Meet the Schlobodkins!!
A party is being held in the house of the Schlobodkins.There were 4 couples present in the house(other than Mr. and Mrs. Schlobodkin),and many,but not all,pairs of people shook hands.Nobody shook hands with anyone twice and nobody shook hands with his/her spouse.Both the host and the hostess shook some hands.
At the end of the party,Mr.Schlobodkin polls each person present to see how many hands each person(other than himself) shook.Interestingly each person gives a different answer.Determine the number of hands Mrs.Schlobodkin must have shaken.
The answer is 4.
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1 solution
Let us write S for the 2 Schlobodkins and denote each of the other couples as A,B,C and D. Nobody shook 9 hands since nobody shook the hands of his/her spouse.Therefore the numbers from 0 to 8 are used in describing the number of handshakes performed by each of the 9 people(excluding Mr.Scholbodkin).
Hence somebody must have shaken 8 hands.Say it is Mr.A.Now we try to find how many hands did Mrs.A shake???Each of the other couples shook hands at least once(since they must have shaken mr.A's hand.But somebody did shake 0 hands...it must be Mrs.A.
Now we eliminate Mr and Mrs A from consideration. Say Mrs.B shook exactly 7 hands.She did not hake hands with Mrs.B since no-one did.But to obtain 7 shakes she must have shaken hands with each of the other couples .But someone must have shaken 1 hand ,it must be Mr.B since each of the other people have shaken hands with both Mr.A and Mrs.B..
Continuing in this fashion,we see that the person who shook 6 is betrothed to the person who shook 2 hands.....5 hands is betrothed to 2 hands which leaves behind only Mrs.Schobodkin....who must have shaken 4 hands as it is the only number and the only number that cannot be paired in this case.....
So Mrs.Schlobokin shook only 4 hands....