document.write( "Question 594344: Prove that for all integers n, that the last digit of n^5 is the same as the last digit of n. \n" ); document.write( "

Algebra.Com's Answer #376707 by richard1234(7193) $\"\"$ $\"About$

You can put this solution on YOUR website!
Just check ten cases, where each case covers a residue class (mod 10). In other words, we can simply check n = 0, 1, ..., 9 and we are done.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "0^5 = 0
\n" ); document.write( "1^5 = 1
\n" ); document.write( "2^5 = 32 ≡ 2
\n" ); document.write( "3^5 = 243 ≡ 3
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "9^5 = 59049 ≡ 9\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "You can fill in the remaining exponents if you wish. Point is, n^5 is congruent to n (mod 10), so they must have the same last digit.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );