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The Third SUNYMathics

Gabriel T. Prˇajiturˇa

ii

Copyright ©2022 Gabriel T. Prˇajiturˇa

iii

In memory of Ioan Serdean

iv

Foreword

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

Bucharest

April 2022

Contents

1 The Problems 1

2 Top Six Students 5

3 Solutions 7

4 Comments 19

v

vi CONTENTS

Chapter 1

The Problems

1

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.

Mihai Barbu

4. Let fbe a continuous function such that

πZ1

0

f(x)dx = 2

Prove that there is a number a∈(0,1) such that

f(a) = sin(πa)

Dumitru Batinet¸u

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.

Marius Munteanu

3

6. Prove that for all n≥2

n

r1 + n!

nn+n

r1−n!

nn<2

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.

Laura Munteanu

9. Prove that there is no natural number nsuch that

2022!

n!

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

and ones.

4CHAPTER 1. THE PROBLEMS

Chapter 2

Top Six Students

1. Nathaniel Paddock, Brockport

2. Joshua Krienke, Bard & Matthew Too, Brockport

3. George Clapper, Geneseo; Hadley Chan, Oneonta & Stacey Do, Brockport

5

6CHAPTER 2. TOP SIX STUDENTS

Chapter 3

Solutions

Problem 1

First solution

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.

Second solution

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.

Then

(10x+y)2= 100x2+ 20yx +y2

7

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

last.

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

digits.

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.

Problem 2

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.

Problem 3

9

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

2−m(∠EAC )

while from 4EAB,

m(∠BAC) = π

2−m(∠EAC )

Hence

∠AEF ≡∠BAC

Since

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.

Problem 4

πZ1

0

f(x)dx = 2 ⇐⇒ Z1

0

f(x)dx =2

π⇐⇒ Z1

0

f(x)dx =Z1

0

sin(πx)dx

⇐⇒ Z1

0

(f(x)−sin(πx))dx = 0

By the mean value theorem for integrals, there is a∈(0,1) such that

Z1

0

(f(x)−sin(πx))dx =f(a)−sin(πa),

which implies the conclusion.

10 CHAPTER 3. SOLUTIONS

Problem 5

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

z=x+y.

Thus, the inequality to be proven becomes

xy(x+y)≤2.

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

xy ≤x+y

22

≤1.

Thus, xy(x+y)≤1·2=2.

11

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

Ptolemy’s theorem.

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 =

AB2= 3.

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

arc AB.

Symmetry considerations imply that the same maximal value is obtained at

the midpoints of arcs AC and BC.

Fourth solution

By symmetry it suﬃces to solve the problem for Pon the small arc BC.

Let Dbe the midpoint of this arc.

Then

||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

√3Area4P BC

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

√3b· ||BC||

=4

√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

which

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.

Problem 6

Let f: [0,1] →R,

f(x) = n

√1 + x+n

√1−x

fis diﬀerentiable on [0,1) and

f0(x) = 1

nn

√1 + x−1

nn

√1−x

For x∈(0,1),

1

nn

√1 + x−1

nn

√1−x<0⇐⇒ 1

nn

√1 + x<1

nn

√1−x

⇐⇒ n

√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

x∈(0,1)

f(x)< f(0) = 2.

To complete the proof we need to notice that for n≥2

0<n!

nn<1

Problem 7

13

First solution

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

Second solution

It can also be proved by induction. Let P(k) : 72 divides 32k+1+42k+1+52k+1

for k≥1.

Then P(1) : 72 divides 33+ 43+ 53= 27 + 64 + 125 = 216 = 72 ·3, which is

true.

Assume that 72 divides 32k+1 + 42k+1 + 52k+1.

Then

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

induction.

Problem 8

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

h·Y Z

2,

and so the square of this area equals

h2·Y Z2

4

In the same time

h2=XU2+h2

1,

based upon the Pythagorean theorem applied to the right triangle 4X UT ,

whereas

Y Z2=Y U 2+UZ 2,

based upon the Pythagorean theorem applied to the right triangle 4Y UZ.

Then

(Area4XY Z )2=(XU2+h2

1)(Y U2+UZ2)

4=XU2·Y U 2

4+XU2·U Z 2

4+h2

1·Y Z2

4

This says that

(Area4XY Z )2= (Area4XU Y )2+ (Area4X U Z )2+ (Area4Y U Z )2,

meaning that

(Area4XY Z )2= (2k)2+ (6k)2+ (2k2−5)2= (2k2+ 5)2

Thus,

Area4XY Z = 2k2+ 5,

which is an integer value.

