Question

Find the last two digits of the number ${14}^{{14}^{14}}.$

Answer 1

${14}^{{14}^{14}}={196}^{\frac{{14}^{14}}{2}}=(200-4{)}^{{7.14}^{13}}$
The last two digits of this number will be the same as the last two digits of
$(-4{)}^{{7.14}^{13}}=4^{{7.14}^{13}}$

Now, notice that the last two digits of the even powers of $4$ have the following $5$ terms occurring periodically.

$\underset{}{16,56,96,36,76}$
So, last two digits of $4^2$ is $16$

So, last two digits of $4^4$ is $56$

So, last two digits of $4^6$ is $96$

So, last two digits of $4^8$ is $36$

And, last two digits of $4^{10}$ is $76$

And this is repeated for $4^{12},4^{14}\dots \dots$

Now, $4^{{7.14}^{13}}$ has an even exponent to $4$. Now we just need to find the last digit in this exponent and match it with one of the five cases listed above.

The last digit in ${7.14}^{13}$ is the last digit in $7.(10+4{)}^{13}$

Which is the same as last digit in $7\times 4^{13}$

Which is the same as last digit in $28\times 4^{12}$

Using the list above,

It is the same as last digit in $28\times 16$ which is $8$

Again using the list above, we have the final result as $36$

Hence, answer is $36$.