how can u solve 17^-1 mod 100?
plz explain the method followed in solving such cases where the exponent is a negative number..
thanks
You are looking for a number x such that
$\displaystyle x \cdot 17 \equiv 1$ (mod 100)
Basically we're down to trial and error, but we can make the following observation: since the base is 100 we can simply start looking at numbers that multiply by 7 to end in a 1. So we start with this list:
3, 13, 23, 33, 43, 53, 63, 73, 83, 93
We also know that in the integer system x is going to have to be larger than 6 to make 17x > 100, so we can count out the 3. (Small savings, but in other problems it might cut the list down significantly.)
So let's try them:
$\displaystyle 13 \cdot 17 \equiv 21$ (mod 100)
$\displaystyle 23 \cdot 17 \equiv 91$ (mod 100)
$\displaystyle 33 \cdot 17 \equiv 61$ (mod 100)
$\displaystyle 43 \cdot 17 \equiv 31$ (mod 100)
$\displaystyle 53 \cdot 17 \equiv 1$ (mod 100)
$\displaystyle 63 \cdot 17 \equiv 71$ (mod 100)
$\displaystyle 73 \cdot 17 \equiv 41$ (mod 100)
$\displaystyle 83 \cdot 17 \equiv 11$ (mod 100)
$\displaystyle 93 \cdot 17 \equiv 81$ (mod 100)
So it looks like $\displaystyle 17^{-1} = 53 $ (mod 100)
-Dan
It means the inverse of a number. Given a positive interger $\displaystyle n$ and any integer $\displaystyle m$, we can write $\displaystyle m^{-1}=k$ (modulo $\displaystyle n$)) iff $\displaystyle mk\equiv 1 (\bmod n)$. Exactly what topsquak said. It is the inverse.
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We want to find $\displaystyle x$ such that:
$\displaystyle 17x\equiv 1 (\bmod 100)$.
Since $\displaystyle 100 = 2^2\cdot 5^2$.
We can simulatenously solve (since $\displaystyle \gcd(2,5)=1$) the congruences:
$\displaystyle 17x\equiv 1 (\bmod 4)$
$\displaystyle 17x \equiv 1(\bmod 25)$
The first one is really simple, add 4 to RHS 4 times to get:
$\displaystyle 17x\equiv 17 (\bmod 4) \Rightarrow x\equiv 1(\bmod 4)$
The second one is also not so bad, add 25 to RHS 2 times to get:
$\displaystyle 17x\equiv 51 (\bmod 25)\Rightarrow x\equiv 3(\bmod 25)$
So you are left with,
$\displaystyle x\equiv 1 (\bmod 4)$
$\displaystyle x\equiv 3 (\bmod 25)$
You can use the Chinese Remainder Theorem.
To solve this modulo 100 (which is what you are looking for).
this is how i have done,plz tell me if there is anything wrong..
m trying to solve
17x = 1 mod 100
using euclidean algorithm this is wat i get
100 = 17*5+15----> eqn 1
17 = 15*1+2---->eqn 2
15 = 7*2 + 1
2 = 2*1
now working bacwards
1 = 15-(7*2)
1 = 15 - (17-15*1)*7
1 = 100-(17*5) - 17*7 + 15*7
1 = 100 - 17 *12 - 15*7
1= 100 - 17*12 + (100-17*5)*7
1 = 100(8) - 17*47
which is of the form
100(8)+17(-47)=1
17* (-47) = 1 mod 100
since we are looking for a positive no hence x = (-47)+100 = 53
m not too sure abt the reasoning of my last step..
plz tell me
cheers
I see you decided not to do it my way.
Using the Euclidean algorithm is another good way to do it.
I just cannot follow what you do!
Here, what an learn frum die meister to how to present mathematical arguments.
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$\displaystyle 17x\equiv 1(\bmod 100)$
By definition it means,
$\displaystyle 17x - 1 = 100y \mbox{ for some }y\in \mathbb{Z}$.
Thus,
$\displaystyle 17x +100z = 1 \mbox{ where }z=-y$.
Now this is a linear diophatine equation.
You can use the Euclidean algorithm to solve this.
First, we need to find $\displaystyle \gcd(17,100)$.
$\displaystyle 100 = 5\cdot 17 +15$
$\displaystyle 17 = 1\cdot 15+2$
$\displaystyle 15 = 7\cdot 2 +1$
Thus, we see that $\displaystyle \gcd(17,100)=1$.
Work backwards,
$\displaystyle 15 - 7\cdot 2 = 1$
$\displaystyle 15 - 7 (17 - 1\cdot 15) = 1$
$\displaystyle 8\cdot 15 - 7\cdot 17 = 1$
$\displaystyle 8 (100 - 5\cdot 17) - 7\cdot 17 = 1$
$\displaystyle 8\cdot 100 -47\cdot 17 =1$
$\displaystyle 17(-47)+100(8)=1$
Thus, we see that $\displaystyle x=-47$ works.
Meaning,
$\displaystyle 17(-47)\equiv 1(\bmod 100)$.
Now you can add 100 to this number because we are working modulo 100 and it makes no difference to get,
$\displaystyle -47 + 100 \equiv 53 (\bmod 100)$.
Thus we get 53.