unc16 solutionsbyrcv
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 1
February 1, 2016
1. Even on the easiest problem on a contest, I often make a sketch and/or copy the information from the
question. I find it helps me to internalize the problem, and to reduce "dumb" mistakes. Since this is a right
triangle, I immediately think of the Pythagorean Theorem.
a + b + c = 56
a + b 2 + c 2 = 1250
2
a2 + b2 = c2
Solution 1: Pythagorean Triplets. (A nice hint from the name of the question.) Combine the last two equations
written above.
c 2 + c 2 = 1250
c 2 = 625
c = 25
The most well-known Pythagorean Triplet is the 3-4-5 triangle. If we scale this triangle by a factor of 5, we
will have the correct hypotenuse. That would give us a 15-20-25 triangle. But the sum of sides is
15+20+25=60. And that doesn't meet the other specification of the problem.
The first series of Pythagorean Triplets continues with odd sides: 3-4-5, 5-12-13, 7-24-25, 9-40-41, ... . The
7-24-25 triangle has the correct hypotenuse. Check the sum of the sides, 7+24+25=56. This meets all the
criteria of the question. Now, we can answer the question: "What is the area?" Using the formula for the area
of a triangle, we get the final answer:
A = 12 base × height = 12 × 24 × 7 = 12 × 7 = 84
Solution 2: Algebra. First, we solve for c. Then we solve for a+b. A technique that sometimes helps with
right triangle problems that don't seem to give enough information is to square the sum or difference of two
sides.
c 2 + c 2 = 1250
c 2 = 625
c = 25
a + b + c = 56
a + b + 25 = 56
a + b = 31
(a + b )2 = 312
a 2 + 2ab + b 2 = 961
We already found a 2 + b 2 = c 2 = 625 . Subtracting the last two equations leaves us with the area of 4 congruent
triangles. 2ab = 961 − 625 = 336 ⇒ 12 base × height = 12 ab = 84 .
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 2
February 1, 2016
2. The factorial of n is the product of all of the positive integers from n down to 1. The first step is to break the
expression down starting from the innermost parentheses.
4!= 4 × 3 × 2 × 1 = 24
(4!)!= 24!= 24 × 23 × 22 × L × 3 × 2 × 1
3!= 3 × 2 × 1 = 6
(3!)!= 6!= 6 × 5 × 4 × 3 × 2 × 1
[6!]4 = (6 × 5 × 4 × 3 × 2 ×1)4
We needed to multiply out 4! and 3!. But we must resist the urge to multiply out 24! and 6!. Why? First,
quotients involving factorials are often easier to divide when they are already factored. Second, this question is
specifically asking for the "prime factorization". So, anything we multiply will have to be factored, later.
Probably the most straightforward way to solve this is to write out all the factors in the numerator and all the
factors in the denominator and "cancel" out factors.
24 × 23 × 22 × 21 × 20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
6 × 5 × 4 × 3 × 2 × 1 × 6 × 5 × 4 × 3 × 2 × 1 × 6 × 5 × 4 × 3 × 2 × 1× 6 × 5 × 4 × 3 × 2 × 1
This is a perfectly fine solution method. A subtle enhancement is to group the numerator into 4 sets of 6
consecutive integer factor. Each group of 6 factors in the numerator is evenly divisible by 6!. (Think about
why that's true, as we cancel factors.)
(24 × 23 × 22 × 21× 20 × 19) × (18 × 17 ×16 × 15 ×14 × 13)× (12 × 11× 10 × 9 × 8 × 7 )× (6 × 5 × 4 × 3 × 2 × 1)
(6 × 5 × 4 × 3 × 2 × 1)× (6 × 5 × 4 × 3 × 2 × 1) × (6 × 5 × 4 × 3 × 2 ×1)× (6 × 5 × 4 × 3 × 2 × 1)
=
24 × 23 × 22 × 21× 20 × 19 18 × 17 × 16 × 15 × 14 × 13 12 × 11× 10 × 9 × 8 × 7 6 × 5 × 4 × 3 × 2 × 1
×
×
×
6 × 5 × 4 × 3 × 2 ×1
6 × 5 × 4 × 3 × 2 ×1
6 × 5 × 4 × 3 × 2 ×1
6 × 5 × 4 × 3 × 2 ×1
The rightmost fraction completely cancels. Rearranging the factors in the other denominators:
× 23 × 22 × 21 × 20 × 19
24
6× 4 ×
18
1
×
2
3
×
5
× 17 × 16 × 15 × 14 × 13
6×3 ×
12
×
×
1
×
4
5
×
2
× 11 × 10 × 9 × 8 × 7
6× 2 ×
1
×
5
× 3 × 4
= 1 × 23 × 11× 7 × 4 × 19
= 1 × 17 × 4 × 3 × 7 × 13
= 1 × 11× 2 × 3 × 2 × 7
Now, everything is completely factored, except 4 = 2 2 , so we can directly write the final answer:
23 × 19 × 17 × 13 × 112 × 7 3 × 32 × 2 6
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 3
February 1, 2016
3. First, if the question had not provided a reasonably nice sketch, that is the very first thing I would draw.
Even with the given perspective sketch, I still need to draw and label a face of the cube and an equilateral
triangle that replaced a corner.
The area of the face of the cube was originally the area of a square of side-length 1. But we shaved off four
isosceles right triangles, each with an unknown side length of x. So, the remaining area of the face of the square
is 12 − 4 × 12 x 2 = 1 − 2 x 2 . The hypotenuse of the isosceles right triangle has the same length as the side-length of
the equilateral triangles we created in place of the corners, s = 2 x .
