document.write( "Question 66493: Please help me solve this problem.\r
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\n" ); document.write( "\n" ); 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?
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Algebra.Com's Answer #47173 by josmiceli(19441)\"\" \"About 
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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
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