Let $X=\lbrace 0,1,2,3,...,99 \rbrace$ For $a,b$ in $X$ we define $a*b$ to be the remainder obtained by dividing the product $ab$ by $100$. For example $9*18=62$ and $7*5=35$. Let $x$ be an element in $X$. An element $y$ in $X$ is called the inverse of $x$ if $x*y=1$ The question is to find out which of the elements $1,2,3,4,5,6,7$ have inverses and also find out their inverses.
We have to find out the number which on multiplication with the given number results in $01$ to be the last two digits.I understand that inverse $1,2,5$ doesn't exist but couldn't figure out others. Any ideas? Thanks
Hint: You're looking for numbers that are relatively prime to $100$ that satisfy the equation $$ax \equiv 1 \space\ (\text{mod} \space\ 100)$$ where $a = \left\{ 1,2,3,4,5,6,7\right\}$ per your question. I will correct your mistake with the element $a=1$, note that: $$(1)(x) \equiv 1 \space\ (\text{mod} \space\ 100)$$ is satisfied when $x=1$, so that the inverse of $1$ is itself, in other words, $a = 1 = a^{-1}$. Then, ask yourself why there would not be an $x$ that satisfies $$2x \equiv 1 \space\ (\text{mod} \space\ 100).$$ Think of plugging in values of $x$, every product $2x$ would be...