# Can x mod (N - a) or x mod (N + a) be calculated just by knowing x mod N??

Can x mod (N - a) or x mod (N + a) be calculated just by knowing x mod N? where a is an arbitrary integer, and N is a prime. e.g. i know 22! mod 23 ≡ 22 using Wilson's theorem. lets say i want to know 22! mod 24, is there a way i can do so without solving the factorial?

• I added an answer which discusses the general case. – Bill Dubuque Jan 8 at 20:21

If $$\,d := \gcd(n,m)=1\,$$ then there is no relationship between $$\,a = x\bmod n\,$$ and $$\, b = x\bmod m\,$$ since CRT $$\,\Rightarrow\,x\equiv a\pmod{\!n},\ x\equiv b\pmod{\! m}\$$ is solvable for all values of $$\,a,b.\,$$ This is true in your example $$\iff 1 = \gcd(n,n\pm a) = \gcd(n,a).$$
However, when $$\,d > 1\,$$a solution exists iff $$\,b\equiv a\pmod{\!d},\,$$ and this does place constraints on the values $$\,b\,$$ that $$\,x\,$$ can take $$\!\bmod m$$. Then $$\,b\,$$ is determined uniquely iff $$\,m\mid d\iff m\mid n\$$ (i.e. iff $$\,n\pm a\mid n\,$$ in your example), hence $$\,x\bmod m = (x\bmod n)\bmod m\,$$ is uniquely determined. This arises frequently, e.g. knowing $$\, u = x\bmod 10 =$$ units digit of $$x$$ we can compute its parity by taking the parity of its units digit, i.e. $$\ x\bmod 2 = (x\bmod 10)\bmod 2 = u\bmod 2,\,$$ e.g. $$\,123\bmod 2 = 3\bmod 2 = 1$$.
For the specific case $$22!$$, the answer is easy: since $$4 \leq 22$$ and $$6 \leq 22$$ you have that $$22! = 0 \pmod{24}$$. However, we didn't use the fact we got from Wilson's Theorem, and it doesn't help us in general.
The answer to your question is no, and here is an illustrative example: suppose you know that $$x = 5 \pmod{10}$$. Then you could have, among many options, that $$x = 5$$, or that $$x = 15$$. In the former case, $$x = 5 \pmod{12}$$. In the latter case, $$x = 3 \pmod{12}$$.
• But the answer can be yes too, e.g. $\,n=10, n−a=5.\$ See my answer for the general case. – Bill Dubuque Jan 8 at 20:24