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At Right Angles | Vol. 3, No. 2, July 2014 71
problem corner
Problems for the
Middle School
Problem Editor : R. Athmaraman
Problem III-2-M.1
What is the least multiple of 9 which has no odd
digits?
Problem III-2-M.2
Which number is larger: 3111 or 1714?
Problem III-2-M.3
What is the remainder when 20152014 is divided
by 2014?
Problem III-2-M.4
Find the least natural number larger than 1 which
is simultaneously a perfect square, a perfect cube,
a perfect fourth power, a perfect fifth power and a
perfect sixth power. How many such numbers are
there?
Problem III-2-M.5
A group of ten people (men and women), sit
side by side at a long table, all facing the same
direction. In this particular group, ladies always
tell the Truth while the men always lie. Each of
the ten people announces: “There are more men
on my left, than on my right.” How many men
are there in the group? (This problem has been
adapted from the Berkeley Math Circle, Monthly
Contests.)
SOLUTIONS OF PROBLEMS IN ISSUE-III-1
Solution to problem III-1-M.1
Show that the following number is a perfect square
for every positive integer n:
Let an denote the given integer; e.g., a1 = 11−2 =
9, a2 = 1111−22 = 1089. Observe that a1 = 32 and
a2 = 332. That gives us a clue to the solution. Let
bn denote the number with n ones, e.g., b4 = 1111.
The proof that an is a perfect square for all n is
illustrated for the case n = 4 (the general case is
written the same way):
Problems for Solution
PROBLEMS FOR THE MIDDLE SCHOOL
ATHMARAMAN R
PROBLEMS FOR SOLUTION
Problem III-2-M.1 What is the least multiple of 9 which has no odd digits?
Problem III-2-M.2 Which number is larger: 3111 or 1714?
Problem III-2-M.3 What is the remainder when 20152014 is divided by 2014?
Problem III-2-M.4 Find the least natural number larger than 1 which is simultaneously a
perfect square, a perfect cube, a perfect fourth power, a perfect fifth power and a
perfect sixth power. How many such numbers are there?
Problem III-2-M.5 A group of ten people (men and women), sit side by side at a long table,
all facing the same direction. In this particular group, ladies always tell the Truth
while the men always lie. Each of the ten people announces: “There are more
men on my left, than on my right.” How many men are there in the group? (This
problem has been adapted from the Berkeley Math Circle, Monthly Contests.)
SOLUTIONS OF PROBLEMS IN ISSUE-III-1
Solution to problem III-1-M.1 Show that the following number is a perfect square for every
positive integer n:
111111...111111
2n digits
−222...222
n digits
.
Let an denote the given integer; e.g., a1 = 11−2 = 9, a2 = 1111−22 = 1089. Observe
that a1 = 32 and a2 = 332. That gives us a clue to the solution. Let bn denote the
number with n ones, e.g., b4 = 1111. The proof that an is a perfect square for all n
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72 At Right Angles | Vol. 3, No. 2, July 2014
2 ATHMARAMAN R
is illustrated for the case n = 4 (the general case is written the same way):
a4 = 11111111−2222
= 11110000−1111 = 1111×10000−1111×1
= 1111×(104 −1) = 1111×9999
= 1111×1111×9 = 1111×1111×3×3 = (3×1111)2.
In general, an is the square of the number 333...333 which has n threes.
Solution to problem III-1-M.2 On a digital clock, the display reads 6 ∶ 38. What will the
clock display twenty-eight digit changes later?
Let us compute the digit-changes, step by step.
From To # digit changes Cumulative total
6 ∶ 38 6 ∶ 39 1 1
6 ∶ 39 6 ∶ 40 2 3
6 ∶ 40 6 ∶ 49 9 12
6 ∶ 49 6 ∶ 50 2 14
6 ∶ 50 6 ∶ 59 9 23
6 ∶ 59 7 ∶ 00 3 26
7 ∶ 00 7.01 1 27
7.01 7 ∶ 02 1 28
The time is 7 ∶ 02 after twenty-eight digit changes are over.
Solution to problem III-1-M.3 The figure shows a hall ABCDEF with right angles at its
corners. Its area is 2520 sq units, and AB = BC, CD = 30 units, AF = 60 units. A
point P is located on EF such that line CP divides the hall into two parts with
equal area. Find the length EP.
Let AB = x; then BC = x. The area of the hall is then 60(30+x)−30x = 1800+30x
which equals 2520 (given information; see Figure 1). Hence x = 720/30 = 24,
which leads to DE = 60−24 = 36.
