**20th Annual Lehigh University / AT&T High
School Math Contest**

March 4, 2000

Solutions

**Please note the corrections
in red.**

2. 0

3. 7 . Find the point of intersection of the lines *5r+7b = 69 *and
*9r+3b=57*,
which is *(r,b) = (4,7)*

4. –*3/4*.A line* L*, between the orgin and the point *(3,4)*
is perpendicular to the line we

want to find. Since* L* has slope *4/3*, the slope of the
given line is *–3/4*.

5. *7*. The sum of any two sides of a triangle must be larger
than the third side. The smallest value satisfying this constraint is *7*.

6. *69* cents.

7. *y = -2x + 6*

8.

10. 21. We will have one of two cases.

12. 2^{2000 }- 1. Every game has exactly 1 loser, so at any
point during the tournament the number of

games played equal the number of losers. There are 2^{2000}
players in the tournament and only one

winner. Hence there are 2^{2000 }- 1 losers.

14. 3. Divide the chips into three groups: *A*: the coins
on the left platform, *B*: the chips on the

right platform and *C*: the coins on neither platform. Start with
*A=5,
B=5* and *C=6*.

If *weight(A) < weight(B)* then the fake coin is in set A,
if *weight(A) > weight(B)* then the

fake coin is in set *B*. If *weight (A) = weight(B)* then
the fake coin is in set *C*. After the

first weighing we are able to eliminate at least *10* coins from
consideration. We then take

*A = 2, B=2*, and the rest in *C*. If *weight (A)<weight(B)
*then
the fake coin is in *A*, if

*weight (A) > weight(B)* the fake coin is in* B*, and if
*weight(A)
= weight(B)* then the fake

coin is in set *C*. After the second weighing we can eliminate
all but two coins, and these

can be placed on scale to determine the fake one.

15. 10. The number of seconds in 6 weeks equals 6 * 7 * 24 * 60 * 60
=

6 * 7 * 24 * 30 * 5! = 24 * 30 * 7 = 10!

19. 5. Going from step 4 to 5 is not valid since it requires dividing
by *(a-b)=0*.

**Note the answer is -1000/3**

21. *12,960 = 3*3!*6!* We must have one of the following
three cases where the

T's represent teachers and the S's represent students:

(i) T S S T S S T S S (ii) S T S S T S S T S (iii) S S T S S
T S S T

In each of these cases there are *3! *ways to arrange the teachers
and *6!* ways to

arrange the students. The answer is then *3*3!*6!*.

22. *5040. 10*9*8*7*

**Note this is ***t = r = 3*, not *t =
r = 2*

The diagram above gives us that

27. *(2,2000)*. (Modification from R1-5, 1985 ARML) Each line
is of the form

*y = mx + b = mx + (2000 - 2m)*, or *y – t = m (x-2)*. This
is the "point-slope" form of a line,

and all these lines go through *(2,2000)*. It is interesting to
consider sets of lines whose slope

and *y *- intercept are related in a given ways. For example what
property has the set of lines

for which *m*^{2} = b?

31. *1,499,500*. Suppose *24y+1= x*^{2}. *24y
+ 1* is not divisible by *2 *or *3*, so* x* is not divisible
by *2* or *3*. *24y - x*^{2} – 1 = (x + 1)( x - 1).
If x is not divisible by 2, then one of x+1 and x-1 is divisible by *2*
and the other is divisible by 4. If x is not divisible by *3*, then
exactly one of* x+1* or *x-1* is. Then *x*^{2}-1
is divisible by *24*, so there is one y for every x not divisible
by *2* or *3.* the *2000th* such *x *is *5999*.
So

35. 3

38.

To win the center of the penny must lie in the

square as shown in the circle on the right.

So the winning space is 16 * (1/16) = 1 square units, and the center
of the penny can

be anywhere on or inside the frame.

If the entire penny must lie on the board which has dimensions:

the center of the penny must lie in a

square.

39. *233.* Let *A*_{n} be the number of different
tilings of an *1 x n *rectangle. So *A*_{1 }= 1 and

*A*_{2} = 2. Any tiling of *A*_{n} where
*n
> 2* must begin with a *1x1* gray square or a *1x2 *black

rectangle. Hence we have that *A*_{n} = A_{n-1}+A_{n-2 }
This gives us the Fibonacci numbers

*A*_{1} = 1, A_{2} = 2, A_{3} = 3, A_{4}=
5,…, A_{12} = 233.