A student may know the area of an equilateral triangle in terms of its side-length. If not, divide the equilateral
triangle vertically into a 30-60-90 triangle. The ratio of the side-lengths of any 30-60-90 triangle is 1 : 3 : 2 .
All of the side-lengths can be found by proportions if any of the three side-lengths are known. Or, if these
proportions are not known, we can use the Pythagorean Theorem. The length of the hypotenuse is s. The
length of the short side is 12 s (because we just cut the equilateral triangle in half). The unknown side (the
height) is found by the Pythagorean Theorem. With the height now known, we can find the area of the
equilateral triangle.
( 12 s )2 + h 2 = s 2
h 2 = (1 − 14 )s 2
h 2 = 34 s 2
h=
3
s
2
1
× base × height
2
1
3
A = ×s×
s
2
2
3 2
A=
s
4
A60−60−60 =
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 4
February 1, 2016
The requested area, A, is equal to the area of eight equilateral triangles. It is also equal to area of the octagonal
face of the cube. Let's find both areas in terms of s2.
A = 1 − 2x 2
3 2
A = 8×
s
s = 2x
4
s 2 = 2x 2
A = 2 3s 2
A = 1− s2
Finally, let's equate the two areas, solve for s2, and then substitute to find A. Although this problem didn't
explicitly ask, some contests prefer that you rationalize the denominator.
2 3s 2 = 1 − s 2
(2
)
A=
1
2 3 +1
A=
3 +1 s2 = 1
s2 =
A = 2 3 × s2
A=
2 3
2 3 +1
A=
2 3
2 3 −1
×
2 3 +1 2 3 −1
4⋅3 − 2 3
(2 3 ) − 1
2
2
12 − 2 3 12 − 2 3
=
4 ⋅ 3 −1
11
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 5
February 1, 2016
4. Solution 1: Number Sieve. (A hint from the name of the question.) If a student has seen how the Sieve of
Eratosthenes is used to quickly and reliably find the prime numbers less than 100, a similar strategy will work
for this problem. First, we write the numbers from 1 to 99 or make a grid representing the numbers from 1 to
99. Then, for each factor given in the question, we put a tick mark on top of (or in the grid) for every number
that is divisible by the specified factor.
0
1
0
2
3
4
5
6
7
8
9
/
/
/
///
/
///
//
////
//
//
//
///
/
/////
/
/
/
//
/////
/
/
/////
/
10
//
20
///
30
////
///
40
////
////
//
///
/
50
//
//
////
/
////
60
/////
/
///
/
///
70
///
//////
/
//
//
/
///
80
////
//
/
/////
/
/
/
///
90
/////
/
//
/
/
/////
//
/
/
///
/
////
//
/
///
/
//
//
/
//
Most of the tricks that make the Sieve of Eratosthenes fast (and clever) don't apply here. You have to
laboriously mark each factor of each number. You have to be careful not to obscure one tick mark with another.
And if you lose track of where you are, you may have to start from the beginning.
Once the tick marks are done, find the numbers that are ticked exactly twice. I would suggest you write down
those numbers (for partial credit). I found 18 such numbers. They are 4, 9, 10, 14, 15, 21, 27, 35, 44, 50, 52,
68, 75, 76, 81, 92, 98, and 99.
Solution 2. Logical Reasoning. This method may be less tedious and more fun. But, I had a difficult time
convincing myself I had not missed any categories.
•
•
•
•
•
•
•
Powers of 3: (9, 27, 81)
Powers of 2: (4)
Odd prime (>9) multiples of 9: 9 × (11)
Odd prime (>9) multiples of 4: 4 × (11, 13, 17, 19, 23)
Odd multiples of 2: 2 × (5, 7, 52, 72)
Odd multiples of 3: 3 × (5, 7, 52)
Odd multiples of 5: 5 × (7)
All together, there are 18 numbers in my lists.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 6
February 1, 2016
5. Solution 1. Probability tree. Probability trees are a safe and effective way to solve this kind of problem.
However, if the problem only wants a tiny bit of information, we sometimes do a lot of extra work to complete
the entire tree. Nevertheless, that's the method we will use for this solution.
There are several different ways to draw and label a probability tree. In my sketch, we start at the top and work
our way to the bottom. The top row represents the start of Day 1. The next row is the start of Day 2. And the
bottom row is the start of Day 3. Horizontally, the left column represents zero rocks in the valley. Each
subsequent column represents one more rock in the valley. The numbers in the nodes (circles) represent the
probability that we are in that state at the start of the specified day. For example, we are guaranteed to have 1
rock in the valley on the start of Day 1, so we put "1" in the circle corresponding to "Day 1, 1 rock". Each edge
(line segment) in our tree shows how the status can change from one day to the next. The numbers that label
each edge represent the probability that the event will occur. So the edge that connects the node at Day 1 to the
leftmost node at Day 2 indicates there is a 1/10 probability that we will go from 1 rock in the valley to zero
rocks in the valley. The sum of the edges that leave any circle must always total 1. The numbers in each node
can be found by summing the products of each probability along an incoming edge by the value in the
immediately preceding node. If we haven't made a mistake, the sum of the probabilities leaving each node on a
given day must total 1.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 7
February 1, 2016
The edge leaving "Start of Day 2 / 0 rocks" shows a probability of 1. Once there are zero rocks in the valley,
there will always be zero rocks in the valley. The edges leaving "1 rock" are given directly by the problem.
