Sorry, I can read most of that, but I'm not sure what the question is. Are you trying to prove that N is the solution to the set of equations? You'll need to assume that m₁, m₂, and m₃ are (pairwise) relatively prime, or else it isn't true. If that's the case, then the Chinese Remainder Theorem states that there is a unique (mod m₁m₂m₃) solution, so you just have to verify that N satisfies those three equations (which isn't very difficult to show).

Thank you

As Mark stated, everything is quite clear, except where is the question.

On the other hand, the derivation is very understandable, except for the notation fi(), probably because of my ignorance.

If I may resume the example, it states that N is a feasible solution, when it equals the sum of three numbers, each of which is a multiple of the other two modulo factors such that the given remainder is satisfied.

For example,
4*72^fi(5) = 72k where 72k (mod 5) = 4, (= 144)
7*45^fi(8) = 45l where 45l (mod 8) = 7, (= 135)
3*40^fi(9) = 40m where 40m (mod 9) = 3, (= 120)

A feasible solution is the sum of the three numbers, namely 399.
Since 399 mod(5*8*9)=39, so 39 is the smallest solution.

A detailed explanation can be found in:
http://www.math.hawaii.edu/~lee/courses/Chinese.pdf

PC, I translated fi to . I don't know if this is common, but it makes the answer correct. 8)

Thanks Mark, I didn't think of that!  
I kept thinking the ^ sign means raised to the fi(5) power.  Too much programming is not good for math health!

It does mean that. And I almost guarantee I do more programming than you do. 8)

 x = 4mod(5); x = 7mod(8); x = 3mod(9)

 Let x = 4+5k, k an integer, then

 4+5k = 7mod(8), 5k = 3mod(8), 15k = 9mod(8), -k = 1mod(8), k = 7mod(8).

Substitute: x = 4 +5[7mod(8)]; x = 4+35mod(40), x = 39mod(40).

Again let x =39+40k, then 39+40k =3mod(9),3+4k = 3mod(9), k = 0mod(9).

Substitute: x=39+40[0mod(9)]. x =39mod(360), x =39+360t, t an element of Z.

Check: 39+360t=4mod(5), 39+360t=7mod(8) and 39+360t = 3mod(9).

This I could do, but the rest is in Spanish or has symbols I don't understand.

 Chinese remainder theorem: 9 thieves steal a bag of gold coins and after dividing

 them up equally, 3 coins are left over. A quarrel resume and one of the thieves is killed.

The remaining 8 thieves redivide the coins and this time 7 coins remain over. Another

quarrel is initiated and in the resulting ruckus 3 more thieves are killed. The remaining

five thieves re-redivide the coins amongst themselves and lo and behold, this time 4 coins

are left over. What is the mimimum number of coins the thieves stole? 39.

 PC: A detailed explanation can be found in:

http://www.math.hawaii.edu/~lee/courses/Chinese.pdf

Going to this URL, I espy the following group of logically equivalent equations:

 x=1mod2; x=2mod3: x=3mod5: and x=1mod7 to solve. Now the solver goes

thru a bunch of hit and miss explanations to solve these equations which

isn't necessary. to wit:

x=1mod2 implies 2|x-1 implies x-1 =2k, k an integer, hence x = 1+2k.

Ergo, let x= 1+2k. 1+2k =2mod3, 2k =1mod3, 4k = 2mod3, k = 2mod3.

Substitute: x=1+2(2mod3),x = 1+4mod6, x = 5mod6, Let x=5+6k

5+6k =3mod5, k = 3mod5

Substitute: x= 5+6(3mod5),x=5+18mod30, x=23mod30, Let x=23+30k

23+30k=1mod7, 2+2k=1mod7, 2k=-1mod7, 8k=-4mod7,k=3mod7

Substitute: x = 23+30(3mod7), x=23+90mod210, x=113mod210 or x = 113 +210t, t an integer.

No guess work or hit or miss. Final solution is x = 113 + 210t, t an integer



  

If I understand correctly, there are two approaches to solve a Chinese remainder theorem, one is the one shown in Spanish, which is along the lines of the reference I quoted.  The other is by substitution, which you explained in detail.  So we have the best of both worlds!

The example you proposed created four integers, 105, 140, 168, 120, which translated to
533 + 210t, which simplifies to 113+210t.