document.write( "Question 842476: A baker packed 32 doughnuts into 3 boxes. She put a different number of doughnuts into each box and there was an even amount in each box.
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document.write( "How many doughnuts could there be in each box? \n" );
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Algebra.Com's Answer #507562 by josgarithmetic(39618)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Three boxes must distribute these 32 doughnuts. . \n" ); document.write( "Start with 10 and 12. That would be 22 doughnuts. Another with 8 will finish the 32. That is about the best way to distribute these 32 doughnuts. \n" ); document.write( "- \n" ); document.write( "Boxes each for 8, 10, 12 doughnuts\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "ANOTHER TRY \n" ); document.write( "16+12=28, and then third box to contain 4. \n" ); document.write( "This is not as neat as the first described combination.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "ANOTHER \n" ); document.write( "14+12=26 and then third box to contain 6. \n" ); document.write( "Still not as uniform as the first described combination. \n" ); document.write( " |