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The Third SUNYMathics
Gabriel T. Prˇajiturˇa
Copyright ©2022 Gabriel T. Prˇajiturˇa
In memory of Ioan Serdean
The pandemic interrupted our competition. The second edition took place in
the spring of 2020 the weekend before all students were sent home and instruc-
tion was switched overnight to 100% online. As neither the campuses nor the
professors were prepared for such a change, that created a lot of issues and took
a lot of energy from everybody. For the time being, all competitions (including
the math ones) were cancelled.
We had a timid restart this spring. It takes time to develop a culture of
problem solving and math competitions, especially after a forced interruption.
Not all competitions restarted this year, but I preferred to have it back as soon
as possible because we all need to feel that we are back to our normal lives.
And we are not back to normal until we do not go oﬀ the geodesic. I was glad
all the SUNY campuses who participated before took part in it again and this
year we welcomed Bard College to the challenge.
I also have a personal reason to start this competition as soon as possible.
Last year I lost a friend who gave up a programmer job and worked for many
years to train middle and high school students for math competitions. An
analogy with Jaime Escalante, of “Stand and Deliver” fame, came to mind.
He did not stop working with the students just because the competitions were
cancelled. And he was right. We need to ﬁnd ways to continue so we can keep
the culture alive.
I would like to thank the Seaway section of MAA for its support and to the
local organizers from each campus.
All comments with no name belong to the author. I welcome any new
approach and comment to any of these problems as well as new problems for
the next competitions.
This work was done while the author was a Fulbright scholar visiting the
Polytechnic University of Bucharest. I would like to thank the Fulbright Scholar
Program, Fulbright Commission Romania, Mathematical Institute of the Ro-
manian Academy, Central University Library of Bucharest and the Department
of Applied Sciences of the Polytechnic University of Bucharest for their support.
Gabriel T. Prˇajiturˇa
1 The Problems 1
2 Top Six Students 5
3 Solutions 7
4 Comments 19
2CHAPTER 1. THE PROBLEMS
1. Find a four - digit square number such that the ﬁrst two digits are identical
and the last two digits are identical.
2. Find the least number starting with 1 such that if we move 1 from the
beginning to the end of the number the resulting number is 3 times bigger.
3. ABCD is a parallelogram and E a point such that EA ⊥AB and EC ⊥BC.
Prove that the angles AEC and DEB have the same bisector.
4. Let fbe a continuous function such that
f(x)dx = 2
Prove that there is a number a∈(0,1) such that
f(a) = sin(πa)
5. Let A, B, C be three points on a circle of radius 1 such that the 4ABC is
equilateral. Find all points Pon the same circle such that the product
||P A|| · ||P B|| · ||P C ||
has the maximum value.
6. Prove that for all n≥2
r1 + n!
7. If n≥3 is an odd number prove that 72 divides
3n+ 4n+ 5n
8. Let XYZU be a tetrahedron such that the faces XYU, XZU, and YZU are
all right triangles with the right angle U. Let T ∈(Y Z) such that XT ⊥YZ.
a) Prove that UT ⊥YZ
b) Prove that if, for some integer k≥2, the areas of the faces XYU, XZU,
and YZU are given by 2k, 6kand 2k2−5 respectively, then the area of XYZ
is also an integer.
9. Prove that there is no natural number nsuch that
is an integer ending with 100 zeros.
10. Prove that in a chess tournament in which each player plays everyone else
there are at least two players with the same number of draws.
11. If we pick up at random 5 numbers from 1 to 1,000 what is the probability
that their sum is divisible by 3?
12. Prove that every natural number has a multiple whose digits are only zeros
4CHAPTER 1. THE PROBLEMS
Top Six Students
1. Nathaniel Paddock, Brockport
2. Joshua Krienke, Bard & Matthew Too, Brockport
3. George Clapper, Geneseo; Hadley Chan, Oneonta & Stacey Do, Brockport
6CHAPTER 2. TOP SIX STUDENTS
Every number of the type aabb is divisible by 11 since a−a+b−b= 0 is
divisible by 11. Being a square it must, in fact, be divisible by 121.
Then we multiply 121 by other squares to see which one leads to a product
of the type we want.
We get 121 ·64 = 7744 = 882.
One can prove that if a square ends with two identical digits the only pos-
sibilities are either 00 or 44.
First, since a square cannot end in 2, 3, 7, or 8, we discard 22, 33, 77, 88.
Next I will show that if a square ends with an odd digit then the digit before
last must be even. This will allow us to discard 11, 55, 99.