The edges leaving "2 rocks" require some additional thought. Each rock is independently vaporized, rolled
back down into the valley, or split into two rocks. The probability that both rocks are vaporized is 0.12=0.01.
The probability that both rocks are duplicated is 0.42=.16. One rock can be vaporized and one rock can be
rolled back down into the valley (resulting in 1 rock in the valley) in 2 different ways. So, the probability this
occurs is 2×0.1×0.5=0.10. One rock can be duplicated and one rock can be rolled back down into the valley
(resulting in 3 rocks in the valley) in 2 different ways. So, the probability this occurs is 2×0.4×0.5=0.40. And
finally, one rock can be vaporized and one rock can be duplicated in two different ways (2×0.1×0.4=0.08) or
both rocks can be rolled back down into the valley (0.5×0.5=0.25). So the probability we go from 2 rocks to 2
rocks is 0.08+0.25=0.33.
Let's check our probabilities of leaving "Start of Day 2 / 2 rocks". 0.01+0.10+0.33+0.40+0.16 = 1.00, which is
the correct total.
Next, we sum the probabilities entering the nodes along "Start of Day 3". For the node with zero rocks, we
have 0.1×1+0.5×0.1+0.4×0.01=0.1+0.05+0.004=0.154. For the node with one rock, we have
0.5×0.5+0.4×0.10=0.25+0.04=0.29. For the node with two rocks, we have
0.5×0.4+0.4×0.33=0.2+0.132=0.332. For the node with three rocks, we have 0.4×0.40=0.16. And for the node
with four rocks, we have 0.4×0.16=0.064.
Let's check the total probabilities inside each node along "Start of Day 3".
0.154+0.290+0.332+.160+0.064=1.000, which is the correct total.
The chance of having 2 rocks in the valley at the start of Day 3 is 0.332 or 33.2%.
If you are confident in your skills with probability trees, we weren't required to compute the complete
probability tree. In fact, we only needed to compute the edges and nodes leading to the node at the start of Day
3 with 2 rocks. (4 edges and 4 nodes.)
Solution 2. Logical Reasoning. We can list all of the ways of ending up with 2 rocks:
Roll the rock into the valley (5/10 probability). Then duplicate the rock (4/10 probability). 20/100.
Duplicate the rock (4/10 probability). Then vaporize one, duplicate the other (4/10*1/10*4/10 * 2 ways) =
32/1000; or roll both rocks back down into the valley (4/10*5/10*5/10=100/1000).
Tallying the above cases, 200/1000 + 32/1000 + 100/1000 = 332/1000 = 83/250.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 8
February 1, 2016
6(a). We can simplify the problem by ignoring Sisyphus and the valley and the days. The question simply
wants to know the probability that the number of rocks never reaches 0, when starting with 1 rock. Zeus picks
up a rock at his leisure and vaporizes it (with probability 1/10), returns it to the pile (with probability 5/10), or
splits the rock in two (with probability 4/10). If there are no rocks remaining, Zeus rolls over and goes to sleep.
If Zeus returns the rock to the pile, nothing has changed, so we can further simplify the problem by ignoring
that action, and considering only the relative probability of the other actions.
Zeus picks up a rock at his liesure. He vaporizes it (with probability 1/5) or splits the rock in two (with
probability 4/5). (The probabilities of vaporization:duplication = 1:4, as before the simplification.)
Solution via State Diagram:
We can represent the situation with a State Diagram. There is a single state for every possible number of rocks
in the pile. Zero rocks is State S0. One rock is State S1. Etc. For each state, we have a variable nk which
represents the total number of times the system has entered that state from the beginning to the end of time. See
attached illustration.
We can write formulas to relate the variables, n, with each other. For example, if you are in State S1, Zeus will
vaporize that one and only rock with probability 1/5. So, the number of times we enter State S0 is exactly equal
to 1/5 the number of times we enter State S1. We enter State S1 once when the system is initialized. We also
enter State S1 when we were in State S2 and Zeus vaporizes one of the two remaining rocks. So, the net
n
number of times we enter State S1 is exactly n1 = 1 + 2 . At each other State, Sk we can enter the State when
5
Zeus splits a rock while we were in State Sk-1 or when Zeus vaporizes a rock while we were in State Sk+1.
We have an infinite number of simple linear equations and an infinite number of unknowns. We could make an
arbitrary guess and solve the system of equations iteratively. Or we could make a good guess and prove that
our guess satisfies the system.
First, note that each value of n is a weighted average of its two neighbors. Also note how perfectly regular the
systems is for States S2 and higher.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 9
February 1, 2016
Let me redraw the State diagram ever so slightly:
Now, notice that state S1 looks almost the same as States S2 and higher. What if we guess that it *really* is the
same as State S2? Let's set the input entering State S2 from the left equal to the input entering State S1 from the
n
n
4
5
5
25 - 20 5
= 1+ 2 ⇒ n2 =
= . Now, it is a
left: 1 = n1 ⇒ n1 = . Let's solve for n2: n1 = 1 + 2 ⇒
5
4
5
4
5
4
4
n 1
5
3
simple matter to confirm nk = , k ≥ 1 . Then we solve for n0 = 1 = . The final answer is 1 − n0 = .
