document.write( "Question 66493: Please help me solve this problem.\r
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document.write( "Toward the end of the principals' meeting for the Drug-Free Community campaign project, a total of 21 handshakes were exchanged. Assuming each principal shakes hands once with the other principals, how many principals were present? \n" );
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Algebra.Com's Answer #47173 by josmiceli(19441)![]() ![]() You can put this solution on YOUR website! Picture the group of principals in a room. \n" ); document.write( "Pick one principal to shake all the others hands. \n" ); document.write( "He will shake n-1 hands, where n = himself plus \n" ); document.write( "all the others. \n" ); document.write( "The next principal doesn't need to shake hands \n" ); document.write( "again with the 1st one, so he shakes n-2 hands. \n" ); document.write( "The next principal shakes n-3 hands, and so on. \n" ); document.write( "The total number of handshakes is 21, so \n" ); document.write( "(n-1) + (n-2) + (n-3) + . . . = 21 \n" ); document.write( "Try different values of n that will not lead to a result \n" ); document.write( "less than 21 or a negative result \n" ); document.write( "How about n = 1? That's 0 handshakes. \n" ); document.write( "How about n=2 \n" ); document.write( "(2-1) + (2-2) = 1 Just 1 handshake \n" ); document.write( "n = 3 \n" ); document.write( "(3-1) + (3-2) + (3-3) = 3 handshakes \n" ); document.write( "n = 4 \n" ); document.write( "(4-1) + (4-2) + (4-3) + (4-4) = 6 handshakes \n" ); document.write( "n = 5 \n" ); document.write( "(5-1) + (5-2) + (5-3) + (5-4) + (5-5) = 10 handshakes \n" ); document.write( "n = 6 \n" ); document.write( "(6-1) + (6-2) + (6-3) + (6-4) + (6-5) + (6-6) = 15 handshakes \n" ); document.write( "n = 7 \n" ); document.write( "(7-1) + (7-2) + (7-3) + (7-4) + (7-5) + (7-6) + (7-7) = 21 handshakes \n" ); document.write( "So, there are 7 principals \n" ); document.write( " |