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 .

2013 Mathcounts State Prep: Counting and Probability

Question #1: Rolling two dice, what is the probability that the product is a multiple of 3?
Solution I :
As long as one of the two numbers turn up as 3 or multiple of 3 (in this case, a "6"), the product of the two numbers will be a multiple of 3.
There are 6 * 6 = 36 ways to get 2 numbers. Out of the 36 pairs, you can have
3 - 1
3 - 2
3 - 3
3 - 4
3 - 5
3 - 6, 6 ways.
However, there are only 5 ways left if the other die has a 3 since 3 - 3 only counts as once.
1 - 3
2 - 3
4 - 3
5 - 3
6 - 3
Next we look at "at least one number is "6".
6 - 1
6 - 2
6 - 4 (We already used 6 - 3)
6 - 5
6 - 6 so 5 ways.
The other way around, we have
1 - 6
2 - 6
4 - 6
5 - 6  Total 4 ways, so the answer is \(\frac{\Large{( 6 + 5 + 5 + 4 )}}{\Large{36}}\) = \(\frac{\Large{ 20}}{\Large{36}}\) = \(\frac{\Large{5}}{\Large{9}}\).

Solution II: 
The easiest way to solve this problem is to use complementary counting, which is 1 (100% or total possible way) - none of the the multiples of 3 showing up, so 1 - \(\frac{\Large{4}}{\Large{6}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{2}}{\Large{3}}\) * \(\frac{\Large{2}}{\Large{3}}\)=  \(\frac{\Large{5}}{\Large{9}}\)

Question #2: [2002 AMC-12B #16] Juan rolls a fair regular eight-sided die. Then Amal rolls a fair regular six-sided die. What isthe probability that the product of the two rolls is a multiple of 3?
Solution:
Using complementary counting (see solution II of the previous question), 1 - \(\frac{\Large{6}}{\Large{8}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{3}}{\Large{4}}\) * \(\frac{\Large{2}}{\Large{3}}\)=  \(\frac{\Large{1}}{\Large{2}}\)