4
5 4
4
6(a). Solution Estimate via Probability Tree
See the adjoining probability tree. The leftmost column shows
the probability the Zeus has vaporized the only remaining rock.
The next column shows the probability that one rock remains.
At each step, Zeus adds a rock or destroys a rock, so the number
of rocks remaining alternates from odd to even. (Unless it was
already zero.)
The total probability at each stage must equal 1. So, before Zeus
picks up the solitary rock, the probability that 1 rock is present is
exactly 1. (Represented by the number in the circle at the top of
the tree.)
After Zeus handles a rock for the first time, we are on the next
row. The total probability is 0.2+0.8=1. There is a 1/5
probability that Zeus destroyed the rock, and there is an 4/5
probability that there are now 2 rocks.
At the next stage, Zeus has either destroyed one of the two rocks
remaining or he has split one of the two rocks remaining, and
now we have three rocks. The total probability is 0.2 (no rocks
remained from the previous step) + 0.16 (one rock remains) +
0.64 (three rocks are in the pile) = 1.00.
After the fourth step, the probabilities for 0, 1, 3, or 5 rocks are
0.232 + 0.0512 + 0.3072 + 0.4096 = 1.0000. After the ninth
step, the probability that there are no rocks remaining is
0.248171008.
Maybe we can guess the answer is converging towards 1–0.25,
and hope the graders will award partial credit.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 10
February 1, 2016
6(a). Solution via Probability Tree
This is going to be an infinitely wide and deep probability tree.
We need a formula. So, I'll begin by counting the number of
paths that get us to each place where Zeus has destroyed the last
rock. See attached diagram. The numbers in the circles
represent the number of paths from the root of the tree to that
particular node. This was performed using standard
MATHCOUNTS path counting methods.
Note the 14s in the circles near the bottom. That indicates there
are 14 paths from the root of the tree to those particular nodes.
The lower-left of the two nodes is where Zeus has just destroyed
the final rock. No matter how we get from the root to this node,
Zeus has created 4 additional rocks and he has vaporized 5 rocks.
Since Zeus has 14 ways to do this, the formula for the probability
4
5
4 1
that we reach this node is 14 × × = 0.001835008 ,
5 5
which number you may recognize from the previous solution.
The sequence we have found begins with 1, 1, 2, 5, 14. Not
surprisingly, these are the Catalan Numbers. (Also known as
mountain numbers. If we rotate the diagram counterclockwise
90 degrees, we can imagine climbing a mountain, possibly with
ups and downs, and eventually descending back to earth.
C(0)=1, C(1)=1, C(2)=2, C(3)=5, C(4)=14.
Now we can write a general formula for each of the nodes where
Zeus has just destroyed the last rock:
n
4 1
C ( n) ⋅
5 5
n +1
n
n
1
4 1 1 C ( n) 4
= C ( n) ⋅ =
⋅
5 25
5 5 5
n
The total probability that all rocks will eventually be destroyed,
is given by the infinite sum:
n
∞
2n 1
C (n) 4
pever = ∑
⋅ , where C (n ) = ⋅
5 25
n =0
n n +1
In general, the contest awards more points for a closed-form solution. However, as this is an infinite sum and
the problem asks for "the probability", I am skeptical that much credit would be awarded without a closed form
solution. Although I am not a grader, I would recommend a student should spend some of their 3 hours
attempting to reduce the infinite sum to a number. At the very least, I think a student should provide a
convincing argument that the sum converges to a valid probability (a number between 0 and 1).
n
1 4
If we let T (n) = C (n) ⋅ ⋅ represent each term in the summation, above, we can find the ratio between
5 25
consecutive terms. Using the formula for C(n), above, a little algebra quickly shows that
C (n) C (n − 1) = (4n − 2 ) (n + 2 ) . Then, we have:
T ( n)
4 4n − 2 16n − 8 16
=
⋅
=
<
T (n − 1) 25 n + 2 25n + 25 25
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 11
February 1, 2016
Since the ratios of every term in our sequence are less than the ratios in a geometric series, we can find the
upper bound on our summation, using the formula for the sum of a geometric series. (Although this may look
too scary to do without a calculator, if you know your powers of 2, only the final term we show really requires
any hand calculation (a power of 2, decimal shifted, times 14, divided by 9).
(
)
16
16
Pever < T (0) + T (1) + T (2) + T (3) + T (4) 1 + 16
25 + ( 25 ) + ( 25 ) + L
2
2
3
3
4
1 4
1
1
1 4
1 4
1 4
Pever < ⋅1 ⋅ ⋅ 1 + 1 ⋅ ⋅
+ 2 ⋅ ⋅ + 5 ⋅ ⋅ + 14 ⋅ ⋅ ⋅
5 25 1 − 16
5
5 25
5 25
5 25
25
1 4 32 320 14 ⋅ 256 25
+ +
+ 7 +
⋅
5 53 55
5
59
9
< 0.2 + 0.032 + 0.01024 + 0.004096 + 0.001835008 ⋅ 25 9
Pever <
Pever
0.248171008 < Pever < 0.25143324
No single step, above, is exceptionally hard. Putting it all together is rather brutal. I think most students who
could put all this together will probably find an easier solution. However, with a few more (non-trivial) steps,
the summation can be reduced to a closed form, but I will leave that as an exercise for the very dedicated reader.
pever = 14 . So, pnever = 1 − 14 =
3
4
Solutions to questions 1 through 6(a), by Rocke Verser
Page 1
February 1, 2016
1. Even on the easiest problem on a contest, I often make a sketch and/or copy the information from the
question. I find it helps me to internalize the problem, and to reduce "dumb" mistakes. Since this is a right
triangle, I immediately think of the Pythagorean Theorem.
a + b + c = 56
a + b 2 + c 2 = 1250
2
a2 + b2 = c2
Solution 1: Pythagorean Triplets. (A nice hint from the name of the question.) Combine the last two equations
written above.
c 2 + c 2 = 1250
c 2 = 625
c = 25
The most well-known Pythagorean Triplet is the 3-4-5 triangle. If we scale this triangle by a factor of 5, we
will have the correct hypotenuse. That would give us a 15-20-25 triangle. But the sum of sides is
15+20+25=60. And that doesn't meet the other specification of the problem.
