Suzuki Kawasaki
AFTER-SCHOOL
MATHS
100 Challenging Maths Problems
c
Copyright �2020
Suzuki Kawasaki
All rights reserved. This book or any portion thereof may not be reproduced or used in any
manner whatsoever without the express written permission of the publisher except for the
use of brief quotations in a book review.
ISBN 978-1-0933-3417-3
Preface
This book contains 100 challenging algebra problems which were carefully selected
from different national mathematical competitions. These problems involve applying
a variety of algebra skills. This kind of after-school problem solving is tantalizing and
incorporates all sorts of mathematical skills. The exercises have been carefully selected
to encourage young students to use their creativity and imagination.The exercises and
their full solutions present clear concepts and provide helpful tips and tricks. The book
can equally be used in the classroom or in extracurricular activities.
Suzuki Kawasaki
i
Contents
Preface
i
1 Winter Challenges
1
2 Spring Challenges
15
3 Summer Challenges
27
4 Autumn Challenges
41
iii
Chapter 1
Winter Challenges
Problem: 1
Let a, b, c be real numbers such that a + b + c = 6. Prove that:
a2 + b2 + c2 ≥ 12
Proof.
We know that
(a − 2)2 ≥ 0
Therefore, we get
a2 ≥ 4a − 4
Similarly, we obtain
b2 ≥ 4b − 4
c2 ≥ 4c − 4
Summing up all these inequalities, we get
a2 + b2 + c2 ≥ 4(a + b + c) − 12
On the other hand, we know that a + b + c = 6, therefore
a2 + b2 + c2 ≥ 4 · 6 − 12 = 24 − 12 = 12
The proof is now complete.
Problem: 2
Find all integer numbers x, y, z such that:
(x − y)3 + (y − z)3 + (z − x)3 = 30
1
After-School Maths
Proof.
If a, b, c are real numbers such that a + b + c = 0, we have
a3 + b3 + c3 = 3abc
Let


