# Find the last two digits of 9^{9^{9}}

Here is a cute example of the power of theory of congruences. Monster numbers can be tamed !!

Question :

Find the last two digits of $9^{9^{9}}$.

Solution:

A famous mathematician, George Polya said that a good problem solving technique is to solve an analagous less difficult problem.

So, for example, if the problem posed was “find the last two digits of 2479”. How do we go about it? Find the remainder upon division by 100. Now, how does it relate to congruences ? Modulo 100 numbers !

So, the problem reduces to — find out $9^{9^{9}} \equiv 9 \pmod {10}$.

Now, what is the stumbling block…the exponent $9^{9}$ makes the whole problem very ugly. But, $9^{9} \equiv 9 \pmod {10}$, which means $9^{9}-9=10k$, that is, $9^{9} = 9 + 10k$,

also, use the fact $9^{9} \equiv 89 \pmod {100}$

Hence, $9^{9^{9}}=9^{9+10k} = 9^{9}.9^{10k}$ $9^{9^{9}} \pmod {100} = 9^{9}.9^{10k} \pmod {100} \equiv 89. 9^{10k} \pmod {100}$

So, now we need to compute $9^{10k} \pmod {100} = (9^{10})^{k} \pmod {100} = (89.9)^{k} \pmod {100} = 89^{k}. 9^{k} \pmod {100}$

Hence, $9^{9^{9}} \pmod {100} \equiv 89^{k+1}.9^{k} \pmod {100} \equiv (89.9)^{k}.89 \pmod {100} \equiv (1)^{k}.89 \pmod {100} \equiv 89 \pmod {100}$.

-Nalin Pithwa.

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