Counting II : Practice Counting Systematically

Counting Coins 
Lots of similar questions appear on Mathcounts tests. Be careful when there are limits, for example, the sum of the coins do not exceed ___ or you have to have at least one for each type, etc...

Partition 

The Hockey Stick Identity from Art of Problem Solving
Same as "Sticks and Stones", or "Stars and Bars" methods

Applicable question: Mathcounts 2008 Chapter #9--During football season, 25 teams are ranked by three reporters (Alice, Bob and Cecil). Each reporter assigned all 25 integers (1 through 25) when ranking the twenty-five teams. A team earns 25 points for each first-place ranking, 24 points for each second-place ranking, and so on, getting one point for a 25th place ranking. The Hedgehogs earned 27 total points from the three reporters. How many different ways could the three reporters have assigned their rankings for the Hedgehogs? One such way to be included is Alice - 14th place, Bob - 17th place and Cecil - 20th place.

Solution I :
Let's see how it could be ranked for Hedgehogs to get 27 points from the three reporters.

A          B         C

1           1        25     1 way for C to get 25 points and the other two combined to get 2 points

1           2        24

2           1        24    2 ways for C to get 24 points and the other two combined to get 3 points.

1           3        23

3           1        23

2           2        23    3 ways for C to get 23 and the other two combined to get 4 points.

.

.

25        1         1     25 ways for C to get 1 point and the other two combined to get 26 points.

1 + 2 + 3 + ...25 = \(\dfrac {25\times 26} {2}=325\)

Solution II:

Use 26C2. Look at this questions as A + B + C = 27 and A, B C are natural numbers. To split the objects into three groups (for Alice, Bob, and Cecil), we must put 2 dividers between the 27 objects. (You can't grant "0" point.) There are 26 places to put the dividers, so 26C2 and the answer is \(\dfrac {26\times 25} {2}=325\)

2013 Mathcounts State Prep: Partition Questions

#24 2001 Mathcounts Sate Sprint Round: The number 4 can be written as a sum of one or more natural numbers in exactly five ways: 4, 3+1, 2 + 1 + 1, 2 +2 and 1 + 1 + 1 + 1; and so 4 is said to have five partitions. What is the number of partitions for the number 7?

#2: Extra: Try partition the number 5 and the number 8. 

Solution: 

#24: You can solve this problem using the same technique as counting coins:

Counting coins questions

7     6    5    4    3    2    1

1                                           1 way

       1                                    1 way

             1                 1           2 ways ( 5 + 2 or 5 + 1 + 1)

                   1    1                  1 way

                   1           1           2 ways ( 4 + 2 + 1 or 4 + 1 + 1 + 1)

                         2     0   1      1 ways

                         1     2           3 ways ( 3 + 2 + 2, 3 + 2 + 1 + 1 and 3 + 1 + 1 + 1 + 1)

                                3           4 ways (2 + 2 + 2 + 1, 2 + 2 + 3 ones, 2 + 5 ones and 7 ones.)

Total 15 ways.

The partitions of 5 are listed below (There are 7 ways total.):

5   4   3   2   1

1                           1 way

     1                      1 way

          1    1           2 ways  (3 + 2 and 3 + 1 + 1)

                2           3 ways  (2 + 2 + 1, 2 + 1 + 1 + 1 and 1 + 1 + 1 + 1 + 1)

There are 22 ways to partition the number 8.

2013 Mathcounts State Prep: Harder State Questions

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

2004 Mathcounts State Sprint #19: The points (x, y) represented in this table lie on a straight line. The point (13, q) lies on the same line. What is the value of p + q? Express your answer as a decimal to the nearest tenth. 

 

Solution: 

#19:  Look at the table and you'll see each time x + 2, y would -3. 

-5 to -14 is (-9), three times (-3) so p = 2 + 3 x 2 = 8

p would = 13 when 2 + 5.5 * 2 = 13 so q = -5  +  (5.5) * (-3) = -21.5

p + q = -13.5

2004 Mathcounts State Sprint #24: The terms x, x + 2, x + 4, ..., x + 2n form an arithmetic sequence, with x an integer. If each term of the sequence is cubed, the sum of the cubes is - 1197. What is the value of n if n > 3?

Solution: 

The common difference in that arithmetic sequence is 2 and the sum of the cubes is -1197, which means that these numbers are all odd numbers. (cubes of odd numbers are odd and the sum of odd terms of odd numbers is odd. )

(-5)³ + (-7)³ + (-9)³ = -1197  However, n is larger than 3 (given) so the sequence will look like this:

 (-9)³+ (-7)³+ (-5)³+ (-3)³+ (-1)³ + (1)³ + (3)³ = -1197

 x = - 9 and x + 2n = 3, plug in and you get -9 + 2n = 3;  n = 6

2013 Mathcounts Basic Concept Review: Some Common Sums/numbers

These are some common sums that appear on Mathcounts often.

\(1+2+3+\ldots +n=\dfrac {n\left( n+1\right) } {2}\)

 \(2+4+6+\ldots .2n=n\left( n+1\right)\)

\(1+3+5+\ldots .\left( 2n-1\right) =n^{2}\)

The above are all arithmetic sequences.

The sum of any arithmetic sequence is average times terms (how many numbers).
Besides, the mean and median are the same in any arithmetic sequence.
Combining these knowledge, along with distributive rules some times
(case in point, sum of multiples of n, etc...) will expedite the calculation.

The "nth" triangular number is \(\dfrac {n\left( n+1\right) } {2}\)

The sum of the first n triangular numbers is \(\dfrac {n\left( n+1\right) \left( n+2\right) } {6}\).

 \(1^{2}+2^{2}+3^{2}+\ldots +n^{2}\) = \(\dfrac {n\left( n+1\right) \left( 2n+1\right) } {6}\)

\(1^{3}+2^{3}+3^{3}+\ldots +n^{3}=\)\(\left[ \dfrac {n\left( n+1\right) } {2}\right] ^{2}\)

Harder Geometry Question from National Mathcounts

Question: Similar to a 96 National Mathcounts question: Circles of the same colors are congruent and tangent to each other. What is the ratio of the area of the largest circle to the area of the smallest circle?

Solution :
The most difficult part might be to find the leg that is "2R - r".
R and r being the radius of the median and the smallest circle.

Using Pythagorean theorem, you have \(\left( R+r\right) ^{2}= R^{2}+\left( 2R-r\right) ^{2}\)
\(R^{2}+2Rr+r^{2}=R^{2}+4R^{2}-4Rr+r^{2}\)
Consolidate and you have 6Rr = \(4R^{2}\)\(\rightarrow\) 3r = 2R \(\rightarrow\) r =\(\dfrac {2R} {3}\) The ratio of the area of the largest circle to the area of the smallest circle is thus \(\rightarrow\)\(\dfrac {\left( 2R\right) ^{2}} {\left( \dfrac {2R} {3}\right) ^{2}}\) = 9 .