When the last digit is odd the number that is squared also ends with an odd
digit. Let this number be 10x+ywhere yis the last digit, odd.
(10x+y)2= 100x2+ 20yx +y2
8CHAPTER 3. SOLUTIONS
Notice that 100x2does not inﬂuence the last two digits of the square while
20yx ends with a 0 and the digit before last is even.
y2is, for odd y, one of 01, 09, 25, 49, or 81, all having an even digit before
Therefore for the square the digit before last is even.
The only case that remains to be discarded is 66.
I will show that if a square ends in 6 then the digit before last is odd.
If the square ends in 6 then the number that is squared ends either with a 4
or with a 6.
In the ﬁrst case
(10x+ 4)2= 100x2+ 80x+ 16
while in the second
(10x+ 6)2= 100x2+ 120x+ 36
As before, 100x2does not inﬂuence the last two digits of the square. Both
80xand 120xend with a 0 and the digit before last is even. In both cases,
added with the odd digit of the third term will result in an odd digit before last.
Therefore if aabb is a square then it is either aa00 or aa44.
In the ﬁrst case
aa00 = aa ·100
which cannot be a square since there is no two digit square number with identical
Therefore for the number we are looking for, the only possibility is aa44.
The rest is just checking which of the 9 possibilities is actually a square.
Let n+ 1 be the number of digits of the number and let xbe the number
made by the last ndigits.
Then the condition in the problem can be expressed by the equation
3(10n+x) = 10x+ 1 ⇐⇒ 7x= 3 ·10n−1 = 299 . . . 9
where the number of 9s is n.
Therefore the problem asks for the smallest nsuch that 7 divides the number
2 followed by n9s. Checking the possibilities we ﬁnd that the least such nis 5
and thus the number is 142857.
Since CD||AB and AB ⊥EA we have that CD ⊥EA.
Since CB||AD and CB ⊥E C we have that EC ⊥AD.
Therefore AD and CD are altitudes in 4AE C. This means that the point
Dis the orthocenter of the triangle and that the third altitude is ED. That is,
ED ⊥AC .
Let Fbe the point where ED intersects AC.
Then, from 4AEF ,
m(∠AEF ) = π
while from 4EAB,
m(∠BAC) = π
m(∠EAB) + m(∠ECB) = π
we conclude that the quadrilateral EABC is cyclic. Thus
∠CE B ≡∠BAC.
Thus we get that
∠AEF ≡∠C EB
which implies the conclusion.
f(x)dx = 2 ⇐⇒ Z1
(f(x)−sin(πx))dx = 0
By the mean value theorem for integrals, there is a∈(0,1) such that
which implies the conclusion.
10 CHAPTER 3. SOLUTIONS
We will show that the maximum is 2.
First solution (Marius Munteanu)
Without loss of generality, we may assume that the circle is centered at the
origin of the plane coordinate system and that A, B, C correspond to the com-
plex numbers 1, ω, ω2,where ωis a third root of unity. With Pcorresponding
to z, the inequality to be proven becomes
|(z−1)(z−ω)(z−ω2)| ≤ 2
for all zwith |z|= 1.
However, given that ωis a third root of unity, the left side of the inequality
is equal to |z3−1|.By the triangle inequality,
|z3−1| ≤ |z|3+ 1 = 2.
Equality above is achieved when z3=−1.This is equivalent to z6= 1 and
z36= 1.If we consider the regular hexagon having A(thus, also Band C)as
a vertex, then zmust correspond to one of the three vertices of this regular
hexagon, diﬀerent from A, B, and C. In other words, Pmust be the midpoint
of one of the arcs AB, B C, or AC.
Second solution (Marius Munteanu)
Let x, y, z denote the lengths of P A, P B , and P C.
Without loss of generality, we may assume that Pis on the small arc AB.
By Ptolemy’s theorem applied to the cyclic quadrilateral AP BC, we then have
Thus, the inequality to be proven becomes
Since P C =z=x+yis at most as long as the diameter of the circle, we
obtain that x+y≤2.By the triangle inequality, we also have
Equality occurs when x=yand z(= x+y=P C) = 2,which means
x=y= 1,corresponding to Pbeing the midpoint of arc AB. Of course, by
symmetry, equality also holds when Pis the midpoint of arcs AC or AB.