a = x − y
b=y−z


c=z−x
We have
a+b+c=x−y+y−z+z−x=0
Therefore, we get
(x − y)3 + (y − z)3 + (z − x)3 = 3(x − y)(y − z)(z − x)
Consequently, we obtain
(x − y)(y − z)(z − x) = 10
Therefore
x − y, y − z, z − x ∈ {±1, ±2, ±5, ±10}
We can easily check that there is no such triplet!
The proof is complete.
Problem: 3
Evaluate the following expression:
�
�
√
√
3
3
E = 27 + 756 + 27 − 756
Proof.
Let a =
�
3
27 +
√
756 and b =
3
3
�
3
27 −
√
��
756. Therefore, we get
�3
√
3
��
3
27 + 756 +
27 −
√
√
= 27 + 756 + 27 − 756
a +b =
√
= 54
Similarly, we have
�
�
√
3
756 · 27 − 756
�
√
√
3
= (27 + 756) · (27 − 756)
�
= 3 (272 − 756)
ab =
3
27 +
√
2
756
�3
CHAPTER 1. WINTER CHALLENGES
�
3
(729 − 756)
√
= 3 −27
=
= −3
On the other hand, we know that
a3 + b3 = (a + b)(a2 − ab + b2 )
Therefore
(a + b)(a2 − ab + b2 ) = 54
Or equivalently
(a + b)((a + b)2 − 3ab) = 54
Let a + b = u. We have
u(u2 + 9) = 54
Or
u3 + 9u − 54 = 0
Now, observe that u = 3 is the unique real root of this equation while the other roots are
not real. Therefore
E=3
Problem: 4
Suppose that a and b are are non-zero real numbers, and that a and b are roots
of the equation x2 + ax + b = 0. Then, find the pair (a, b).
Proof.
We know that a and b are the roots of x2 + ax + b = 0.
formulas, we obtain
�
a + b = −a
ab = b
Or equivalently
�
b = −2a
(a − 1)b = 0
Furthermore, we know that b �= 0, therefore
�
a=1
b = −2
Conclusion
(a, b) = (1, −2)
3
According to Vieta’s
After-School Maths
Problem: 5
Factorise:
y4 + 4
Proof.
We have
y 4 + 4 = y 4 + 4y 2 + 4 − 4y 2
= (y 2 + 2)2 − 4y 2
= (y 2 − 2y + 2)(y 2 + 2y + 2)
The proof is complete.
Problem: 6
√
If x = 2 + 1, find the value of:
�
�
√
√ �
4
2
4
2 (x + 1) (x + 1) (x + 1) + x − 2
Proof.
If x =
√
2 + 1, we get
�
√
x−1= 2
√
x− 2=1
Now, Let
A=
We have
�
4
�
√ �
(x − 1) (x + 1) (x2 + 1) (x4 + 1) + x − 2
A=
=
=
=
=
=
�
�
√ �
(x2 − 1) (x2 + 1) (x4 + 1) + x − 2
�
�
√ �
4
4
4
(x − 1) (x + 1) + x − 2
�
�
√ �
4
(x8 − 1) + x − 2
�
√
4
x8 − 1 + x − 2
�
4
x8 − 1 + 1
√
4
x8
4
= x2
4
CHAPTER 1. WINTER CHALLENGES
√
2)2
√
=3+2 2
= (1 +
Finally, we get
√
A=3+2 2
The proof is now complete.
Problem: 7
Find all values of y such that:
3
4
√ +
√ =1
2+ y 2+ y
Proof.
If we consider a = 2 +
√
y, a �= 0, we get
3 4
+ =1
a a
Or equivalently
7
=1
a
Therefore, we get
a=7
√
On the other hand, we know that a = 2 + y. Therefore, we get
√
2+ y =7
which means that
√
y=5
Finally, we get
y = 25
Problem: 8
Find all values of x such that:
7
3
3x + 21
+
= 2
x + 21 x − 7
x + 14x − 147
Proof.
Writing them with a common denominator, we get
7x − 49 + 3x + 63
3x + 21
= 2
2
x + 14x − 147
x + 14x − 147
5
After-School Maths
Or equivalently
x2
Cross multiplying, we get
10x + 14
3x + 21
= 2
+ 14x − 147
x + 14x − 147
10x + 14 = 3x + 21
7x = 7
Finally, we get
x=1
The proof is complete.
Problem: 9
In the quadratic ax2 + bx + c = 0, a, b �= 0 and c = 0. Let r and s be the roots,
with r ≥ s ≥ 0. The sum of the cubes of the roots is −a, and the bigger root
minus the smaller root is 2. What is (a + b)2 ?
Proof.
Let the two roots be r and s. Now, from Vieta’s formulas, we have
c
= rs
a
Since c = 0, we get
rs = 0
On the other hand, we have r ≥ s ≥ 0, therefore
s=0
Furthermore, given that r − s = 2, we get
r=2
Thus, the roots of the equation are 2 and 0. Since
−a = r3 + s3 = 8
We get
a = −8
Now, with any quadratic equation,
r+s=−
Thus
2=
b
8
Therefore
b = 16
6
b
a
CHAPTER 1. WINTER CHALLENGES
Consequently
(a + b)2 = (−8 + 16)2
= 82
= 64
Finally, we obtain
(a + b)2 = 64
The proof is complete.
Problem: 10
If x +
1
x
= 8, evaluate the following expression:
x2 +
1
x2
Proof.
We have
x2 +
1
1
= x2 + 2 + 2 − 2
2
x
x
�2
�
1
= x+
−2
x
= 82 − 2
= 64 − 2
= 62
Finally, we obtain
x2 +
1
= 62
x2
The proof is complete.
Problem: 11
Find the sum of the roots of the equation:
�
4
x2 − 6x + 9 − 4 x2 − 6x + 6 = 0
Proof.
Let x2 − 6x + 6 = t4 . Therefore, we get
t4 + 3 = 4t
Or equivalently
t4 − 4t + 3 = 0
7
After-School Maths
Or
On the other hand, we have
(t2 + 2t + 3)(t − 1)2 = 0
t2 + 2t + 3 = (t + 1)2 + 2 > 0
Therefore 1 is the only root of (t2 + 2t + 3)(t − 1)2 = 0, and as such
t=1
On the other hand, we have x2 − 6x + 6 = t4 , therefore
x2 − 6x + 6 = 1
Or equivalently
x2 − 6x + 5 = 0
which is a quadratic with two roots x1 = 1 and x2 = 5. Finally, we obtain
x1 + x2 = 1 + 5 = 6
The proof is complete.
Problem: 12
Solve the following equation in R:
(x − 1)2 + (x − 2)2 + (x − 3)2 + ... + (x − 2010)2 = 0
Proof.
Obviously, we have
(x − 1)2 + (x − 2)2 + (x − 3)2 + ... + (x − 2010)2 > 0
The equation does not have any real solution.
Problem: 13
Factor:
x4 − 2x3 + 3x2 − 2x + 1
Proof.
We have
4
3
2
x − 2x + 3x − 2x + 1 =
�
2 2
2
(x ) − 2(x)(x ) + x
2
�
+ 2x2 − 2x + 1
= (x2 − x)2 + 2(x2 − x) + 1
= (x2 − x + 1)2
The proof is complete.
8
CHAPTER 1. WINTER CHALLENGES
Problem: 14
Solve the following equation in R:
4x + 4x+1 = 160
Proof.
We have
Or equivalently
4x + 4x+1 = 4x (1 + 4) = 4x · 5 = 160
4x =
160
5
Therefore
4x = 32
Or equivalently
22x = 25
Consequently, we get
2x = 5
Finally, we obtain
x=
5
2
The proof is complete.
Problem: 15
Find y such that:
�
y+
�
y+
√
y + . . . = 16
Proof.
We have
Therefore, we obtain
Solving, we get
�
y+
�
�
y+
√
y + ... = 16
y + 16 = 16
y = 240
The proof is complete.
Problem: 16
If the roots of the equation x2 + px + q = 0 are α and β, then find the equation
with the roots α12 and β12 .
9
After-School Maths
Proof.
Since α and β are the roots of the equation x2 + px + q = 0, we obtain
�
α + β = −p
αβ = q
The roots of the second equation are
1
α2
and
1
β2 ,
and we have
1
1
(α + β)2 − 2αβ
p2 − 2q
+
=
=
α2
β2
(αβ)2
q2
and
1
1
1
1
× 2 =
= 2
2
2
α
β
(αβ)
q
Plugging in the values of α and β we get the sum and product of roots to be
1
q 2 respectively. Therefore the new equation is
x2 +
p2 −2q
q2
and
1
(p2 − 2q)x
+ 2 =0
2
q
q
Multiplying throughout by q 2 , we get
q 2 x2 + (p2 − 2q)x + 1 = 0
The proof is complete.
Problem: 17
Compute:
√
√
√
√
� 1� + � 2� + � 3� + · · · + � 16�
Proof.
We know that for any positive integers n and k such that n2 ≤ k < (n + 1)2 , we
have
√
� k� = n
In particular, for 1 ≤ k < 4, we have
For 4 ≤ k < 9, we have
For 9 ≤ k < 16, we have
For k = 16, we have
Therefore, the sum is
The proof is complete.
√
� k� = 1
√
� k� = 2
√
� k� = 3
√
� k� = 4
1 · 3 + 2 · 5 + 3 · 7 + 4 · 1 = 38
10
CHAPTER 1. WINTER CHALLENGES
Problem: 18
Find all positive integers (x, y) such that:
1
1
1
+ =
x y
7
Proof.
We have
x+y
1
=
xy
7
Or
7x + 7y = xy
Or
xy − 7x − 7y = 0
Or
xy − 7x − 7y + 49 = 49
Or
(x − 7)(y − 7) = 49
Therefore, x − 7 = 1, y − 7 = 49 or x − 7 = 7, y − 7 = 7 or x − 7 = 49, y − 7 = 1.
Finally, we obtain
(x, y) = (8, 56), (14, 14), (56, 8)
The proof is now complete.
Problem: 19
Solve in R:
�
x4 + y 4 = 706
x+y =8
Proof.
We have
x4 + y 4 = (x + y)4 − 4xy(x + y)2 + 2(xy)2
Therefore, we get
(xy)2 − 128(xy) + 1695 = 0
Solving this quadratic in xy, we get
xy ∈ {15, 113}
But for real x and y, we have
xy ≤
(x + y)2
= 16
4
11
After-School Maths
Therefore, we conclude
xy = 15
Hence, we get the following system of equations
�
x+y =8
xy = 15
Therefore x, y are the roots of
t2 − 8t + 15 = 0
which implies that
(x, y) ≡ (5, 3), (3, 5)
The proof is complete.
Problem: 20
Find the value of:
1+
6
1+
6
1+
6
1+ 6
···
Proof.
Let n be this number. We have
n=1+
6
n
Therefore
n2 − n − 6 = 0
Or equivalently
(n − 3)(n + 2) = 0
Thus n = −2 or 3. Since n is positive,we conclude that
n=3
This ends the proof.
Problem: 21
Given that x +
1
x
= 6, evaluate:
x4 +
1
x4
Proof.
We have
�
1
x+
x
�2
12
= 62
CHAPTER 1. WINTER CHALLENGES
Or
x2 + 2 +
Or
x2 +
Similarly, we have
�
1
= 36
x2
1
= 34
x2
1
x + 2
x
2
Or
x4 + 2 +
�2
= 342
1
= 1156
x4
Finally, we conclude that
x4 +
1
= 1154
x4
The proof is now complete.
Problem: 22
What is the smallest possible value of x for the equation
2
�
4x − 16
3x − 4
�2
+ 102 = 29
�
4x − 16
3x − 4
?
Proof.
First of all, let
4x−16
3x−4
= t. Our equation then becomes
2t2 + 102 = 29t
which we can change to
2t2 − 29t + 102 = 0
which we can factor into
(t − 6)(2t − 17) = 0
Then, we have our roots
t = 6, t =
Now, plugging back in
4x−16
3x−4
17
2
for t, we can solve for x.
Case 1: t = 6
In this case, we get
4x − 16
=6
3x − 4
13
�
After-School Maths
Therefore
4
7
x=
Case 2: t =
17
2
In this case, we get
Therefore
17
4x − 16
=
3x − 4
2
x=
36
43
x=
4
7
The smallest is
The proof is now complete.
Problem: 23
Given that a + b = 4 and ab = 6, evaluate:
a
b
+
b
a
Proof.
We have
b
a 2 + b2
a
+ =
b
a
ab
(a + b)2 − 2ab
=
ab
42 − 2 · 6
=
6
4
=
6
2
=
3
The proof is complete.
Problem: 24
Find all x ∈ R such that:
x4 + 16x − 12 = 0
Proof.
We have
x4 + 16x − 12 = 0
14
CHAPTER 1. WINTER CHALLENGES
Or
x4 − 12 + 16 + 16x − 16 = 0
Or
x4 + 4x2 + 4 − 4x2 + 16x − 16 = 0
Or
(x2 + 2)2 − (2x − 4)2 = 0
Or
(x2 − 2x + 10)(x2 + 2x − 2) = 0
Finally, we get
x = −1 ±
√
3
The proof is complete.
Problem: 25
Find all x ∈ R such that:
�
x+
√
1−x=3
Proof.
After squaring both sides, we get
√
1−x=9−x
Now, squaring again, we get
1 − x = 81 − 18x + x2
Or equivalently
80 − 17x + x2 = 0
which does not have any real roots.
15
After-School Maths
16
Chapter 2
Spring Challenges
Problem: 26
Find the minimum possible value of:
x2 + y 2 + 4x + 6y
Proof.
The equation can be rewritten as
(x + 2)2 + (y + 3)2 − 13
Because (x + 2)2 and (y + 3)2 must be positive, the minimum possible value is −13 which
is attained for x = −2 and y = −3.
The proof is now complete.
Problem: 27
Factorise the following polynomial:
x5 + 10x4 + 35x3 + 50x2 + 24x
Proof.
Trying the small numbers, we can easily check that 0, −1, −2, −3, −4 are all different roots of the polynomial. Therefore, we can factorise it as follows
x5 + 10x4 + 35x3 + 50x2 + 24x = x(x + 1)(x + 2)(x + 3)(x + 4)
17
After-School Maths
Problem: 28
Solve for (a,b,c):