The first series of Pythagorean Triplets continues with odd sides: 3-4-5, 5-12-13, 7-24-25, 9-40-41, ... . The
7-24-25 triangle has the correct hypotenuse. Check the sum of the sides, 7+24+25=56. This meets all the
criteria of the question. Now, we can answer the question: "What is the area?" Using the formula for the area
of a triangle, we get the final answer:
A = 12 base × height = 12 × 24 × 7 = 12 × 7 = 84
Solution 2: Algebra. First, we solve for c. Then we solve for a+b. A technique that sometimes helps with
right triangle problems that don't seem to give enough information is to square the sum or difference of two
sides.
c 2 + c 2 = 1250
c 2 = 625
c = 25
a + b + c = 56
a + b + 25 = 56
a + b = 31
(a + b )2 = 312
a 2 + 2ab + b 2 = 961
We already found a 2 + b 2 = c 2 = 625 . Subtracting the last two equations leaves us with the area of 4 congruent
triangles. 2ab = 961 − 625 = 336 ⇒ 12 base × height = 12 ab = 84 .
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 2
February 1, 2016
2. The factorial of n is the product of all of the positive integers from n down to 1. The first step is to break the
expression down starting from the innermost parentheses.
4!= 4 × 3 × 2 × 1 = 24
(4!)!= 24!= 24 × 23 × 22 × L × 3 × 2 × 1
3!= 3 × 2 × 1 = 6
(3!)!= 6!= 6 × 5 × 4 × 3 × 2 × 1
[6!]4 = (6 × 5 × 4 × 3 × 2 ×1)4
We needed to multiply out 4! and 3!. But we must resist the urge to multiply out 24! and 6!. Why? First,
quotients involving factorials are often easier to divide when they are already factored. Second, this question is
specifically asking for the "prime factorization". So, anything we multiply will have to be factored, later.
Probably the most straightforward way to solve this is to write out all the factors in the numerator and all the
factors in the denominator and "cancel" out factors.
24 × 23 × 22 × 21 × 20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
6 × 5 × 4 × 3 × 2 × 1 × 6 × 5 × 4 × 3 × 2 × 1 × 6 × 5 × 4 × 3 × 2 × 1× 6 × 5 × 4 × 3 × 2 × 1
This is a perfectly fine solution method. A subtle enhancement is to group the numerator into 4 sets of 6
consecutive integer factor. Each group of 6 factors in the numerator is evenly divisible by 6!. (Think about
why that's true, as we cancel factors.)
(24 × 23 × 22 × 21× 20 × 19) × (18 × 17 ×16 × 15 ×14 × 13)× (12 × 11× 10 × 9 × 8 × 7 )× (6 × 5 × 4 × 3 × 2 × 1)
(6 × 5 × 4 × 3 × 2 × 1)× (6 × 5 × 4 × 3 × 2 × 1) × (6 × 5 × 4 × 3 × 2 ×1)× (6 × 5 × 4 × 3 × 2 × 1)
=
24 × 23 × 22 × 21× 20 × 19 18 × 17 × 16 × 15 × 14 × 13 12 × 11× 10 × 9 × 8 × 7 6 × 5 × 4 × 3 × 2 × 1
×
×
×
6 × 5 × 4 × 3 × 2 ×1
6 × 5 × 4 × 3 × 2 ×1
6 × 5 × 4 × 3 × 2 ×1
6 × 5 × 4 × 3 × 2 ×1
The rightmost fraction completely cancels. Rearranging the factors in the other denominators:
× 23 × 22 × 21 × 20 × 19
24
6× 4 ×
18
1
×
2
3
×
5
× 17 × 16 × 15 × 14 × 13
6×3 ×
12
×
×
1
×
4
5
×
2
× 11 × 10 × 9 × 8 × 7
6× 2 ×
1
×
5
× 3 × 4
= 1 × 23 × 11× 7 × 4 × 19
= 1 × 17 × 4 × 3 × 7 × 13
= 1 × 11× 2 × 3 × 2 × 7
Now, everything is completely factored, except 4 = 2 2 , so we can directly write the final answer:
23 × 19 × 17 × 13 × 112 × 7 3 × 32 × 2 6
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 3
February 1, 2016
3. First, if the question had not provided a reasonably nice sketch, that is the very first thing I would draw.
Even with the given perspective sketch, I still need to draw and label a face of the cube and an equilateral
triangle that replaced a corner.
The area of the face of the cube was originally the area of a square of side-length 1. But we shaved off four
isosceles right triangles, each with an unknown side length of x. So, the remaining area of the face of the square
is 12 − 4 × 12 x 2 = 1 − 2 x 2 . The hypotenuse of the isosceles right triangle has the same length as the side-length of
the equilateral triangles we created in place of the corners, s = 2 x .
