how do you find all the solutions of:?

Question

x = 3 mod 49 and x = 5 mod 99

Answer 1

Not being crash-hot on modular arithmetic,
I would tackle it this way -

x = 3 mod 49 means (x - 3) / 49 = J (J an integer)
x = 5 mod 99 means (x - 5) / 99 = K (K an integer)

From these two equations, we find that:
x = 49J + 3 and x = 99K + 5

Combining gives: 49J + 3 = 99K + 5
or, J = (99K + 2) / 49 = 2K + (K + 2) / 49.

{ EDIT:
(Note here, that I've arranged the equation so
that the smaller number of 49 and 99 is to be
the divisor. That way, I can extract an integer
(2K) to simplify the later process) }

Now J and 2K are integers, then so must be (K + 2) / 49.
Let (K + 2) / 49 = N (N an integer)
Therefore, K = 49N - 2

Substituting back, we find that:
x = 99K + 5 = 99(49N - 2) + 5 = 4851N - 193

Now we can substitute any integer for N
and get an appropriate integer for x.

Answer 2

i think 4658 would satisfy both equations

or x would be of the form 99k +5

where k is an integer of the form M(49) -2

M(49) refers to integral multiple of 49

note: i am rather rusty in number theory