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.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

Find all numbers r for which the system of congruences

x = r (mod 6),

x = 9 (mod 20),

x = 4 (mod 45)

has a solution.

*Sorry I didn't know how to put the sign with 3 straight horizontal lines so I just made them equal signs.

Guest Aug 3, 2017

edited by
Guest
Aug 3, 2017

#1**+1 **

x mod 20=9, x mod 45=4, x mod 6 =r, solve for x, r

By simple iteration, the smallest positive integer that satisfies all the congruences is**=49**

**Check:**

**49 / 20 = 2 with a remainder of 9, and:**

**49 /45 =1 with a remainder of 4, and:**

**49 / 6 =8 with a remainder of 1. Therefore:**

**r=1 and x=49**

Guest Aug 3, 2017