When working modulo m, the notation a^-1 is used to denote the residue b for which \(ab\equiv 1\pmod{m}\), if any exists. For how many integers a satisfying 0 < a < 100 is it true that \(a(a-1)^{-1} \equiv 4a^{-1} \pmod{20}\)?
Please hurry up guys!!
+26367 When working modulo m, the notation \(a^{-1}\) is used to denote the residue \(b\) for which \(ab\equiv 1\pmod{m}\),
if any exists.
For how many integers a satisfying \(0 < a < 100\) is it true that \(a(a-1)^{-1} \equiv 4a^{-1} \pmod{20}\) ?
\(\begin{array}{|rcll|} \hline a(a-1)^{-1} &\equiv& 4a^{-1} \pmod{20} \quad & | \quad \cdot a \\ a^2(a-1)^{-1} &\equiv& 4a^{-1}\cdot a \pmod{20} \quad & | \quad a^{-1}\cdot a = \dfrac{a}{a} = 1 \\ a^2(a-1)^{-1} &\equiv& 4 \pmod{20} \quad & | \quad a^2(a-1)^{-1} = \dfrac{a^2}{a-1} = a+1 + \dfrac{1}{a-1} \\ a+1 + \underbrace{\dfrac{1}{a-1}}_{\text{is integer, only if }a=2} &\equiv& 4 \pmod{20} \quad & | \quad a = 2 \\\\ 2+1 + \dfrac{1}{2-1} &\equiv& 4 \pmod{20} \quad & | \quad \\ 3 + 1 &\equiv& 4 \pmod{20} \\ 4 &\equiv& 4 \pmod{20} \ \checkmark \\ \hline \end{array}\)
One Integer solution \(\mathbf{a=2}\)
+26367 When working modulo m, the notation \(a^{-1}\) is used to denote the residue \(b\) for which \(ab\equiv 1\pmod{m}\),
if any exists.
For how many integers a satisfying \(0 < a < 100\) is it true that \(a(a-1)^{-1} \equiv 4a^{-1} \pmod{20}\) ?
\(\begin{array}{|rcll|} \hline a(a-1)^{-1} &\equiv& 4a^{-1} \pmod{20} \quad & | \quad \cdot a \\ a^2(a-1)^{-1} &\equiv& 4a^{-1}\cdot a \pmod{20} \quad & | \quad a^{-1}\cdot a = \dfrac{a}{a} = 1 \\ a^2(a-1)^{-1} &\equiv& 4 \pmod{20} \quad & | \quad a^2(a-1)^{-1} = \dfrac{a^2}{a-1} = a+1 + \dfrac{1}{a-1} \\ a+1 + \underbrace{\dfrac{1}{a-1}}_{\text{is integer, only if }a=2} &\equiv& 4 \pmod{20} \quad & | \quad a = 2 \\\\ 2+1 + \dfrac{1}{2-1} &\equiv& 4 \pmod{20} \quad & | \quad \\ 3 + 1 &\equiv& 4 \pmod{20} \\ 4 &\equiv& 4 \pmod{20} \ \checkmark \\ \hline \end{array}\)
One Integer solution \(\mathbf{a=2}\)
Er... I don't think that it is...
For one, it says integers
Second, my friend has seen this before and he says it's not 1
+2440 Solving it this way gives:
\(a(a-1)^{-1} \equiv 4a^{-1} \pmod{20}\\ a \equiv 4a^{-1}(a-1) \pmod{20}\\ a^2 \equiv 4(a-1) \pmod{20}\\ a^2-4a+4 \equiv 0 \pmod{20}\\ (a-2)^2 \equiv 0 \pmod{20} \\ \text { }\\ a= 2+10n \;|\text { for n = (0, 1, 2, … 9), so 10 integers satisfy 0 < a < 100}\)
Well blimey! You’re right. (I’m surprised because Heureka can do mod problems while napping.)
GA