Third solution (Marius Munteanu)
With xand ydenoting the lengths of P A and P B, we get P C =x+yby
Now, under the assumption that Pis on the small arc AB, since angle AP B
has measure 120◦,by the Law of Cosines in triangle AP B, we have x2+y2+xy =
Thus, we have to maximize the function f(x, y) = xy(x+y) = x2y+xy2
subject to the constraints x2+y2+xy = 3, x ≥0, y ≥0.So, by using Lagrange
multipliers, we obtain
2xy +y2=λ(2x+y), x2+ 2xy =λ(2y+x).
Subtracting the two equations shows that y2−x2=λ(x−y),which implies
that either x=yor λ=−x−y.
In the ﬁrst case, by x2+xy +y2= 3,we get that x=y= 1.
The second case is actually impossible due to the nonnegativity of xand y,
the nonpositivity of λand the equations above (x=y= 0 is also impossible
due to the constraint).
Since f(1,1) = 2 and since f(0,√3) = f(√3,0) = 0,the largest value of f
is 2 and is achieved when x=y, meaning that Pis the midpoint of the minor
Symmetry considerations imply that the same maximal value is obtained at
the midpoints of arcs AC and BC.
By symmetry it suﬃces to solve the problem for Pon the small arc BC.
Let Dbe the midpoint of this arc.
||P A|| ≤ ||DA|| = 2
because DA is a diameter.
In the same time
||P B|| · ||P C || = Area4P BC ·1
2 sin ∠BP C =1
Let abe the distance from Pto BC and bthe distance from Dto BC. Then
12 CHAPTER 3. SOLUTIONS
||P B|| · ||P C || =2
√3a· ||BC|| ≤ 2
√3Area4DBC = 1
Therefore ||P A|| · ||P B|| · ||P C || ≤ 2 = ||DA|| · ||DB|| · ||DC||.
As a byproduct, here we also solved the problem of ﬁnding the point Pfor
Area4P BC ·Area4P AC ·Area4P B A
is maximal as well as the one where the product of the distances from Pto the
sides of the 4ABC is maximal.
Let f: [0,1] →R,
f(x) = n
√1 + x+n
fis diﬀerentiable on [0,1) and
f0(x) = 1
√1 + x−1
√1 + x−1
√1 + x<1
√1−x < n
√1 + x⇐⇒ 1−x < 1 + x⇐⇒ −2x < 0
which is true.
Therefore f0(x)<0 and thus the function is strictly decreasing. Hence for
f(x)< f(0) = 2.
To complete the proof we need to notice that for n≥2
Since 72 = 8·9 and gcd(8,9) = 1 it suﬃces to show that 8 divides 3n+4n+ 5n
and that 9 divides 3n+ 4n+ 5n.
Let n= 2k+ 1, with k≥1. Then
32k+1 + 42k+1 + 52k+1 = 16 ·42k−1+ 32k+1 + 52k+1
16 ·42k−1+ (3 + 5)(32k+ 32k−1·5 + · ·· + 3 ·52k+ 52k)
which is divisible by 8.
In the same time,
32k+1 + 42k+1 + 52k+1 = 9 ·32k−1+ (4 + 5)(42k+ 42k−1·5 + · ·· + 4 ·52k−1+ 52k)
which is divisible by 9
It can also be proved by induction. Let P(k) : 72 divides 32k+1+42k+1+52k+1
Then P(1) : 72 divides 33+ 43+ 53= 27 + 64 + 125 = 216 = 72 ·3, which is
Assume that 72 divides 32k+1 + 42k+1 + 52k+1.
32k+3 + 42k+3 + 52k+3 = 9 ·33k+1 + 16 ·42k+1 + 25 ·52k+1
= 25(32k+1 + 42k+1 + 52k+1)−16 ·32k+1 −9·42k+1
= 25(32k+1 + 42k+1 + 52k+1)−144 ·32k−1−72 ·42k−1
which is divisible by 72 because each part of the sum is. This completes the
First solution (Laura Munteanu)
a) By hypothesis, triangles 4XUY and 4XU Z are right triangles at U. As
such, XU ⊥U Y and XU ⊥UZ. Therefore X U is perpendicular to the entire
plane determined by the points U, Y, Z and so XU will be perpendicular on
any line or segment included in this plane. Since Y Z ⊂(Y UZ), it follows that
XU ⊥Y Z.
14 CHAPTER 3. SOLUTIONS
On the other hand, XT ⊥Y Z, by hypothesis, and so Y Z is perpendicular
to both XT and XU , which implies that
Y Z ⊥(X U T )
Since UT ⊂(XU T ), it follows that
UT ⊥Y Z
b) Let us ﬁrst note that, from part a), XU ⊥(Y UZ),and so XU ⊥U T . As
such, triangle 4XU T is a right triangle at U.