A student may know the area of an equilateral triangle in terms of its side-length. If not, divide the equilateral
triangle vertically into a 30-60-90 triangle. The ratio of the side-lengths of any 30-60-90 triangle is 1 : 3 : 2 .
All of the side-lengths can be found by proportions if any of the three side-lengths are known. Or, if these
proportions are not known, we can use the Pythagorean Theorem. The length of the hypotenuse is s. The
length of the short side is 12 s (because we just cut the equilateral triangle in half). The unknown side (the
height) is found by the Pythagorean Theorem. With the height now known, we can find the area of the
equilateral triangle.
( 12 s )2 + h 2 = s 2
h 2 = (1 − 14 )s 2
h 2 = 34 s 2
h=
3
s
2
1
× base × height
2
1
3
A = ×s×
s
2
2
3 2
A=
s
4
A60−60−60 =
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 4
February 1, 2016
The requested area, A, is equal to the area of eight equilateral triangles. It is also equal to area of the octagonal
face of the cube. Let's find both areas in terms of s2.
A = 1 − 2x 2
3 2
A = 8×
s
s = 2x
4
s 2 = 2x 2
A = 2 3s 2
A = 1− s2
Finally, let's equate the two areas, solve for s2, and then substitute to find A. Although this problem didn't
explicitly ask, some contests prefer that you rationalize the denominator.
2 3s 2 = 1 − s 2
(2
)
A=
1
2 3 +1
A=
3 +1 s2 = 1
s2 =
A = 2 3 × s2
A=
2 3
2 3 +1
A=
2 3
2 3 −1
×
2 3 +1 2 3 −1
4⋅3 − 2 3
(2 3 ) − 1
2
2
12 − 2 3 12 − 2 3
=
4 ⋅ 3 −1
11
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 5
February 1, 2016
4. Solution 1: Number Sieve. (A hint from the name of the question.) If a student has seen how the Sieve of
Eratosthenes is used to quickly and reliably find the prime numbers less than 100, a similar strategy will work
for this problem. First, we write the numbers from 1 to 99 or make a grid representing the numbers from 1 to
99. Then, for each factor given in the question, we put a tick mark on top of (or in the grid) for every number
that is divisible by the specified factor.
0
1
0
2
3
4
5
6
7
8
9
/
/
/
///
/
///
//
////
//
//
//
///
/
/////
/
/
/
//
/////
/
/
/////
/
10
//
20
///
30
////
///
40
////
////
//
///
/
50
//
//
////
/
////
60
/////
/
///
/
///
70
///
//////
/
//
//
/
///
80
////
//
/
/////
/
/
/
///
90
/////
/
//
/
/
/////
//
/
/
///
/
////
//
/
///
/
//
//
/
//
Most of the tricks that make the Sieve of Eratosthenes fast (and clever) don't apply here. You have to
laboriously mark each factor of each number. You have to be careful not to obscure one tick mark with another.
And if you lose track of where you are, you may have to start from the beginning.
Once the tick marks are done, find the numbers that are ticked exactly twice. I would suggest you write down
those numbers (for partial credit). I found 18 such numbers. They are 4, 9, 10, 14, 15, 21, 27, 35, 44, 50, 52,
68, 75, 76, 81, 92, 98, and 99.
Solution 2. Logical Reasoning. This method may be less tedious and more fun. But, I had a difficult time
convincing myself I had not missed any categories.
•
•
•
•
•
•
•
Powers of 3: (9, 27, 81)
Powers of 2: (4)
Odd prime (>9) multiples of 9: 9 × (11)
Odd prime (>9) multiples of 4: 4 × (11, 13, 17, 19, 23)
Odd multiples of 2: 2 × (5, 7, 52, 72)
Odd multiples of 3: 3 × (5, 7, 52)
Odd multiples of 5: 5 × (7)
All together, there are 18 numbers in my lists.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 6
February 1, 2016
5. Solution 1. Probability tree. Probability trees are a safe and effective way to solve this kind of problem.
However, if the problem only wants a tiny bit of information, we sometimes do a lot of extra work to complete
the entire tree. Nevertheless, that's the method we will use for this solution.
There are several different ways to draw and label a probability tree. In my sketch, we start at the top and work
our way to the bottom. The top row represents the start of Day 1. The next row is the start of Day 2. And the
bottom row is the start of Day 3. Horizontally, the left column represents zero rocks in the valley. Each
subsequent column represents one more rock in the valley. The numbers in the nodes (circles) represent the
probability that we are in that state at the start of the specified day. For example, we are guaranteed to have 1
rock in the valley on the start of Day 1, so we put "1" in the circle corresponding to "Day 1, 1 rock". Each edge
(line segment) in our tree shows how the status can change from one day to the next. The numbers that label
each edge represent the probability that the event will occur. So the edge that connects the node at Day 1 to the
leftmost node at Day 2 indicates there is a 1/10 probability that we will go from 1 rock in the valley to zero
rocks in the valley. The sum of the edges that leave any circle must always total 1. The numbers in each node
can be found by summing the products of each probability along an incoming edge by the value in the
immediately preceding node. If we haven't made a mistake, the sum of the probabilities leaving each node on a
given day must total 1.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 7
February 1, 2016
The edge leaving "Start of Day 2 / 0 rocks" shows a probability of 1. Once there are zero rocks in the valley,
there will always be zero rocks in the valley. The edges leaving "1 rock" are given directly by the problem.