Let PE = y. Then the area of the trapezium CDEP is 1
2 ×2520 = 1260. Hence:
1
2
(30+y)×36 = 1260, ∴ 30+y = 2520
36 = 70, ∴ y = 40.
PROBLEMS FOR THE MIDDLE SCHOOL 3
A B
C D
F P E
30
60
x
x
y
FIGURE 1.
a
4
x
y 8
FIGURE 2.
Solution to problem III-1-M.4 In a circle with radius 4 units, a rectangle and an equilateral
triangle are inscribed. If their areas are equal, find the dimensions of the rectangle.
Let the side of the equilateral triangle be a, and let the rectangle have dimensions
x,y (see Figure 2). The radius of the circle is 4 units. The height of the equilateral
triangle is a×
√3/2, and since the radius of the circle is 2/3 of the height, we get:
4 = 2
3 ×a×
√3
2 , ∴ a = 4
√
3.
FIGURE 1
FIGURE 2
�� � �������� � ����
� �������� � ���� � ���� × ����� � ���� × �
� ���� × (��� � �x � ���� × ����
� ���� × ���� × � � ���� × ���� × � × �
� (� × ����x�.
In general, aa� is the square of the number
333 ... 333 which has nn threes.
Solution to problem III-1-M.2 On a digital clock,
the display reads 6 ∶ 38. What will the clock display
twenty-eight digit changes later?
Let us compute the digit-changes, step by step.
The time is 7 ∶ 02 after twenty-eight digit changes
are over.
Solution to problem III-1-M.3 �he �igure shows a
hall AAAAAAAAAAAA with right angles at its corners. Its
area is 2520 sq units, and AAAA A AAAA, CCCCCCC units,
AAAA A AA units. A point PP is located on EEEE such that
line CCCC divides the hall into two parts with equal
area. Find the length EEEE.
Let AAAA A AA; then BBBBBBB. The area of the hall is
then 60(30 + xxx x xxxxx xxxxxxxxx which
equals 2520 (given information; see Figure 1).
Hence xxxxxxxxxxxx, which leads to
DDDDDDDDDDDDD.
Let PPPPP PP. Then the area of the trapezium CCCCCCCC
is �
� × 2520 =1260. Hence:
1
2
(30 + yyyyyyyyyyyy
∴ 30 + yyy
2520
36 = 70, ∴ yyyyyy
2
�� � �������� � ����
� �������� � ���� � ���� × ����� � ���� × �
� ���� × (��� � �x � ���� × ����
� ���� × ���� × � � ���� × ���� × � × �
� (� × ����x�.
In general, aa� is the square of the number
333 ... 333 which has nn threes.
Solution to problem III-1-M.2 On a digital clock,
the display reads 6 ∶ 38. What will the clock display
twenty-eight digit changes later?
Let us compute the digit-changes, step by step.
The time is 7 ∶ 02 after twenty-eight digit changes
are over.
Solution to problem III-1-M.3 �he �igure shows a
hall AAAAAAAAAAAA with right angles at its corners. Its
area is 2520 sq units, and AAAA A AAAA, CCCCCCC units,
AAAA A AA units. A point PP is located on EEEE such that
line CCCC divides the hall into two parts with equal
area. Find the length EEEE.
Let AAAA A AA; then BBBBBBB. The area of the hall is
then 60(30 + xxx x xxxxx xxxxxxxxx which
equals 2520 (given information; see Figure 1).
Hence xxxxxxxxxxxx, which leads to
DDDDDDDDDDDDD.
Let PPPPP PP. Then the area of the trapezium CCCCCCCC
is �
� × 2520 =1260. Hence:
1
2
(30 + yyyyyyyyyyyy
∴ 30 + yyy
2520
36 = 70, ∴ yyyyyy
2
�� � �������� � ����
� �������� � ���� � ���� × ����� � ���� × �
� ���� × (��� � �x � ���� × ����
� ���� × ���� × � � ���� × ���� × � × �
� (� × ����x�.
In general, aa� is the square of the number
333 ... 333 which has nn threes.
Solution to problem III-1-M.2 On a digital clock,
the display reads 6 ∶ 38. What will the clock display
twenty-eight digit changes later?
Let us compute the digit-changes, step by step.
The time is 7 ∶ 02 after twenty-eight digit changes
are over.
Solution to problem III-1-M.3 �he �igure shows a
hall AAAAAAAAAAAA with right angles at its corners. Its
area is 2520 sq units, and AAAA A AAAA, CCCCCCC units,
AAAA A AA units. A point PP is located on EEEE such that
line CCCC divides the hall into two parts with equal
area. Find the length EEEE.
