Question 360515
What are the final two digits of 7<sup>1997</sup>?
<pre>
7<sup>0</sup> = <font color = "red">1</font> (consider that as <font color = "red">01</font>)

7<sup>1</sup> = <font color = "red">7</font> (consider that as <font color = "red">07</font>)

7<sup>2</sup> = <font color = "red">49</font>

7<sup>3</sup> = 3<font color = "red">43</font>

7<sup>4</sup> = 24<font color = "red">01</font>

7<sup>5</sup> = 168<font color = "red">07</font>

7<sup>6</sup> = 1176<font color = "red">49</font>

7<sup>7</sup> = 8235<font color = "red">43</font>

7<sup>8</sup> = 57648<font color = "red">01</font>

...


So we see that the last two digits of powers of 7 are

01, 07, 49, 43, 01, 07, 49, 43, 

ad the last two digits continue to repeat in groups of 4 forever. 

So to find the last two digits of any positive integral power of

7, we can just divide the power by 4 and take only the remainder.

If the remainder is 0, the last two digits are 01,
If the remainder is 1, the last two digits are 07,
If the remainder is 2, the last two digits are 49,
If the remainder is 3, the last two digits are 43,

To find 7<sup>1997</sup>, we divide 1997 by 4

        <u>  499</u>  
       4)1997
         <u>16</u>
          39 
          <u>36</u>
           37 
           <u>36</u>
            1

The remainder is 1, so the last two digits of 7<sup>1997</sup> are 07.

Edwin</pre>