The edges leaving "2 rocks" require some additional thought. Each rock is independently vaporized, rolled
back down into the valley, or split into two rocks. The probability that both rocks are vaporized is 0.12=0.01.
The probability that both rocks are duplicated is 0.42=.16. One rock can be vaporized and one rock can be
rolled back down into the valley (resulting in 1 rock in the valley) in 2 different ways. So, the probability this
occurs is 2×0.1×0.5=0.10. One rock can be duplicated and one rock can be rolled back down into the valley
(resulting in 3 rocks in the valley) in 2 different ways. So, the probability this occurs is 2×0.4×0.5=0.40. And
finally, one rock can be vaporized and one rock can be duplicated in two different ways (2×0.1×0.4=0.08) or
both rocks can be rolled back down into the valley (0.5×0.5=0.25). So the probability we go from 2 rocks to 2
rocks is 0.08+0.25=0.33.
Let's check our probabilities of leaving "Start of Day 2 / 2 rocks". 0.01+0.10+0.33+0.40+0.16 = 1.00, which is
the correct total.
Next, we sum the probabilities entering the nodes along "Start of Day 3". For the node with zero rocks, we
have 0.1×1+0.5×0.1+0.4×0.01=0.1+0.05+0.004=0.154. For the node with one rock, we have
0.5×0.5+0.4×0.10=0.25+0.04=0.29. For the node with two rocks, we have
0.5×0.4+0.4×0.33=0.2+0.132=0.332. For the node with three rocks, we have 0.4×0.40=0.16. And for the node
with four rocks, we have 0.4×0.16=0.064.
Let's check the total probabilities inside each node along "Start of Day 3".
0.154+0.290+0.332+.160+0.064=1.000, which is the correct total.
The chance of having 2 rocks in the valley at the start of Day 3 is 0.332 or 33.2%.
If you are confident in your skills with probability trees, we weren't required to compute the complete
probability tree. In fact, we only needed to compute the edges and nodes leading to the node at the start of Day
3 with 2 rocks. (4 edges and 4 nodes.)
Solution 2. Logical Reasoning. We can list all of the ways of ending up with 2 rocks:
Roll the rock into the valley (5/10 probability). Then duplicate the rock (4/10 probability). 20/100.
Duplicate the rock (4/10 probability). Then vaporize one, duplicate the other (4/10*1/10*4/10 * 2 ways) =
32/1000; or roll both rocks back down into the valley (4/10*5/10*5/10=100/1000).
Tallying the above cases, 200/1000 + 32/1000 + 100/1000 = 332/1000 = 83/250.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 8
February 1, 2016
6(a). We can simplify the problem by ignoring Sisyphus and the valley and the days. The question simply
wants to know the probability that the number of rocks never reaches 0, when starting with 1 rock. Zeus picks
up a rock at his leisure and vaporizes it (with probability 1/10), returns it to the pile (with probability 5/10), or
splits the rock in two (with probability 4/10). If there are no rocks remaining, Zeus rolls over and goes to sleep.
If Zeus returns the rock to the pile, nothing has changed, so we can further simplify the problem by ignoring
that action, and considering only the relative probability of the other actions.
Zeus picks up a rock at his liesure. He vaporizes it (with probability 1/5) or splits the rock in two (with
probability 4/5). (The probabilities of vaporization:duplication = 1:4, as before the simplification.)
Solution via State Diagram:
We can represent the situation with a State Diagram. There is a single state for every possible number of rocks
in the pile. Zero rocks is State S0. One rock is State S1. Etc. For each state, we have a variable nk which
represents the total number of times the system has entered that state from the beginning to the end of time. See
attached illustration.
We can write formulas to relate the variables, n, with each other. For example, if you are in State S1, Zeus will
vaporize that one and only rock with probability 1/5. So, the number of times we enter State S0 is exactly equal
to 1/5 the number of times we enter State S1. We enter State S1 once when the system is initialized. We also
enter State S1 when we were in State S2 and Zeus vaporizes one of the two remaining rocks. So, the net
n
number of times we enter State S1 is exactly n1 = 1 + 2 . At each other State, Sk we can enter the State when
5
Zeus splits a rock while we were in State Sk-1 or when Zeus vaporizes a rock while we were in State Sk+1.
We have an infinite number of simple linear equations and an infinite number of unknowns. We could make an
arbitrary guess and solve the system of equations iteratively. Or we could make a good guess and prove that
our guess satisfies the system.
First, note that each value of n is a weighted average of its two neighbors. Also note how perfectly regular the
systems is for States S2 and higher.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 9
February 1, 2016
Let me redraw the State diagram ever so slightly:
Now, notice that state S1 looks almost the same as States S2 and higher. What if we guess that it *really* is the
same as State S2? Let's set the input entering State S2 from the left equal to the input entering State S1 from the
n
n
4
5
5
25 - 20 5
= 1+ 2 ⇒ n2 =
= . Now, it is a
left: 1 = n1 ⇒ n1 = . Let's solve for n2: n1 = 1 + 2 ⇒
5
4
5
4
5
4
4
n 1
5
3
simple matter to confirm nk = , k ≥ 1 . Then we solve for n0 = 1 = . The final answer is 1 − n0 = .
4
5 4
4
6(a). Solution Estimate via Probability Tree
See the adjoining probability tree. The leftmost column shows
the probability the Zeus has vaporized the only remaining rock.