Second solution for part b)

15

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

Then

a2=ab ·ac

bc =48k2

4k2−10 =24k2

2k2−5

In a similar way we get

b2=4k2−10

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

3(2k2−5) q2=40k2−100

3r2=24k4−96k2+ 150

2k2−5

In the same time, using Heron’s formula, the area of 4XY Z is

1

4p(p+q+r)(p+q−r)(p−q+r)(−p+q+r)

=1

4p((p+q)2−r2)(r2−(p−q)2)

=1

4p(p2+q2+ 2pq −r2)(r2−p2−q2+ 2pq)

=1

4p4p2q2−(p2+q2−r2)2

=1

4r160(4k4+ 16k2+ 25)

9−16(2k2−5)2

9

=1

9p36k4+ 180k2+ 225 = p4k4+ 20k2+ 25 = p(2k2+ 5)2

= 2k2+ 5

Problem 9

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

2022

5+2022

25 +2022

125 +2022

625

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

1624

5+1624

25 +1624

125 +1624

625 = 324 + 64 + 12 + 2 = 402

zeros, while

1625

5+1625

25 +1625

125 +1625

625 = 325 + 65 + 13 + 2 = 405

zeros.

Problem 10

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

to n−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

draws.

Problem 11

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.

17

There are 7 possibilities for a sum of ﬁve numbers to be divisible by 3, as a

function of the three cases above:

I

5·3k

II

3·3k+ (3p+ 1) + (3q+ 2)

III

2·3k+ 3(3p+ 1)

IV

2·3k+ 3(3q+ 2)

V

3k+ 2(3p+ 1) + 2(3q+ 2)

VI

4(3p+ 1) + (3q+ 2)

VII

(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

333

5= 33,108,454,071

choices, for Case II

333

3334

1333

1= 678,343,645332

choices, for Case III

333

2334

3= 340,196,510,952

choices, for Case IV

333

2333

3= 337,140,853,668

choices, for Case V

333

1334

2333

2= 1,023,663,597,714

choices, for Case VI

334

4333

1= 169,585,911,333

18 CHAPTER 3. SOLUTIONS

choices, while for Case VII

334

1333

4= 168,058,110,330

choices.

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

1000

5= 8,250,291,250,200

and hence the probability is

2,750,097,083,400

8,250,291,250,200 =1

3

Problem 12

Let nbe the number.

We consider the numbers 1, 11, 111 . . . 111 . . . 1, where the last one has n+ 1

digits.

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.

Chapter 4

Comments

19

20 CHAPTER 4. COMMENTS

Problem 1

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.

Problem 2

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-

ginning.

Problem 3

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.

Problem 4

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

21

to the integral of the diﬀerence rather than to each integral separate.

If we have

Zb

a

f(x)dx =Zb

a

g(x)dx

and we apply the theorem for each integral we get that there are numbers

c1, c2∈(a, b) such that

Zb

a

f(x)dx = (b−a)f(c1)

and

Zb

a

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

Zb

a

(f(x)−g(x))dx = 0

when we apply the Mean Value Theorem we get the same cfor both functions.

Problem 5

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||)

4

√3Area4P AB ·4

√3Area4P BC ·4

√3Area4P CA

=64

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

64

3√3·1

2x||AB|| · 1

2y||BC|| · 1

2z||CA||

Since the side of the equilateral triangle is √3 the product equals

8xyz

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.

Problem 6

The proof is independent of the quantity

n!

nn

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

n

r1 + n!

nn+n

r1−n!

nn>n

√2.

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.

Problem 7

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

odd powers.

Problem 8

23

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.

Problem 9

The theory used here is the following: if pis a prime number then the

exponent of pin the prime number decomposition of n! is

n

p+n

p2+n

p3+n

p4+. . .

Notice that even though the sum is inﬁnite, except for a ﬁnite number of

terms at the beginning, the sum consists of zeros.

Problem 10

The solution is a more careful application of the pigeonhole principle.

Problem 11

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.

Problem 12

One fact from number theory used here is that if adivides xy and gcd(a, y) =

1 then adivides x.