Question 6

The last two digits of the expression $$1(1!)^{1!} + 2(2!)^{2!} + 3(3!)^{3!} + .... + 121(121!)^{121!}$$

Solution

From the 5th term onwards the last two of all the terms will 00.

The last two digits of 24 to the power of an even number will be 76 always.

So, last two digits of the above expression will be the last two digits of $$1+2\left(2\right)^2+3\left(6\right)^6+4\left(76\right)$$

=1+8+139968+304= 140281

So last two digits is 81

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