a + b + c = 10
a − b + c = 20


a + b − c = 30
Proof.
If we subtract the first two equations, we get
2b = −10
Therefore
b = −5
If we subtract the first and third equations, we get
2c = −20
Therefore
c = −10
If we substitute both b = −5 and c = −10 in the first equation, we get
a − 5 − 10 = 10
Therefore
a = 25
Finally, we conclude
(a, b, c) = (25, −5, −10)
The proof is now complete.
Problem: 29
The sum of two numbers is two more than their positive difference. What is the
product of the least two positive integers that will satisfy these conditions?
Proof.
Let a and b be the two numbers. Then:
a + b = 2 + (a − b)
Or equivalently
b=2−b
Therefore
b=1
18
CHAPTER 2. SPRING CHALLENGES
Because a must be a positive integer, a = 1. Therefore
ab = 1
The proof is complete.
Problem: 30
If the area of a rectangle is 14, and the perimeter of the rectangle is 18, then what
is the length of the diagonal of the rectangle?
Proof.
We set a as one of the sides and b as the other. We wish to find
�
a 2 + b2
From the given conditions, we have
�
ab = 14
a+b=9
Square the second one
(a + b)2 = 81
Or equivalently
a2 + 2ab + b2 = 81
Or
a2 + b2 = 81 − 28 = 53
Therefore Hence, the length of the diagonal is
�
√
a2 + b2 = 53
The proof is complete.
Problem: 31
Find all positive integers (x, y) such that:

