Finding last two digits of $14^{ 14^{\large 14}}\!$ without mod method

Question

For $ 14^{\large 14^{\Large 14}}\!$ using mod method I know the solution to find the last two digits. Even on this site I have found the solution but with mod method. But without using mod method can we solve this?

Answer 1

Hint:

The number is clearly divisible by $4$

So, let us find the reminder when divided by $25$

$a=14^{14^{14}}=(1-15)^{14^{14}}=1-\binom{14^{14}}115+$ terms divisible by $25$

Now as $5$ divides $15,$ let us the reminder of $14^{14}$ when divided by $5$

$14^{14}=(1-15)^{14}=1+$ terms divisible by $5$

Let $14^{14}=1+5u$ for some integer $u$

$15\cdot14^{14}=15(1+5u)=15+25(3u)$

So, $a=1-15-25(3u)=11+25(-3u-1)$

and $a$ is divisible by $4$

Hence $a=25+11$ when divided by $100$