We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Modulo inverse help please

0
303
3

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!!

Aug 28, 2018
edited by Guest  Aug 28, 2018

### Best Answer

#1
+22172
+5

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 $$\mathbf{a=2}$$

Aug 29, 2018

### 3+0 Answers

#1
+22172
+5
Best Answer

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 $$\mathbf{a=2}$$

heureka Aug 29, 2018
#2
+1

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

Aug 29, 2018
#3
+1414
+3

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

Aug 30, 2018