${\overline{...y1}}^{\overline{y1}!}$

Number Theory Level 3

What is the value of the last two digits of $n^{n!}$ , where $n$ is a positive number greater than $2^{2016}$ , the last digit being $1$ , and the second last digit being an integer $y$ ?

Please select the best answer.

10y+1

0

01

\overline{y1}

(10y+1)\:mod\: 100

1

1 solution

Efren Medallo
Jul 18, 2016

The general form for the last two digits of a number ending in 1 raised to any number is as follows:

${\overline {...ab1}}^{\overline{...xyz}} = (10bz + 1) \: mod\: 100$

For a less mathematical explanation, it basically means that the units digit is 1, and the tens digit is the units digit of the product of the tens' digit of the base and the units' digit of the exponent.

For $n^{n!}$ with $n$ ending in 1, assuming the tens' digit of $n$ to be $y$ , we know that the exponent would have many trailing zeroes, thus we know that at least its last digit is a zero. That makes the last two digits of $n^{n!}$ to be $01$ .

However , as far as the question is concerned, the value of the last two digits is demanded. That is,

$0\: \times \: 10 \: + \: 1 \: \times \: 1 = \boxed{1}$

. . . y 1 ‾ y 1 ‾ ! {\overline{...y1}}^{\overline{y1}!} . . . y 1 ​ y 1 ​ !

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${\overline{...y1}}^{\overline{y1}!}$