Now, let XT =hand U T =h1.The area of 4XY Z equals
and so the square of this area equals
In the same time
based upon the Pythagorean theorem applied to the right triangle 4X UT ,
Y Z2=Y U 2+UZ 2,
based upon the Pythagorean theorem applied to the right triangle 4Y UZ.
(Area4XY Z )2=(XU2+h2
4=XU2·Y U 2
4+XU2·U Z 2
This says that
(Area4XY Z )2= (Area4XU Y )2+ (Area4X U Z )2+ (Area4Y U Z )2,
(Area4XY Z )2= (2k)2+ (6k)2+ (2k2−5)2= (2k2+ 5)2
Area4XY Z = 2k2+ 5,
which is an integer value.
Second solution for part b)
Let a, b, c be the lengths of the line segments U X, U Y, U Z respectively. Then
ab = 4k, ac = 12k bc = 4k2−10
In a similar way we get
3and c2= 12k2−30.
Let p, q, r be the lengths of the line segments XY , Y Z, Z X respectively.
Using the Pythagorean theorem we have
p2=8k4+ 32k2+ 50
In the same time, using Heron’s formula, the area of 4XY Z is
4p(p2+q2+ 2pq −r2)(r2−p2−q2+ 2pq)
4r160(4k4+ 16k2+ 25)
9p36k4+ 180k2+ 225 = p4k4+ 20k2+ 25 = p(2k2+ 5)2
= 2k2+ 5
For the result to be a natural number we need n≤2022.
The number of zeros at the end of a factorial equals the power of 5 in the
prime factor decomposition of the factorial. That is because the power of 2 is
always greater than the power of 5 and a zero at the end comes from of factor
of 10, that is, from a 2 ·5.
16 CHAPTER 3. SOLUTIONS
The power of 5 in 2022! is
because for bigger powers of 5 the ﬂoor will be 0.
This sum is
404 + 80 + 16 + 3 = 503.
Hence, to complete the problem, we need to show that there is no number
n≤2022 such that n! ends with 403 zeros.
Notice that the number of zeros at the end of n! is an increasing function of
n. The reason there is no factorial with 403 zeros is because 1624! ends with
625 = 324 + 64 + 12 + 2 = 402
625 = 325 + 65 + 13 + 2 = 405
Let nbe the number of players. Thus, the number of draws of a player is
some number between 0 and n−1.
However, if some player has n−1 draws this means that he had a draw with
each of the other players, so nobody got 0 draws, while if somebody got 0 draws
nobody could get n−1.
This implies that the numbers of draws are either from 0 to n−2 or from 1
In both cases there are only n−1 options. Since we have nplayers, the
pigeonhole principle implies that two players must have the same number of
I will denote by 3knumbers divisible by 3, by 3p+ 1 numbers 1 more than
a multiple of 3 and by 3q+ 2 numbers 2 more than a multiple of 3.
There are 7 possibilities for a sum of ﬁve numbers to be divisible by 3, as a
function of the three cases above:
3·3k+ (3p+ 1) + (3q+ 2)
2·3k+ 3(3p+ 1)
2·3k+ 3(3q+ 2)
3k+ 2(3p+ 1) + 2(3q+ 2)
4(3p+ 1) + (3q+ 2)
(3p+ 1) + 4(3q+ 2)
From 1 to 1000 we have 333 −3knumbers, 334 −3p+ 1 numbers, and
333 −3q+ 2 numbers.
Therefore for Case I we have
choices, for Case II
choices, for Case III
choices, for Case IV
choices, for Case V
choices, for Case VI
18 CHAPTER 3. SOLUTIONS
choices, while for Case VII
Therefore the total number of choices resulting in a sum divisible by 3 is
33,108,454,071 + 678,343,645332 + 340,196,510,952 + 337,140,853,668
+1,023,663,597,714+169,585,911,333+168,058,110,330 = 2,750,097,083,400
The total number of choices is
and hence the probability is
Let nbe the number.
We consider the numbers 1, 11, 111 . . . 111 . . . 1, where the last one has n+ 1
When we divide a number by nthere are npossible remainders. Since we
have n+ 1 numbers, the pigeonhole principle implies that two of them must
have the same remainders when divided by n.
Therefore their diﬀerence will be divisible by n. Their diﬀerence is a number
with only 1s and 0s.