xy = 360
x y
412
 + =
y
x
360
Proof.
We have
x2 + y 2 = 412 = 1681
19
After-School Maths
Therefore, we get
(x + y)2 = x2 + y 2 + 2xy = 492
Similarly, we obtain
(x − y)2 = x2 + y 2 − 2xy = 312
Consequently, we get the following system of equations
�
x + y = 49
|x − y| = 31
Finally, we conclude that
(x, y) = (9, 40), (40, 9)
The proof is now complete.
Problem: 32
Find y such that:
y =2+
2+
−4
2+
−4
2+
−4
−4
2+....∞
Proof.
This is an infinite sequence, so we can substitute a value of y and solve.
say
−4
y =2+
y
Now just solve
y 2 = 2y − 4
Or
y 2 − 2y = −4
Or
y 2 − 2y + 1 = −4 + 1 = −3
Or
(y − 1)2 = −3
There are no real solutions to this problem!
Problem: 33
If x +
1
= 3, find the value of:
x
x4 +
1
1
1
2
+
x
+
+
x
+
x4
x2
x
20
We can
CHAPTER 2. SPRING CHALLENGES
Proof.
If we square x +
1
, we get
x
�
1
x+
x
�2
= x2 + 2 +
Therefore
x2 +
1
=9
x2
1
=7
x2
If we square this again, we obtain
�
1
x + 2
x
2
�2
= x4 + 2 +
Therefore
x4 +
1
= 49
x4
1
= 47
x4
Finally, we obtain
x4 +
1
1
1
+ x2 + 2 + x + = 47 + 7 + 3 = 57
4
x
x
x
The proof is now complete.
Problem: 34
Given a + b = 1 and a2 + b2 = 2. Find the value of:
a 4 + b4
Proof.
We know that a + b = 1, so
a2 + 2ab + b2 = 12 = 1
Now we can substitute since we know a2 + b2 .
2 + 2ab = 1
Or
2ab = −1
Therefore
ab = −
1
2
Now we are trying to find a4 + b4 , so we square a2 + b2 to get
a4 + 2a2 b2 + b4 = 22 = 4
Or
(a4 + b4 ) + 2(ab)2 = 4
21
After-School Maths
Since ab = − 12 , we get
�
�2
(a4 + b4 ) +
1
=4
2
1
(a + b ) + 2 −
2
4
4
Or
=4
Finally, we obtain
a 4 + b4 =
7
2
The proof is now complete.
Problem: 35
√
If x = 2 + 1, find the value of
√
(x + 2)(x2 + 2)(x4 + 4) − (x8 − 20)
Proof.
Notice that
Since x = 1 +
√
x8 − 16 = (x4 + 4)(x2 + 2)(x +
√
2)(x −
√
2)
2, we get
x8 − 16 = (x4 + 4)(x2 + 2)(x +
√
2)
Substitute this into the original equation
(x8 − 16) − (x8 − 20) = 4
This ends the proof.
Problem: 36
Find the sum of the coefficients of the expansion of (x + y)10 .
Proof.
Using the Binomial-theorem, we get
� �
� �
� �
10 10
10 9
10 10
10
(x + y) =
x +
x y + ··· +
y
0
1
10
Therefore the coefficients are going to be
� � � � � � � �
� � � � � �
10
10
10
10
10
10
10
+
+
+
+ ··· +
+
+
0
1
2
3
8
9
10
22
CHAPTER 2. SPRING CHALLENGES
We know that
n � �
�
n
k=0
k
= 2n
Therefore
� � � � � �
� � � � � � � �
10
10
10
10
10
10
10
= 210 = 1024
+
+
+ ···
+
+
+
10
9
8
3
2
1
0
Problem: 37
Find all possible solution(s) to the equation |a| + |b| + |c| = 0, where a, b, and c
are real numbers.
Proof.
The only thing that works is (a, b, c) = (0, 0, 0). This is because the only way
three non-negative real numbers can add to zero is if they are all equal to zero.
Problem: 38
What is the sum of the digits of 99992 ?
Proof.
We have
99992 = (104 − 1)2
= 108 − 2 · 104 + 1
= 100000000 − 20000 + 1
= 9998000 + 1
= 99980001
Therefore, the sum of digits is
9 + 9 + 9 + 8 + 0 + 0 + 0 + 1 = 27 + 8 + 1 = 36
This ends the proof.
Problem: 39
Compute the product:
(19982 − 19962 )(19982 − 19952 )...(19982 − 02 )
(19972 − 19962 )(19972 − 19952 )...(19972 − 02 )
23
After-School Maths
Proof.
By the difference of squares, this is equivalent to
3994 · 2 · 3993 · 3... · 1998 · 1998
3993 · 1 · 3992 · 2... · 1997 · 1997
This is the same as
3994! 1998
1998
7980012
·
= 3994 ·
=
3993! 1997
1997
1997
Problem: 40
Find all values of x such that the value of the expression below is an integer:
x2 + 3x + 7
x+1
Proof.
We have
5
x2 + 3x + 7
=x+2+
x+1
x+1
Since x is an integer, and x + 1 divides 5, we conclude
x + 1 ∈ {1, 5}
Consequently
x ∈ {0, 4}
Thus, the values of x are 0 and 4.
Problem: 41
Solve the equation:
(x2 + x − 1)2 − 6(x2 + x − 2) − 1 = 0
Proof.
Expanding that, we find it equal to
x4 + 2x3 − x2 − 2x + 1 − 6x2 − 6x + 12 − 1 = 0
Or
x4 + 2x3 − 7x2 − 8x + 12 = 0
24
CHAPTER 2. SPRING CHALLENGES
Or
(x − 2)(x − 1)(x + 2)(x + 3) = 0
Finally, the roots of this equation are
x ∈ {−3, −2, 1, 2}
The proof is now complete.
Problem: 42
Let a, b be two integers such that:
�
√
√
252 + 144 3 = a + b 3
Find the sum a + b.
Proof.
Square both sides to get
√
√
252 + 144 3 = a2 + 3b2 + 2ab 3
This means that
�
a2 + 3b2 = 252
ab = 72
Therefore, we conclude that
�
252
< 10
3
Or b divides 72, therefore by trying the possibilities, we get
b≤
(a, b) = (12, 6)
Finally
a + b = 12 + 6 = 18
The proof is complete.
Problem: 43
Prove that N = 1 + n + 2n2 + n3 + n4 | n ∈ N is not a Prime Number.
Proof.
We know that
n4 + n3 + 2n2 + n + 1 = (n4 + 2n2 + 1) + n(n2 + 1)
= (n2 + 1)(n2 + 1) + n(n2 + 1)
= (n2 + 1)(n2 + n + 1)
Therefore, N is never a prime number for n ≥ 1.
25
After-School Maths
Problem: 44
If two positive integers m, n, both bigger than 1, satisfy the equation
20052 + m2 = 20042 + n2
Find the value of m + n.
Proof.
We know that
20052 + m2 = 20042 + n2
Therefore
Or equivalently
n2 − m2 = 20052 − 20042
(n − m)(n + m) = 4009
We will check two cases:
Case 1: n − m = 1, and n + m = 4009. Such that we get n = 2005, and m = 2004.
Case 2: n − m = 19, and n + m = 211. Such that we obtain n = 115, and m = 96.
Therefore, the solution are
(m, n) ∈ {(2004, 2005), (96, 115)}
Finally, we conclude
m + n ∈ {211, 4009}
Problem: 45
a, b and c are three distinct real numbers such that a + b + c = 0. Prove that:
a3 + b3 + c3 = 3abc
Proof.
We have
Therefore
Or
Or
Finally, we get
a + b = −c
(a + b)3 = −c3
a3 + 3ab(a + b) + b3 = −c3
a3 − 3abc + b3 = −c3
a3 + b3 + c3 = 3abc
The proof is complete.
26
CHAPTER 2. SPRING CHALLENGES
Problem: 46
Find x, y and z so that:
√
x+
�
√
1
y − 1 + z − 2 = (x + y + z)
2
Proof.
We have
�
�
√
√
√
√
1
x + y − 1 + z − 2 = (x + y + z) ⇐⇒ 2 x + 2 y − 1 + 2 z − 2 = x + y + z
2
Rearranging all in left side we get
�
√
√
x−2 x+y−2 y−1+z−2 z−2=0
Or equivalently
�
√
√
( x − 1)2 + ( y − 1 − 1)2 + ( z − 2 − 1)2 = 0
Therefore, we conclude that
√
�
√
x − 1 = 0 =⇒ x = 1
y − 1 − 1 = 0 =⇒ y = 2
z − 2 − 1 = 0 =⇒ z = 3
Problem: 47
What is 13 + 23 + ... + 103 ?
Proof.
We know that
n
�
k=1
Then for n = 10, we get
k3 =
1 2
