Chinese remainders : ichinrem, ichrem

ichinrem([a,p],[b,q]) or ichrem([a,p],[b,q]) returns a list [c,lcm(p,q)] of 2 integers.
The first number c is such that

$$\forall$$k $$\in$$ $$\mathbb {Z}$$,    d = c + k×lcm(p, q)

has the properties

d = a(mod p),    d = b(mod q)

If p and q are coprime, a solution d always exists and all the solutions are congruent modulo p*q.
Examples :
Solve :

$$\tt\left \{ \begin{array}{rcl} x&=&3\ (\bmod\ 5)\\ x&=&9\ (\bmod\ 13) \end{array}\right.$$

Input :

ichinrem([3,5],[9,13])

or input :

ichrem([3,5],[9,13])

Output :

[-17,65]

so x=-17 (mod 65)
we can also input :

ichrem(3%5,9%13)

Output :

-17%65

Solve :

$$\tt\left \{ \begin{array}{rcl} x&=&3\ (\bmod\ 5)\\ x&=&4\ (\bmod\ 7) \\ x&=&1\ (\bmod\ 9)\end{array}\right.$$

First input :

tmp:=ichinrem([3,5],[4,7])

or input :

tmp:=ichrem([3,5],[4,7])

output :

[-17,35]

then input :

ichinrem([1,9],tmp)

or input :

ichrem([1,9],tmp)

Output :

[-17,315]

hence x=-17 (mod 315)
Alternative :

ichinrem([3%5,4%7,1%9])

Output :

-17%315

Remark
ichrem (orichinrem)may be used to find coefficients of polynomial which class are known modulo several integers, for example find ax + b modulo 315 = 5×7×9 under the assumptions:

$$\tt\left \{ \begin{array}{rl} a=&3\ (\bmod\ 5)\\ a=&4\ (\bmod\ 7) \\ a=&1\ (\bmod\ 9) \end{array}\right.$$,    $$\tt\left \{ \begin{array}{rl} b=&1\ (\bmod\ 5)\\ b=&2\ (\bmod\ 7) \\ b=&3\ (\bmod\ 9) \end{array}\right.$$

Input :

ichrem((3x+1)%5,(4x+2)%7,(x+3)%9)

Output :

(-17%315× x+156%315

hence a=-17 (mod 315) and b=156 (mod 315).