2000 High State Mathematics Contest

10:29 AM 1/12/01 2000 High State Mathematics Contest

Integer Answer Solutions

Of course you could list all of the pairs of factors of 2000 and go through the list to see that 40 and 50 give you the smallest sum. If you use the inequality that relates the geometric mean to the arithmetic mean, you see that the geometric mean

. This tells us that the sum must be greater that 89. With this in mind, it is easy to pick the 40 and 50, the two closest factors to the square root of 2000.

3142

Let S be the side of the square and R the radius of the circle. Since

, we know that

, so

. The smallest integer value for the area then is 3142.

The amount in John's account will be

while Mary will have

in her account. We need to find when

, or

. Since

, this would be a good place to start checking. A table of values in your calculators, starting a 10, shows that by the 14^th month, John's total has surpassed Mary's.

The parabola is

. We are looking for ordered pairs (m,n), where

and both m and n are integers between -2000 and 2000 (inclusive). Since the n-coordinate will grow quickly and will never be negative, we will use it to find possible integer values for n. First,

. This implies that

, which implies that

. So we know that

, so

. There are 63 integers, beginning with -29 and ending with 33.

705

Let Q, D, and N represent the number of quarters, dimes and nickels, respectively. First we know that

. The correct amount would be

. If the number of dimes and quarters were exchanged, the amount would be

and if the number of nickels and dimes were exchanged, the amount would be

. Solving this system of 4 equations in variables give 15 quarters, 21 dimes and 24 nickels for a total to 705 cents. [This problem is an extension of a problem that appeared in a Peanuts strip and appears in memory of Charles M. Schulz.]

To be divisible by 11, the different between the sum of the 1^st, 3^rd and 5^th digits and the sum of the 2^nd and 4^th digits must be divisible by 11. In this case, the only possible difference is 0, which results when both sums are 10. Try 11 and -11 to see why they can't work. To get the second and fourth digit to sum to 10, there are only 2 possibilities, both using 6 and 4. There are 2 ways to place these digits and 3!=6 ways to place the remaining 3 digits for a total of 2!3!=12.

This is another system of equations. (1)

, and

. Solving this system we find the solutions T=30, D=18, and H=25, as well as the extraneous solution T=50, D=8, and H=15, which does not work since T<40.

861

Let d, q, and h represent the number of dimes, quarters, and half dollars respectively. The total value of these coins is

, which simplifies to

. This implies that q must be even, so let

. Now

, so

. This now implies that d must be divisible by 5, so let

. Now we have

. The problem now is reduced to finding the number of ways to pick three non-negative integers that sum to 40. This problem can quickly be solved using combinatorics. Think of 40 red markers representing the sum and 2 blue markers which we will use to separate the 40 into three subsets. We now have a total of 42 objects, with one group of 40 undistinguishable objects and one group of 2 undistinguishable objects. There are a total of 42! Permutations, but because there are groups of undistinguishable objects, we must divide that by 2! And 40!.

. Notice that this generalizes to C(42,2).

16384

Look first at a simpler case. For 2 cards, (0,1), it only takes on step to get (0,0). For 3 cards (0,1,2) it take 3 steps (0,0,2), (0,0,0,1), and (0,0,0,0) to get 4 cards. For 4 cards (0,1,2,3) we see that it takes 7 steps to get 8 zeros. Using Induction, we can generalize and prove that the formula for the number of steps is

and the number of zeros will be

. For N = 15, this would be

10.

In the grid the numbers indicate the possible paths to each lattice point.

You could also consider the paths by the number of steps it takes to reach the point (3,3). The following table summarizes these results.

Steps	3	4	5	6
Paths	C(3,3) = 1	C(4,2)C2,1)=12	5C(4,2)=30	C(6,3)=20