The next column shows the probability that one rock remains.
At each step, Zeus adds a rock or destroys a rock, so the number
of rocks remaining alternates from odd to even. (Unless it was
already zero.)
The total probability at each stage must equal 1. So, before Zeus
picks up the solitary rock, the probability that 1 rock is present is
exactly 1. (Represented by the number in the circle at the top of
the tree.)
After Zeus handles a rock for the first time, we are on the next
row. The total probability is 0.2+0.8=1. There is a 1/5
probability that Zeus destroyed the rock, and there is an 4/5
probability that there are now 2 rocks.
At the next stage, Zeus has either destroyed one of the two rocks
remaining or he has split one of the two rocks remaining, and
now we have three rocks. The total probability is 0.2 (no rocks
remained from the previous step) + 0.16 (one rock remains) +
0.64 (three rocks are in the pile) = 1.00.
After the fourth step, the probabilities for 0, 1, 3, or 5 rocks are
0.232 + 0.0512 + 0.3072 + 0.4096 = 1.0000. After the ninth
step, the probability that there are no rocks remaining is
0.248171008.
Maybe we can guess the answer is converging towards 1–0.25,
and hope the graders will award partial credit.
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 10
February 1, 2016
6(a). Solution via Probability Tree
This is going to be an infinitely wide and deep probability tree.
We need a formula. So, I'll begin by counting the number of
paths that get us to each place where Zeus has destroyed the last
rock. See attached diagram. The numbers in the circles
represent the number of paths from the root of the tree to that
particular node. This was performed using standard
MATHCOUNTS path counting methods.
Note the 14s in the circles near the bottom. That indicates there
are 14 paths from the root of the tree to those particular nodes.
The lower-left of the two nodes is where Zeus has just destroyed
the final rock. No matter how we get from the root to this node,
Zeus has created 4 additional rocks and he has vaporized 5 rocks.
Since Zeus has 14 ways to do this, the formula for the probability
4
5
4 1
that we reach this node is 14 × × = 0.001835008 ,
5 5
which number you may recognize from the previous solution.
The sequence we have found begins with 1, 1, 2, 5, 14. Not
surprisingly, these are the Catalan Numbers. (Also known as
mountain numbers. If we rotate the diagram counterclockwise
90 degrees, we can imagine climbing a mountain, possibly with
ups and downs, and eventually descending back to earth.
C(0)=1, C(1)=1, C(2)=2, C(3)=5, C(4)=14.
Now we can write a general formula for each of the nodes where
Zeus has just destroyed the last rock:
n
4 1
C ( n) ⋅
5 5
n +1
n
n
1
4 1 1 C ( n) 4
= C ( n) ⋅ =
⋅
5 25
5 5 5
n
The total probability that all rocks will eventually be destroyed,
is given by the infinite sum:
n
∞
2n 1
C (n) 4
pever = ∑
⋅ , where C (n ) = ⋅
5 25
n =0
n n +1
In general, the contest awards more points for a closed-form solution. However, as this is an infinite sum and
the problem asks for "the probability", I am skeptical that much credit would be awarded without a closed form
solution. Although I am not a grader, I would recommend a student should spend some of their 3 hours
attempting to reduce the infinite sum to a number. At the very least, I think a student should provide a
convincing argument that the sum converges to a valid probability (a number between 0 and 1).
n
1 4
If we let T (n) = C (n) ⋅ ⋅ represent each term in the summation, above, we can find the ratio between
5 25
consecutive terms. Using the formula for C(n), above, a little algebra quickly shows that
C (n) C (n − 1) = (4n − 2 ) (n + 2 ) . Then, we have:
T ( n)
4 4n − 2 16n − 8 16
=
⋅
=
<
T (n − 1) 25 n + 2 25n + 25 25
Twenty-fourth Annual UNC Math Contest Final Round January 2016
Solutions to questions 1 through 6(a), by Rocke Verser
Page 11
February 1, 2016
Since the ratios of every term in our sequence are less than the ratios in a geometric series, we can find the
upper bound on our summation, using the formula for the sum of a geometric series. (Although this may look
too scary to do without a calculator, if you know your powers of 2, only the final term we show really requires
any hand calculation (a power of 2, decimal shifted, times 14, divided by 9).
(
)
16
16
Pever < T (0) + T (1) + T (2) + T (3) + T (4) 1 + 16
25 + ( 25 ) + ( 25 ) + L
2
2
3
3
4
1 4
1
1
1 4
1 4
1 4
Pever < ⋅1 ⋅ ⋅ 1 + 1 ⋅ ⋅
+ 2 ⋅ ⋅ + 5 ⋅ ⋅ + 14 ⋅ ⋅ ⋅
5 25 1 − 16
5
5 25
5 25
5 25
25
1 4 32 320 14 ⋅ 256 25
+ +
+ 7 +
⋅
5 53 55
5
59
9
< 0.2 + 0.032 + 0.01024 + 0.004096 + 0.001835008 ⋅ 25 9
Pever <
Pever
0.248171008 < Pever < 0.25143324
No single step, above, is exceptionally hard. Putting it all together is rather brutal. I think most students who
could put all this together will probably find an easier solution. However, with a few more (non-trivial) steps,
the summation can be reduced to a closed form, but I will leave that as an exercise for the very dedicated reader.
pever = 14 . So, pnever = 1 − 14 =
3
4