Question

Please explain your answer properly.

Answer 1

Answer:

$\boxed{\boxed{\mathsf{299999}}}$

I won't explain, this will do my job:

Binomial Theorem

Any expansion of $(a-b)^n$ can be written as :

$(a-b)^n=a^n+^nC_1a^{n-1}b+.......^nC_{n-1}ab^{n-1}+b^n$

$29^2=841$ which is obviously odd.We can hence come to the conclusion that the power of 299999 is odd.Let us say the whole thing $29999^{2999^{299^{29^{2}}}}$ that is in the power as N.

The whole given question gets transformed to :

$299999^N$ which can be written as :

$(300000-1)^N$

By Binomial theorem we have :

$300000^N+........300000\times 1^{N-1}-1^N$

Observe that all the numbers except the last two will be of tremendously high value and will exceed 10 digits (I am not kidding with you okay?)

So,with a little bit of smartwork,you can easily say that the last digit will be :

$300000 - 1\\\\= \boxed{299999}$

That,my friend is the answer.