2026/078) Find the remainder when we divide $3^{33}-2$ by 18

We need to factor 18 into product of co-primes $18 = 9 *2 $

Now let us find  $3^{33}-2$ mod 9 and $3^{33}-2$ mod 2 then we shall combine both

Let us proceed one by one 

As $3^{33}$ is  divisible by $3^2=9$ we have 

  $3^{33} \equiv 0 \pmod 9$

So $3^{33} -2 \equiv -2 \pmod 9$ 

Adding 9 on RHS to make is positive

so $3^{33} -2 \equiv 7 \pmod 9$ 

$3^{33} -2$ is odd

So $3^{33} -2 \equiv 1 \pmod 2$ 

To find the remainder when divided by 18 we to find the number $x \lt 18$ such that 

$x \equiv 7 \pmod 9$ 

$x \equiv 1 \pmod 2$ 

This can be solved using Chinese Remainder Theorem but as 2 and 9 are small numbers we can solve

Simply by taking numbers which are 7 mod 9 and checking it is 1 mod 2 or odd.

The number should be less than 18. the 2 numbers are 7 and 16 and 7 is odd

So $3^{33} -2 \equiv 7 \pmod {18}$