Permutation of numbers and function

Question

Let $N(x)$ denote the number of terminating zeroes of the number x. If A,B,C is a permutation of the numbers $211^{19}+9$, $9^{101}-9$, and $19^{111}-9$ such that $N(A)<N(B)<N(C)$, determine the ordered triple $(A,B,C)$.

Answer 1

$$211^{19}+9=1^{19}+9=0\mod 10$$ Try again: $$211^{19}+9=11^{19}+9=11\cdot (11^2)^{9}+9=11\cdot 21^{9}+9=(11\cdot 21)(21^2)^4+9=31\cdot 41^4+9=31\cdot 81^2+9=31\cdot 61+9=0\mod 100$$

Here, I repeatedly squared, reduced modulo $100$ before squaring again in order to never work with too big numbers.

So it has at least 2 zeroes. Continue until you get nonzero. It's not the fastest solution, but in this situation it works.

Answer 2

Hint : $211^{19}=(1+210)^{19}=1+ \binom{19}{1}210+\binom{19}{2}210^2...$

Google Binomial Theorem if u didn't get that step.

Last Digit : $1$ // All other terms have at least one zero at end

Second Last : $9$ // see second term. last is 0. second last is 9 as u multiply 19 by 210. all other terms have at least 2 zeros at end.

Third Last : $9+1$=$0$ //third last in second term is 9(19*210) and we also have to look for third term as it had only 2 zeroes and we want third last digit. Rest terms have more than three zeroes

You need not compute full 2nd and 3rd terms only sufficient last digits.

You should stop here as when you add 9 that will become 1.

Can you proceed for other numbers?