How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?

The equivalence 

means that n-5 is a multiple of 12.

that is

n-5=12k, for some integer k

and so

n=12k+5

for k=-1, n=-12+5=-7

for k= 0, n=0+5=5 (the first positive integer n, is for k=0)

we solve 5000=12k+5 to find the last k

12k=5000-5=4995

k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009

The last k which produces n<5000 is 416

For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.

417

Not sure what the question is but i hope this x = years100 * x