Question 66493: Please help me solve this problem.
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?
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! Picture the group of principals in a room.
Pick one principal to shake all the others hands.
He will shake n-1 hands, where n = himself plus
all the others.
The next principal doesn't need to shake hands
again with the 1st one, so he shakes n-2 hands.
The next principal shakes n-3 hands, and so on.
The total number of handshakes is 21, so
(n-1) + (n-2) + (n-3) + . . . = 21
Try different values of n that will not lead to a result
less than 21 or a negative result
How about n = 1? That's 0 handshakes.
How about n=2
(2-1) + (2-2) = 1 Just 1 handshake
n = 3
(3-1) + (3-2) + (3-3) = 3 handshakes
n = 4
(4-1) + (4-2) + (4-3) + (4-4) = 6 handshakes
n = 5
(5-1) + (5-2) + (5-3) + (5-4) + (5-5) = 10 handshakes
n = 6
(6-1) + (6-2) + (6-3) + (6-4) + (6-5) + (6-6) = 15 handshakes
n = 7
(7-1) + (7-2) + (7-3) + (7-4) + (7-5) + (7-6) + (7-7) = 21 handshakes
So, there are 7 principals
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