Do the following congruence equations have solutions? If so how many? Do not solve the equations.

Question

a. 4𝑥 ≡ 7(𝑚𝑜𝑑 16)

b. 5𝑥 ≡ 12(𝑚𝑜𝑑 32)

Answer 1

(a) 4x ≡ 7 mod 16 means 4x = 16 y + 7 for some integer y.

Or 4x-16y = 7. Is this possible. Well clearly the left side is divisible by 4. But the right side isnt. So that gives a contradiction. So no solution.

(b) 5x ≡ 12 mod 32

Well 5x = 32 y + 12. Clearly 32y + 12 has to be even, so that means x has to be even. So that means 32y + 12 will be a multiple of 10. If we find one solution of this 10r = 32y+12 then 32 (y+5) + 12 is also a multiple of 10. y = 1, 5x = 44, y = 2, 5x = 76, y=3, 5x=108, y=4 5x=140, giving x= 28. So x = 28 is one solution. y = 9 gives 32y+12 = 300 and x = 60, y = 14 gives x =92. So x = 28, 60, 92 ,,,, are solutions.

Mike