Use the Chinese remainder theorem. Suppose we set [tex]N=5\cdot13+11\cdot12[/tex]. Then clearly taken modulo 12, the second term vanishes, and incidentally [tex]5\cdot13\equiv65\equiv5\pmod{12}[/tex]; taken modulo 13, the first term vanishes, but the second term leaves a remainder of 2. To counter this, we can multiply the second term by the inverse of 12 modulo 13, which is 12 since [tex]12^2\equiv144\equiv11\cdot13+1\equiv1\pmod{13}[/tex].
So, we found that [tex]N=5\cdot13+11\cdot12^2=1649[/tex], but the least positive solution is [tex]1649\equiv89\pmod{\underbrace{156}_{12\cdot 13}}[/tex], and in general we can have [tex]N=89+156k[/tex] for any integer [tex]k[/tex].
Now, since [tex]5000=32\cdot156+8[/tex], or [tex]4992=32\cdot156[/tex], we know that there are 32 possible integers [tex]N[/tex] that satisfy the congruences.
