{-1}\mod{11}$ is the same as $20^{-1}\mod{11}$. – 2'5 9'2 Oct 03 '16 at 17:41

4 Answers4

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I think the best thing to do in the general case is take

$$11k+1$$

for $k$ starting at $0$ and increasing, figure out whether $9$ divides it, and then just find the quotient of that number and $9$. For instance:

$$11\cdot 0+1 = 1\equiv 1\mod 9$$

$$11\cdot 1+1 = 12\equiv 3\mod 9$$

$$11\cdot 2+1 = 23\equiv 5\mod 9$$

$$11\cdot 3+1 = 34\equiv 7\mod 9$$

$$11\cdot 4+1 = 45\equiv 0\mod 9$$

Since $45\equiv 0\mod 9$, we have that

$$9^{-1}\mod 11 = \frac{45}{9} = 5$$

Carl Schildkraut
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Hint $\ {\rm mod}\ 11\!:\,\ \dfrac{1}9\,\equiv\, \dfrac{-10}{-2}\equiv\, 5.\ $ The idea is: $ $ first choose the least magnitude rep for the denominator (to increase probability of exact division), and then tweak the numerator so the denominator divides it (adding or subtracting multiples of the modulus).

Beware $ $ Modular fraction arithmetic is valid only for fractions with denominator coprime to the modulus. See here for further discussion.

Community
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Bill Dubuque
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$$9=-2$$ modulo 11, then what times -2 is $-(-1)=1$ or in other words what times $2$ is $12$?

$6$, so the inverse is $-6$

operatorerror
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For a problem as small as $9^{-1}$ mod $11$, the easiest mental approach is simply to run through the multiples of $9$ and look for one that's $1$ more than a multiple of $11$. You find $2\cdot9=18=11+7$, $3\cdot9=27=22+5$, $4\cdot9=36=33+3$, and -- bingo! -- $5\cdot9=45=44+1$. So $9^{-1}\equiv5$ mod $11$.

This approach works in principle for any (coprime) pair of numbers, but it gets tedious and taxing as the numbers get large. The Euclidean algorithm is eventually the way to go, but not many people can run the Euclidean algorithm in their head.

Barry Cipra
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