n (n + 1)2
4
13 + 23 + ... + 103 =
1 2 2
10 .11 = 3025
4
Problem: 48
Let x and y be positive real numbers. Prove that:
x y
+ ≥2
y
x
27
After-School Maths
Proof.
We know that
��
Or equivalently
x
−
y
� �2
y
≥0
x
x y
+ −2≥0
y
x
Finally, we conclude that
x y
+ ≥2
y
x
The proof is now complete.
Problem: 49
If the sum of N consecutive integers is 276, then what is N ?
Proof.
We know that
1 + 2 + ... + n =
n(n + 1)
2
Therefore, we get
n(n + 1)
= 276
2
Or equivalently
n(n + 1) = 552
which we can factorise as follows
(n − 23)(n + 24) = 0
Consequently, we obtain
n = 23
The proof is complete.
Problem: 50
Find x such that:
x=1+
1
1+
1
1
1+ x
Proof.
We have
x=1+
28
1
x
CHAPTER 2. SPRING CHALLENGES
Therefore, we get
x2 = x + 1
Or equivalently
x2 − x − 1 = 0
Solving this quadratic equation, we obtain
√
1± 5
x=
2
The proof is now complete.
29
After-School Maths
30
Chapter 3
Summer Challenges
Problem: 51
Let x and y be the roots of the quadratic ax2 + bx + c, where a, b, and c are
constants. If (x − 2)(y − 2) = 8, find 2b+c
a .
Proof.
Since the given quadratic equation has two roots x and y, we have
�
x + y = −b
a
xy = ac
Expand the given equation
(x − 2)(y − 2) = 8
Or
xy − 2(x + y) = 4
Or
c
b
+ 2. = 4
a
a
Finally, we conclude
2b + c
=4
a
The proof is now complete.
Problem: 52
Solve the following equation in R:
x6 + 14x3 + 49 = 0
31
After-School Maths
Proof.
Let t = x3 . Therefore, we get
t2 = (x3 )2 = x6
Hence, we can rewrite the original equation in the following form
t2 + 14t + 49 = x6 + 14x3 + 49 = 0
Consequently, we only need to solve
t2 + 14t + 49 = 0
which is a quadratic equation in t. On the other hand, we have
t2 + 14t + 49 = t2 + 2 · 7 · t + 72
= (t + 7)2
Thus, we get
(t + 7)2 = 0
Or equivalently
x3 = t = −7
Finally, we conclude
√
3
x=− 7
The proof is now complete.
Problem: 53
Simplify:
A=
�
3
√
√
9 3 − 11 2
Proof.
We have
√
√
√
√
√
√
9 3 − 11 2 = 3 3 − 9 2 + 6 3 − 2 2
� √ �3
� √ �2 √
√ �√ � 2 � √ � 3
=
3 −3
3
2+3 3
2 −
2
�√
√ �3
=
3− 2
Finally, we conclude that
�
3
√
√
√
√
9 3 − 11 2 = 3 − 2
32
CHAPTER 3. SUMMER CHALLENGES
Problem: 54
Find x, y such that:
�
(x − 1)(y 2 + 6) = y(x2 + 1)
(y − 1)(x2 + 6) = x(y 2 + 1)
Proof.
Subtracting both equation from each other
2x2 y + 7y − x2 − 2xy 2 − 7x + y 2 = 0
This equation can be factorized to
(2xy − x − 7 − y)(x − y) = 0
There are two cases to consider:
Case 1: x = y.
Plugging this in equation (1), we get
x2 − 5x + 6 = 0
Therefore
x1 = 2, y1 = 2
x2 = 3, y2 = 3
Case 2: 2xy − x − 7 − y = 0.
In this case, we obtain
y=
Plugging this in equation (1), we obtain
x+7
2x − 1
(x2 − x + 4)(x − 2)(x − 3) = 0
From here, we get
x3 = 2, y3 = 3
x4 = 3, y4 = 2
We found all the solutions by this method.
Problem: 55
Find the largest real number x such that :
�
x
x−1
�2
+
�
x
x+1
33
�2
=
325
144
After-School Maths
Proof.
We need to solve the following equation
�
�2 �
�2
x
x
325
+
=
x−1
x+1
144
which we can factorise further as follows
(x − 5)(x + 5)(37x2 − 13)
=0
144(x − 1)2 (x + 1)2
Therefore, we get
x ∈ {−5, 5, −
This ends the proof.
�
13
,
37
�
13
}
37
Problem: 56
In eight years Henry will be three times the age that Sally was last year. Twenty
five years ago their ages added to 83. How old is Henry now?
Proof.
If we note by x, y the ages of Henry and Sally respectively.
ing system
�
x + 8 = 3(y − 1)
(x − 25) + (y − 25) = 83
Or equivalently
�
Solving this system, we obtain
We have the follow-
3y − x = 11
x + y = 133
�
x = 97
y = 36
Therefore, Henry is 97 years old and Sally is 36 years old.
Problem: 57
�
�
If
(x − 1)
(x − 2)
�
(x − 1)
�
(x − 2)... = x − 5, then what is x2 ?
Proof.
We have
�
(x − 1)
�
(x − 2)(x − 5) = x − 5
34
CHAPTER 3. SUMMER CHALLENGES
Or equivalently
(x − 1)2 (x − 2)(x − 5) = (x − 5)4
Or
(x − 5)(11x2 − 70x + 123) = 0
which admits a unique solution
x=5
Finally, we conclude
x2 = 25
The proof is complete.
Problem: 58
√
√
If x + y + x − y = xy , then what is the least possible value of x2 − y 2 ?
Proof.
Squaring both sides we get
Rearranging, we get
�
x2
2x + 2 (x + y)(x − y) = 2
y
x2 − y 2 =
x4 − 4x3 y 2 + 4x2 y 4
4y 4
Or equivalently
x2 − y 2 =
x2 (x − 2y)2
≥0
4y 4
The least possible value is then 0
Problem: 59
What is the coefficient of x3 in the expansion of (2x + 3)5 ?
Proof.
Using the binomial formula, we get
(2x + 3)5 = 32x5 + 240x4 + 720x3 + 1080x2 + 810x + 243
Therefore, the coefficient of x3 is: 720.
Problem: 60
Find the value of positive integer k such that
12 + 22 + 32 + · · · + k 2 = 105k
35
After-School Maths
Proof.
We know that
12 + 22 + 32 + ... + n2 =
Therefore
n(n + 1)(2n + 1)
6
k(k + 1)(2k + 1)
= 105k
6
Hence
(k + 1)(2k + 1) = 630
Or equivalently
(k − 17)(2k + 37) = 0
Therefore, we obtain
k = 17
The proof is complete.
Problem: 61
Given that r1 , r2 , r3 are the roots of x3 − 2x2 + 5x + 1, find:
r1 r2 + r 1 r 3 + r 2 r 3
Proof.
We have
(x − r1 )(x − r2 )(x − r3 ) = x3 − 2x2 + 5x + 1
Or equivalently
x3 − (r1 + r2 + r3 )x2 + (r1 r2 + r2 r3 + r3 r1 )x − r1 r2 r3 = x3 − 2x2 + 5x + 1
Therefore, we conclude that
r 1 r 2 + r2 r 3 + r3 r 1 = 5
The proof is complete.
Problem: 62
Find x in terms of a, if a �= 0, 1, −1
a3 − 1
a(x − 1) + a2 − x
=
a3 + 1
a(x − 1) − a2 + x
Proof.
We have
(a − 1)(a2 + a + 1)
ax − x + a2 − a
=
(a + 1)(a2 − a + 1)
ax + x − a2 − a
36
CHAPTER 3. SUMMER CHALLENGES
Therefore, we get
Or equivalently
(x + a)(a − 1)
(a − 1)(a2 + a + 1)
=
(a + 1)(a2 − a + 1)
(x − a)(a + 1)
(x − a)(a2 + a + 1) = (x + a)(a2 − a + 1)
After simplification, we get
a3 − ax + a = 0
Or
ax = a3 + a
Finally, we get
x = a2 + 1
Problem: 63
Find the sum:
A=
�
2+
�
�
√
2 + 2 + ... + 2
Proof.
We have
A=
√
2+A
After squaring both sides, we get
A2 = 2 + A
Or equivalently
(A − 2)(A + 1) = 0
Or A ≥ 0, therefore, we get
A=2
This ends the proof.
Problem: 64
Alfred is 2 times older than Billy, who is 4 years older than Charlie. The sum of
their ages is 20. How old is Alfred?
Proof.
Let a, b, c be their ages respectively. Thus, we get the following system


