To solve: The system of congruence x=2(\bmod 3), x=1(\bmod 4),\ and\ x=3(\bmod 5) using the method of back substitution.

CMIIh

CMIIh

Answered question

2021-02-22

To solve: The system of congruence x=2(bmod3),x=1(bmod4), and x=3(bmod5) using the method of back substitution.

Answer & Explanation

2abehn

2abehn

Skilled2021-02-24Added 88 answers

Given information:
x=2(bmod3),x=1(bmod4), and x=3(bmod5)
Calculation:
By definition, the first congruence can be written as x = 3t +2 where t is an integer.
Substituting this expression for x into the second congruence tells us that 3t+2=1(bmod4), which can easily be solved to show that t=1(bmod4).
From this we can write t = 4u + 1 for some integer u.
Thus x= 3t+2
=3(4u+1)+2
=12u+5.
We plug this into the third congruence to obtain 12u+5=3(bmod5),
which we easily solve to give
u=4(bmod5).
Hence u =5v +4, and so
x=12u+5
=12(5v+4)+5
= 60v +53.
We check our answer by confirming that 53=2(bmod3),53=1(bmod4), and 53=3(bmod5).
Thus, x= 60k +53.
This is the solution for the system of congruences for every k that is an integer.

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