Check out

Mathcounts here--the best competition math for middle school mathletes.

Problem: (Solution below)

#1: 1993 Mathcounts National Team Round #4 :The teacher whispers positive integer A to Anna, B to Brett, and C to Chris. The students don't know one another's numbers but they do know that the sum of their numbers is 14. Anna says: "I know that Brett and Chris have different numbers." Then Brett says: "I already knew that all three of our numbers were different." Finally, Chris announces: "Now I know all three of our numbers." What is the product ABC?

#2: 2000 AMC10 #22: One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

Solution:

#1: For Anna to know that Brett and Chris have different numbers, she must have an odd number

because 14 - odd = odd and you can only get odd sum of two numbers if they are different, one odd

one even.

For Brett to say that he already know all three have different numbers, he not only must have an odd number, but also his number has to be larger or equal to 7 and is not the same as what A have.

Otherwise, A-B-C could be 1-1-12; 3-3-8; or 5-5-4.

It would exceed 14 if you have 7-7-__. (All numbers are positive) so Brett"s and Anna's numbers must

be different.

If Brett has 7, then the numbers could be A-B-C = 1-7-6 ; 3-7-4 or 5-7-2.

If Brett has 9, then the numbers could be A-B-C = 1-9-4; or 3-9-2.

If Brett has 11, then then numbers could be A-B-C = 1-11-2.

From the above possibilities you know Chris has to have 6 for him to be sure he knows all the numbers.

So A-B-C = 1-7-6 and the product of ABC = 1 x 7 x 6 = **42**
#2: Let there be m cups of mild, c cups of coffee. and x people in Angela's family.
According to the given, you can set up the following equation:
m +

c =

= 8. Two ways to solve this equation.

Solution I:

m +

c =

Both sides times 12

* x*

Both m and c need to be positive so the only *x* that works is* *when* ***x**** = 5**.

Solution II:

m +

c = 8, Both sides times 12 and you have

3m + 2c = 96; m + c is a multiple of 8.

30 3

28 6

26 9

.

.

.

2 45; from (30 + 3 ) = 33 to (2 +45) = 47 only 40 is a multiple of 8

and 40 divided by 8 = 5 so **5** is the answer.