Crack The Number Code

The number can be written as

${ x }_{ 1 }\cdot { 99 }^{ 999 }+{ x }_{ 2 }\cdot { 99 }^{ 998 }+\dots +{ x }_{ 995 }\cdot { 99 }^{ 6 }+1\cdot { 99 }^{ 5 }+2\cdot { 99 }^{ 4 }+3\cdot { 99 }^{ 3 }+5\cdot { 99 }^{ 2 }+6\cdot 99+9\cdot { 99 }^{ 0 }$

The thing to realize here is that simple division is just repeated subtraction . It can be easily found out that every term other than the last one is divisible by 33 because $99 = 3 \cdot 33$ .

So to find the remainder, we just need to remove all the multiples of 33 from the number. We can write the number again but this time in a simplified way: $33n+9$ , for some integer $n$ . Since we just want the remainder, it is 9.

99 is divisible by 33. A number in base 99 with the last digit as 0 should be divisible by 33. Since the last number is 9, the closest number divisible by 33 in base 99 is the code with the last digit as 0 instead of 9. Thus the remainder is 9.