Here we actually proved something more speciﬁc, that every number has a
multiple of the type
1. . . 10 . . . 0=1. . . 1·10k
Using this we can prove now that if a number nis such that gcd(n, 10) = 1
than nhas a multiple made entirely of 1s.
In particular this will be true if nis a prime number other than 2 and 5,
which is a 1915 problem of Traian Lalescu.
20 CHAPTER 4. COMMENTS
The problem can be solved by trial and error but the theory used in the
solutions helps reduce the number of trials.
The use of the test of divisibility by 11 is incidental. However, it provides
the simplest solution.
The test says that a number is divisible by 11 if and only if the alternating
sum of its digits is divisible by 11.
This problem can also be solved by trial and error trying to ﬁgure out the
digits of the number one by one, starting either from the end or from the be-
The two main theorems about cyclic quadrilaterals play a part in the proof.
One says that a quadrilateral is cyclic if and only if the sum of the measures
of two opposite angles is π.
The other one says that a quadrilateral is cyclic if and only if the angle made
by a side and a diagonal is congruent to the angle made by the opposite side
and the other diagonal.
The main idea of the proof is that when one angle is in the interior of another
one, having the same bisector means that the two angles outside the interior
one are congruent.
The ﬁrst step of the proof is to recognize that 2/π is equal to the integral
that replaced it in the proof. That could be guessed from the fact that the
function sin(pix) comes up in the conclusion of the problem so it has to come
in the problem from somewhere.
Another important part of the proof is to apply the Mean Value Theorem
to the integral of the diﬀerence rather than to each integral separate.
If we have
and we apply the theorem for each integral we get that there are numbers
c1, c2∈(a, b) such that
f(x)dx = (b−a)f(c1)
g(x)dx = (b−a)g(c2)
which implies that
(b−a)f(c1)=(b−a)g(c2)⇐⇒ f(c1) = g(c2)
which, although is correct, it is not the conclusion we want.
However, if we ﬁrst write the equation as
(f(x)−g(x))dx = 0
when we apply the Mean Value Theorem we get the same cfor both functions.
The use of complex numbers in the ﬁrst solution is just a more elegant version
of analytic geometry. It gives us a simple algebraic version of the inequality.
Ptolemy’s Theorem, used in the second and the third solution says that in
a cyclic quadrilateral the sum of the products of the opposite sides equals the
product of the diagonals.
The fourth solution relates the product we need to maximize to certain areas.
In fact maximizing
||P A|| · ||P B|| · ||P C ||
is equivalent to maximize
||P A||2· ||P B||2· ||P C ||2= (||P A|| · ||P B||)·(||P B || · ||P C||)·(||P C || · ||P A||)
√3Area4P AB ·4
√3Area4P BC ·4
3√3Area4P AB ·Area4P BC ·Area4P C A
22 CHAPTER 4. COMMENTS
If we denote by x, y, z the distances from Pto AB,BC, and CA respectively,
the last product equals
2x||AB|| · 1
2y||BC|| · 1
Since the side of the equilateral triangle is √3 the product equals
Therefore we also solved the problem of maximizing the product of the areas
and the problem of maximizing the product of the distances from the point to
the sides of the triangle.
The proof is independent of the quantity
The only relevant part is that this quantity belongs to the interval (0,1).
We can obtain an inﬁnite number of similar inequality replacing it by other
quantities in the same interval.
The proof also shows that
r1 + n!
One thing to remember from here is that inequalities are always maximum
or minimum problems and tools from calculus may be used to prove them.
The main ingredient of the ﬁrst proof is the fact that if adivides c,bdivides
cand gcd(a, b) = 1 then ab divides c.
This is no longer true if gcd(a, b)>1 (try to ﬁnd an example).
The rest is just an application of the decomposition formula for the sum of
Part a) use the fact that a line is perpendicular to a plane if and only if
it is perpendicular on two non parallel lines in the plane if and only if it is
perpendicular on every line in the plane.
The theory used here is the following: if pis a prime number then the
exponent of pin the prime number decomposition of n! is
p4+. . .
Notice that even though the sum is inﬁnite, except for a ﬁnite number of
terms at the beginning, the sum consists of zeros.
The solution is a more careful application of the pigeonhole principle.
The answer is both surprising and expected.
Check if it remains true for 1001 and 1002.
Check if it remains true if 4 numbers are chosen.
One fact from number theory used here is that if adivides xy and gcd(a, y) =
1 then adivides x.