a = 2b
b=4+c


a + b + c = 20
37
After-School Maths
Therefore, we get
2b + b + b − 4 = 20
Or equivalently
b=6
Furthermore, we get
a = 2b = 2 · 6 = 12
and
c=b−4=6−4=2
Consequently, we conclude that
Alfred is 12 years old.
Billy is 6 years old.
Charlie is 2 years old.
Problem: 65
Solve for k:
22k+1 + 2k+2 − 2k+1 = 22
Proof.
Let x = 2k . We can rewrite the original equation as follows
2x2 + 4x − 2x = 4
Or equivalently
x2 + x = 2
Or
(x − 1)(x + 2) = 0
Therefore, we obtain
x=1
which means that
k=0
The proof is complete.
Problem: 66
If f (x − 5) = x2 + 5x − 10 and f (x) = ax2 + bx + c, then find the value of:
a+b+c
38
CHAPTER 3. SUMMER CHALLENGES
Proof.
We know that
f (1) = a + b + c
Therefore, we can directly put x = 6 in f (x − 5) = x2 + 5x − 10. Thus, we get
a + b + c = f (1)
= 62 + 5 · 6 − 10
= 36 + 30 − 10
= 56
The proof is now complete.
Problem: 67
If a3 +
1
= 110, find:
a3
a4 +
1
a4
Proof.
We have
a3 +
Therefore, we get
�
Let m = a + a1 . We have
1
a+
a
�3
1
= 110
a3
�
1
−3 a+
a
�
= 110
m3 − 3m − 110 = 0
Or equivalently
(m − 5)(m2 + 5m + 22) = 0
We know that a > 0, therefore
m2 + 5m + 22 > 0
As such, we conclude that
m=a+
1
=5
a
On the other hand, we have
�
�4
�
�
a2
1
1
1
4
2
a+
= a + 4 + 6 2 + 4 a + 2 = 625
a
a
a
a
Therefore, we obtain
a4 +
1
= 625 − 6 − 4(25 − 2) = 527
a4
This ends the proof.
39
After-School Maths
Problem: 68
Find all values of x such that:
√
x+5−
√
x=1
Proof.
√
√
Multiplying both sides with x + 5 + x, we get
√
√
√ √
√
√
( x + 5 − x)( x + 5 + x) = x + 5 + x
Or
√
√
√
√
( x + 5)2 − ( x)2 = x + 5 + x
Or
5=
√
x+5+
√
x
Consequently, we have to solve the following system
�√
√
x+5− x=1
√
√
x+5+ x=5
Summing up the two equations, we obtain
√
2 x+5=6
Or equivalently
√
x+5=3
Finally, we get
x=4
The proof is complete.
Problem: 69
Two positive numbers add up to 8, and the sum of their squares is 34. What is
the sum of their reciprocals?
Proof.
Let a, b be these two numbers. We have
�
a+b=8
a2 + b2 = 34
Therefore, we get
2ab = (a + b)2 − (a2 + b2 ) = 64 − 34 = 30
Therefore
ab = 15
40
CHAPTER 3. SUMMER CHALLENGES
Consequently, we obtain
1 1
a+b
8
+ =
=
a b
ab
15
The proof is now complete.
Problem: 70
Find x such that:
53x − 21 · 5x − 20 = 0
Proof.
Let t = 5x . Therefore, we get the following equation
t3 − 21t − 20 = 0
Or equivalently
(t − 5)(t + 1)(t + 4) = 0
But, we know that t ≥ 0, therefore, we get
t=5
On the other hand, we have t = 5x = 5, therefore
x=1
The proof is now complete.
Problem: 71
Find x if:
3
1+
2
x
= 3x
Proof.
We can simplify the equation
x(x + 1)
=0
x+2
Or x �= 0, therefore the only solution to this equation is
x = −1
This ends the proof.
41
After-School Maths
Problem: 72
Let x, y be real numbers such that |x| �= |y| and:
�
x3 = 13x + 3y
y 3 = 3x + 13y
Evaluate:
(x2 − y 2 )2
Proof.
We will just use sum and difference of cubes factorizations.
nal equations
x3 + y 3 = 16(x + y)
Or
(x + y)(x2 − xy + y 2 ) = 16(x + y)
Therefore
x2 − xy + y 2 = 16
Subtracting the two original equations
x3 − y 3 = 10(x − y)
Or
(x − y)(x2 + xy + y 2 = 10(x − y)
Therefore
x2 + xy + y 2 = 10
Adding the two new equations we get
x2 + y 2 = 13
Subtracting the two new equations we get
xy = −3
We note that
(x2 − y 2 )2 = (x2 + y 2 )2 − 4(xy)2
Therefore, the desired quantity is
(x2 − y 2 )2 = 132 − 4 · (−3)2 = 133
The proof is now complete.
Problem: 73
Evaluate:
1 + 2 + 3 + ... + 2019
42
Adding the two origi-
CHAPTER 3. SUMMER CHALLENGES
Proof.
Let
S = 1 + 2 + ... + 2019
We have
�
S = 1 + 2 + ... + 2018 + 2019
S = 2019 + 2018 + ... + 2 + 1
Adding up the two equations, we get
2S = (1 + 2019) + (2 + 2018) + (3 + 2017) + ... + (2018 + 2) + (2019 + 1)
Or equivalently
2S = 2020 + 2020 + ... + 2020 + 2020
Or
2S = 2019 · 2020
Consequently
S=
2019 · 2020
= 2039190
2
The proof is now complete.
Problem: 74
Let’s consider the following equation:
x4 − 18x3 + kx2 + 200x − 1984 = 0
Determine the value of k if the product of two of the four roots is −32.
Proof.
Let two the factors of the LHS be
(x2 + ax + c)(x2 + bx + d)
We know that c = −32 and as a result
d = 62
So we have
(x2 + ax − 32)(x2 + bx + 62) = x4 + (a + b)x3 + (ab + 30)x2 + (−32b + 62a)x − 1984
So we have two equations
�
a + b = −18
62a − 32b = 200
Solving, we get (a, b) = (−4, −14). Therefore, we conclude
k = (−4) · (−14) + 30 = 56 + 30 = 86
The proof is now complete.
43
After-School Maths
Problem: 75
Find x and y such that:
�
x + y = 16
1
1
1
x + y = 3
Proof.
We have
16
1
x+y
=
=
xy
xy
3
Therefore, we get
xy = 48
We have
�
x + y = 16
xy = 48
Therefore, x, y are the roots of the following quadratic
t2 − 16t + 48 = 0
Or
(t − 12)(t − 4) = 0
Finally, we get
(x, y) ∈ {(4, 12), (12, 4)}
The proof is now complete.
44
Chapter 4
Autumn Challenges
Problem: 76
Let r and s be the roots of the quadratic x2 + bx + c, where b and c are constant.
If (r − 1)(s − 1) = 7, find b + c.
Proof.
We have r and s are the roots of x2 + bx + c = 0. Therefore, we get
�
r + s = −b
rs = c
On the other hand, we have
(r − 1)(s − 1) = rs − r − s + 1
We are given that this equals 7. Hence, we obtain
rs − (r + s) + 1 = 7
Thus
rs − (r + s) = 6
Substituting back in, we get
c − (−b) = 6
Finally, we conclude that
c+b=6
45
After-School Maths
Problem: 77
Find all positive real numbers x, y, z such that:


