document.write( "Question 27147: Formulate a conjecture for which natural number n the formula
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document.write( "(a+b)^(n) = a^n + b^n holds in clock n. I don't have to prove the conjecture, just state it and suffices to consider clock n for n< and equal to 10 \n" );
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Algebra.Com's Answer #15271 by venugopalramana(3286)![]() ![]() You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLE \n" ); document.write( "----------------------------------------- \n" ); document.write( "(a+b)^7 = a^7 + b^7 in clock 7 \n" ); document.write( "IN CLOCK 7 THE NUMBERS REPEAT CHANGE DIGITS AFTER 7 \n" ); document.write( "THAT IS 12=5(MOD7)...NUMBER TWELVE WILL BE REPRESENTED BY 5 THE REMAINDER AFTER DIVIDING 12 BY 7. \n" ); document.write( "LET A=X(MOD7)......................I \n" ); document.write( "AND B=Y(MOD7).......................II.......WHERE X AND Y ARE LESS THAN 7. \n" ); document.write( "A+B=(X+Y)(MOD7).............FROM I+II \n" ); document.write( "(A+B)^7={(X+Y)+(X+Y)+....7 TIMES}(MOD7)=7(X+Y)(MOD7)=(X+Y)(MOD7)............III \n" ); document.write( "A^7=(X+X+X...7 TIMES)(MOD7)=(7X)(MOD7)=X(MOD7)......FROMM I ................IV \n" ); document.write( "SIMILARLY..B^7=Y(M0D7).............FROM II..................................V \n" ); document.write( "ADDING IV AND V WE GET \n" ); document.write( "A^Y+B^7=X(M0D7)+Y(MOD7)=(X+Y)(MOD7).................................VI \n" ); document.write( "HENCE FROM III AND VI WE GET \n" ); document.write( "(a+b)^7 = a^7 + b^7 in clock 7 \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |