ACE, BEA AND CEC ARE EACH GIVEN A POSITIVE INTEGER. THEY...

6. Ace, Bea and Cec are each given a positive integer. They do not know the

numbers given to the others, but are told that the sum of the three numbers is 15.

Ace announces that he can deduce that the other two have di ff erent numbers,

while Bea independently announces that she can deduce that no two of the three

numbers are the same. Hearing both announcement, Cec announces that he

knows all three numbers. What are they?

【Solution】

Let the numbers given to Ace, Bea and Cec be a, b and c respectively. Either all three

are odd, or exactly one is odd. From the announcement of Ace, a must be even as

otherwise Bea and Cec could have the same number. Similarly, b must also be even.

Moreover, it has to be at least 8 as otherwise Bea may have the same number as either

Ace or Cec. At this point, the possible combinations are (a, b, c)=(2, 8, 5), (4,8,3),

(6,8,1), (2,10,3), (4,10,1) and (2,12,1). We cannot have c=1 or 3 since Cec will not be

able to determine all three numbers. Hence c=5, b=8 and a=2.

ANS: a=2, b=8, c=5