x = yz
y = z(x + 1)


z = y(x − 1)
Proof.
Multiplying the second and third equations, we get
yz = yz(x − 1)(x + 1)
Or yz = x, therefore we get
x = x(x2 − 1)
But we know that x > 0, hence
x=
√
2
On the other hand, we have
y 2 = yz(x + 1)
= x(x + 1)
√ √
= 2( 2 + 1)
√
=2+ 2
Therefore
y=
Similarly, we get
�
2+
√
2
z 2 = zy(x − 1)
= x(x − 1)
√ √
= 2( 2 − 1)
√
=2− 2
Therefore
�
√
z = 2− 2
Finally, we conclude
(x, y, z) =
�
√
�
�
�
√
√
2, 2 + 2, 2 − 2
The proof is now complete.
46
CHAPTER 4. AUTUMN CHALLENGES
Problem: 78
Let a, b be positive real numbers. Prove that:
a4
b
b2
a2
a
b4
+
+
+
+
+
≥6
b4
b2
b
a a2
a4
Proof.
We know that
�
a2
b2
−
b2
a2
�2
≥0
Or equivalently
a4
b4
+
−2≥0
b4
a4
Therefore
b4
a4
+
≥2
b4
a4
Similarly, we can prove
a2
b2
+
≥2
b2
a2
And
a
b
+ ≥2
b
a
Adding up all these inequalities, we get
a4
b
b2
a2
a
b4
+
+
+
+
+
≥2+2+2=6
b4
b2
b
a a2
a4
The proof is now complete.
Problem: 79
Bob is 24. He is twice as old as Alice was when Bob was as old as Alice is now.
How old is Alice?
Proof.
Let x, y be Bob’s and Alice’s current ages respectively.
system
�
x = 24
x = 2(y − (x − y))
Solving the system, we get
y = 18
Therefore, Alice is 18 years old.
47
We have the following
After-School Maths
Problem: 80
Charlene’s age in years is 16 more than the number of years in the sum of Betty’s
and Ashley’s ages. The square of Charlene’s age is 1632 more than the number if
years in the square of the sum of Betty’s and Ashley’s age. How many years are
in the sum of the ages of the three women?
Proof.
Let a, b, c be Betty’s, Ashley’s and Charlene’s ages respectively.
lowing system
�
c = a + b + 16
c2 = (a + b)2 + 1632
Therefore, we obtain
Or equivalently
(c − a − b)(c + a + b) = 1632
16(a + b + c) = 1632
Consequently, we conclude that
a + b + c = 102
The proof is now complete.
Problem: 81
Let a, b be positive real numbers. Prove that:
a+b+
1 1
+ ≥4
a b
Proof.
We know that
Or equivalently
�
√
1
a− √
a
a+
�2
≥0
1
−2≥0
a
Therefore
a+
1
≥2
a
Similarly, we get
1
≥2
b
Summing up all these inequalities, we obtain
b+
a+b+
1 1
+ ≥2+2=4
a b
The proof is complete.
48
We get the fol-
CHAPTER 4. AUTUMN CHALLENGES
Problem: 82
If a + b + c = 7, ab + bc + ca = 14, abc = 8, what is (1 + a)(1 + b)(1 + c)?
Proof.
We have
(1 + a)(1 + b)(1 + c) = 1 + a + b + c + ab + bc + ca + abc
= 1 + 7 + 14 + 8
= 30
The proof is now complete.
Problem: 83
Find x:
4x(x+4) = 64
Proof.
We know that 64 = 43 . Thus, we get
x(x + 4) = 3
Solving this quadratic, we obtain
x ∈ {−2 −
√
7, −2 +
√
7}
The proof is now complete.
Problem: 84
Twinbrian and Twinemma are siblings. Twinbrian is 4 years old and Twinemma
is 13 years old. What age will each sibling be when Twinemma is twice as old as
Twinbrian?
Proof.
In t years from now, Twinbrian and Twinemma will be 4 + t and 13 + t respectively. We are looking for t such that
13 + t = 2(4 + t)
Therefore, we obtain
t=5
Finally, Twinbrian and Twinemma will be 9 and 18 respectively.
49
After-School Maths
Problem: 85
If x2 − (2p + 3)x + 4 has two roots, and one of them is 2, determine the value of p.
Proof.
Let a be be the other root of this equation. Therefore, we get
�
a + 2 = 2p + 3
2a = 4
Therefore, we obtain
a=2
Consequently
2p + 3 = 4
Finally, we conclude
p=
1
2
The proof is now complete.
Problem: 86
Mr. Jones owns a farm with 25 animals. If there are only chickens and cows, how
many of each are there if there are 70 feet total?
Proof.
Let x and y be the numbers of chickens and cows respectively.
mals in the farm, therefore
x + y = 25
There are 25 ani-
Each chicken has two legs and each cow has four; together the 25 animals have 70 legs.
Hence, we conclude that
2x + 4y = 70
Consequently, we get the following system
�
x + y = 25
2x + 4y = 70
Multiply the first equation by 2 to obtain
�
2x + 2y = 50
2x + 4y = 70
Subtract the first one from the second
2x + 4y − (2x + 2y) = 70 − 50
50
CHAPTER 4. AUTUMN CHALLENGES
Or equivalently
2y = 20
Therefore
y = 10
On the other hand, we know that x + y = 25 and y = 10, thus
x = 25 − 10
= 15
Finally, Mr. Jones has 10 cows, and 15 chickens.
Problem: 87
Evaluate:
√
1
1
1
√ +√
√ + ... + √
√
1+ 2
2+ 3
2018 + 2019
Proof.
Let n be a positive integer. We have
√
n+
Let
S=√
1
√
n+1
=
√
√
√
n+1− n √
= n+1− n
n+1−n
1
1
1
√ +√
√ + ... + √
√
1+ 2
2+ 3
2018 + 2019
Therefore, we get
S=
√
2−
√
1+
√
3−
√
2 + ... +
√
2019 −
√
2018 =
√
2019 − 1
Finally, we conclude
√
√
1
1
1
√ +√
√ + ... + √
√
= 2019 − 1
1+ 2
2+ 3
2018 + 2019
The proof is complete.
Problem: 88
Bob draws a right triangle with integer side lengths all less than 10. He then
inscribes a circle in it. What is the area of the circle?
Proof.
We know that the only right triangle with integer side lengths less than 10 is
(3, 4, 5). Therefore, the area of the triangle is
S=
3·4
=6
2
51
After-School Maths
Let r be the inradius if the triangle. We know that
S=
(5 + 4 + 3)r
=6
2
Therefore, we get
r=1
Finally, the area of the incircle is
Sincircle = π · r2 = π
The proof is complete.
Problem: 89
Let a, b be positive real numbers. Prove that:
�
�
1 1
+
(a + b)
≥4
a b
Proof.
We know that
√
√
( a − b)2 ≥ 0
Therefore, we obtain
Similarly, we know that
Therefore
√
a + b ≥ 2 ab
�
1
1
√ −√
a
b
�2
≥0
1 1
2
+ ≥√
a b
ab
By multiplying the two inequalities, we
�
1
+
(a + b)
a
get
�
√
1
2
≥ 2 ab · √ = 4
b
ab
The proof is now complete.
Problem: 90
Let a, b, c be positive real numbers. Prove that:
a2 + b2 + c2 ≥ ab + bc + ca
52
CHAPTER 4. AUTUMN CHALLENGES
Proof.
We know that
(a − b)2 ≥ 0
Therefore, we get
a2 + b2 − 2ab ≥ 0
Or equivalently
a2 + b2 ≥ 2ab
Similarly, we have
b2 + c2 ≥ 2bc
c2 + a2 ≥ 2ca
Summing up all these inequalities, we obtain
2(a2 + b2 + c2 ) ≥ 2(ab + bc + ca)
Or equivalently
a2 + b2 + c2 ≥ ab + bc + ca
This ends the proof.
Problem: 91
Find all real numbers x, y, z such that:
�
|x − 1| + y − 2x + (z − 3y)2 = 0
Proof.
We have
|x − 1| +
�
y − 2x + (z − 3y)2 = 0
Or all the term are non-negative, thus we get


− 1| = 0
|x
√
y − 2x = 0


(z − 3y)2 = 0
Consequently, we obtain
This ends the proof.


x = 1
y=2


z=6
53
After-School Maths
Problem: 92
Let a, b, c positive integers such that 2016 = 2a · 3b · 7c . Evaluate the following
expression:
a 2 + b2 + c 2
Proof.
We know that
2016 = 25 · 32 · 7
Therefore


a = 5
b=2


c=1
Consequently
a2 + b2 + c2 = 52 + 22 + 12
= 25 + 4 + 1
= 30
The proof is complete.
Problem: 93
Evaluate the following sum:
S = 1 + 2 + 22 + 23 + ... + 22019
Proof.
We have
S = 1 + 2 + 22 + 23 + ... + 22019
Multiply the sum by 2, we get
2S = 2 + 22 + 23 + 24 + ... + 22020
Subtract the first equation from the second, we obtain
2S − S = (2 + 22 + 23 + 24 + ... + 22020 ) − (1 + 2 + 22 + 23 + ... + 22019 )
Or equivalently
S = 22020 − 1
The proof is now complete.
54
CHAPTER 4. AUTUMN CHALLENGES
Problem: 94
Evaluate the following sum:
S=
1
1
1
1
+
+
+ ... +
1·2 2·3 3·4
2018 · 2019
Proof.
Let n be a positive integer. We have
n+1−n
1
1
1
=
= −
n(n + 1)
n(n + 1)
n n+1
Therefore, we get
1
1
1
1
+
+
+ ... +
1·2 2·3 3·4
2018 · 2019
1
1
1 1 1 1
−
= − + − + .... +
1 2 2 3
2018 2019
1
=1−
2019
S=
This ends the proof.
Problem: 95
If the roots of the polynomial x4 − 10x3 + 35x2 − 50x + 24 = 0 are m, n, o, and p,
evaluate the value of:
�
�2mnop
1
1
1 1
+ + +
m n o p
Proof.
According to Vieta’s formulas, we get
�
mnop = 24
mno + mnp + mop + nop = 50
Therefore, we conclude that
�
�2mnop � �2·24 � �48
50
25
1
1
1 1
+ + +
=
=
m n o p
24
12
The proof is now complete.
Problem: 96
Simplify the expression:
√
3
1
1
1
1
√
√
√
√
√
√
√
√
+√
+√
+· · ·+ √
3
3
3
3
3
3
3
3
3
3
1+ 2+ 4
4+ 6+ 9
9 + 12 + 16
100 + 110 + 3 121
55
After-School Maths
Proof.
Let a, b be positive real numbers. We have
√
√
3
a− 3b
1
√
√
√
= �√
√ � �√ 2 √
√ 2�
3
3
3
3
a2 + 3 ab + b2
a− 3b
a + 3 ab + 3 b
√
√
3
a− 3b
=
a−b
Let
S= √
3
1
1
1
1
√
√
√
√
√
√
√
√
+√
+√
+· · ·+ √
,
3
3
3
3
3
3
3
3
3
3
1+ 2+ 4
4+ 6+ 9
9 + 12 + 16
100 + 110 + 3 121
is equivalent to
S=
10
�
n=1
√
3
n2 +
�
3
1
n(n + 1) +
We set n = a and n + 1 = b. Our expression becomes
�
3
(n + 1)2
√
10 √
10
3
�
√
n− 3n+1 �√
3
S=
=
n + 1 − 3 n.
n − (n + 1)
n=1
n=1
Evaluating, we see that our desired sum is
S=
√
3
11 − 1
The proof is complete.
Problem: 97
Solve for x:
�
4+
√
4+
√
8 + 4x +
�
2+
√
√
2+x=2+2 2
8 + 4x +
�
2+
√
√
2+x=2+2 2
Proof.
We have
�
On the other hand, we have
�
√ �
√
√
4 + 8 + 4x = 2 2 + 2 + x
Therefore
Or equivalently
√
√ �
√
(1 + 2) 2 + 2 + x = 2(1 + 2)
�
√
2+ 2+x=2
56
CHAPTER 4. AUTUMN CHALLENGES
Squaring both sides, we get
2+
Therefore
√
√
x+2=4
x+2=2
Now, it is easy to get
x=2
The proof is complete.
Problem: 98
Find x, y such that:
�
2x + 3y+1 = 11
3y + 2x+1 = 7
Proof.
Let a = 2x and b = 3y . Therefore, we get the following system
�
a + 3b = 11
b + 2a = 7
Solving this system of equations, we obtain
�
a=2
b=3
Or equivalently
�
2x = 2
3y = 3
Consequently, we conclude that
(x, y) = (1, 1)
The proof is complete.
Problem: 99
Prove that:
1
1
1
1
+ 2 + 2 + ... +
<2
2
1
2
3
20192
Proof.
Let n > 1 be a positive integer. We know that
1
1
1
1
=
−
<
2
n
n(n − 1)
n−1 n
57
After-School Maths
Let
S=
1
1
1
1
+ 2 + 2 + ... +
2
1
2
3
20192
Therefore, we get
S <1+
1 1
1
1
1
− + ... +
−
=2−
<2
1 2
2018 2019
2019
The proof is complete.
Problem: 100
Find all real numbers x such that:
� 4
�2
�
� 4
x +1
x +1
−3
+2=0
x2
x2
Proof.
Let t =
x4 +1
x2
> 0. We get the following equation
t2 − 3t + 2 = 0
Or equivalently
(t − 2)(t − 1) = 0
There are two cases:
Case 1: t = 1.
In this case, we get
x4 + 1
=1
x2
Which doesn’t have any real solution.
Case 2: t = 2.
In this case, we get
x4 + 1
=2
x2
Which we can rewrite as follows
(x2 − 1)2 = 0
Therefore
x ∈ {−1, 1}
Finally, we conclude that −1, 1 are the only solutions of this equation.
58