List all integers x satisfying x 2 mod 10 x 4 mod 11 and 100

List all integers x satisfying x 2 (mod 10), x 4 (mod 11), and 100 x 100

Solution

-2(mod10) = x

4(mod11) = x

then

10p + x = -2

11q + x = 4

then

x = -2 - 10p

x = 4 - 11q

then

4 - 11q = -2 - 10p

11q - 10p = 6

q = (6 + 10p)/11

x = -2(5p + 1)

and

-100 <= x <=100

-100 <= -2(5p + 1) <= 100

50 >= (5p + 1) >= -50

49 >= 5p >= -51

49/5 >= p >= -51/5

-10 <= p <= 9

from

q = (6 + 10p)/11

we calculate that p will be integer only when,

-10 <= p <= 9

-100 <= 10p <= 90

-94 <= 6 + 10p <= 96

-94/11 <= (6 + 10p)/11 <= 96/11

for integer solution

-8 <= q <= 8

-10 <= p <= 9

q = (6 + 10p)/11

Now, from hit and trial

p = -5 then q = -4 and x = 48

and

p = 6 then q = 6 and x = -62