document.write( "Question 1144827: Letter A substitutes a nonzero digit and AA is a two-digit number. The product of AA and A is always.
\n" ); document.write( "F) even
\n" ); document.write( "G)divisible by 3
\n" ); document.write( "H)divisible by 11
\n" ); document.write( "J) a perfect square \n" ); document.write( "
Algebra.Com's Answer #766047 by richwmiller(17219)\"\" \"About 
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The problem ask about the product of AA*A?
\n" ); document.write( "1*11=11 11/11=1
\n" ); document.write( "2*22=44 44/11=4
\n" ); document.write( "3*33=99 99/11=9
\n" ); document.write( "4*44=176 176/11-16
\n" ); document.write( "5*55=275 275/11=25
\n" ); document.write( "6*66=396 396/11=36
\n" ); document.write( "7*77=539 539/11=49
\n" ); document.write( "8*88=704 704/11=64
\n" ); document.write( "9*99=891 891/11=81\r
\n" ); document.write( "\n" ); document.write( "F) even
\n" ); document.write( "G)divisible by 3
\n" ); document.write( "H)divisible by 11
\n" ); document.write( "J) a perfect square
\n" ); document.write( "F is not a solution since 11 is not even
\n" ); document.write( "G is not a solution since 11 is prime
\n" ); document.write( "H IS A SOLUTION since all can be divided by 11
\n" ); document.write( "I is not a solution since 11 is prime
\n" ); document.write( "This should be easily understood from Edwin's solution. He showed that all AA are multiples of 11
\n" ); document.write( "Since all AA are divisible by 11 then AA multiplied by anything else is also divisible by 11.
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