Let AAAA A AA; then BBBBBBB. The area of the hall is
then 60(30 + xxx x xxxxx xxxxxxxxx which
equals 2520 (given information; see Figure 1).
Hence xxxxxxxxxxxx, which leads to
DDDDDDDDDDDDD.
Let PPPPP PP. Then the area of the trapezium CCCCCCCC
is �
� × 2520 =1260. Hence:
1
2
(30 + yyyyyyyyyyyy
∴ 30 + yyy
2520
36 = 70, ∴ yyyyyy
2
Solution to problem III-1-M.4 In a circle with
radius 4 units, a rectangle and an equilateral triangle
are inscri�ed. If their areas are equal, �ind the
dimensions of the rectangle. Let the side of the
equilateral triangle be aa, and let the rectangle have
dimensions xxx xx (see Figure 2). The radius of the
circle is 4 units. The height of the equilateral triangle
is aa a √3/2, and since the radius of the circle is 2/3
of the height, we get:
4 =
2
3
× aa a
√3
2 , ∴ aa aa√3.
Hence the area of the triangle is
√�
� aa� = √�
� × 48 = 12√3. This is also the area of the
rectangle. Since the diagonal of the rectangle has
length 8, we have: xxxx xxx√3 and xx� + yy� = 8�. We
must solve these equations for xxx xx. The second
equation yields yy� = 64 − xx�. Substituting into the
�irst one and squaring, we get:
xx�(64 − xx�) = 432, ∴ xx� − 64xx� + 432 = 0.
Treating this as a quadratic equation in xx�, we get:
�� � �� � √��� � � a ���
� � �� � √����
� � �� � �√��.
So the sides of the rectangle are �32 + 4√37 and
�32 − 4√37.
Solution to problem III-1-M.5 Find the value of
� 2014�
2012 × 2013�−� 2012�
2013 × 2014� .
Let aa aaaaa. The expression within the �irst ‘⌊ ⌋’
then equals:
(aa aaa�
(aa aaaaa = aa� + 3aa� + 3aa aa
aa� − aa
= aa aaa
8
aa aa − 1
aa,
∴ � (aa aaa�
(aa aaaaa � = aa + 4x
since 1 > �
��� > �
� . Similarly, the expression
within the second ‘⌊ ⌋’ equals:
(aa aaa�
aaaaa aaa = aa� − 3aa� − 3aa aa
aa� + aa
= aa aaa
8
aa aa − 1
aa,
∴ � (aa aaa�
aaaaa aaa � = aa − 4x
since 1 > �
��� > �
� . Therefore the given quantity
equals (aa aaaaaaa aaaaa.
3
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At Right Angles | Vol. 3, No. 2, July 2014 73
�� � �������� � ����
� �������� � ���� � ���� × ����� � ���� × �
� ���� × (��� � �x � ���� × ����
� ���� × ���� × � � ���� × ���� × � × �
� (� × ����x�.
In general, aa� is the square of the number
333 ... 333 which has nn threes.
Solution to problem III-1-M.2 On a digital clock,
the display reads 6 ∶ 38. What will the clock display
twenty-eight digit changes later?
Let us compute the digit-changes, step by step.
The time is 7 ∶ 02 after twenty-eight digit changes
are over.
Solution to problem III-1-M.3 �he �igure shows a
hall AAAAAAAAAAAA with right angles at its corners. Its
area is 2520 sq units, and AAAA A AAAA, CCCCCCC units,
AAAA A AA units. A point PP is located on EEEE such that
line CCCC divides the hall into two parts with equal
area. Find the length EEEE.
Let AAAA A AA; then BBBBBBB. The area of the hall is
then 60(30 + xxx x xxxxx xxxxxxxxx which
equals 2520 (given information; see Figure 1).
Hence xxxxxxxxxxxx, which leads to
DDDDDDDDDDDDD.
Let PPPPP PP. Then the area of the trapezium CCCCCCCC
is �
� × 2520 =1260. Hence:
1
2
(30 + yyyyyyyyyyyy
∴ 30 + yyy
2520
36 = 70, ∴ yyyyyy
2
Solution to problem III-1-M.4 In a circle with
radius 4 units, a rectangle and an equilateral triangle
are inscri�ed. If their areas are equal, �ind the
dimensions of the rectangle. Let the side of the
equilateral triangle be aa, and let the rectangle have
dimensions xxx xx (see Figure 2). The radius of the
circle is 4 units. The height of the equilateral triangle
is aa a √3/2, and since the radius of the circle is 2/3
of the height, we get:
4 =
2
3
× aa a
√3
2 , ∴ aa aa√3.
Hence the area of the triangle is
√�
� aa� = √�
� × 48 = 12√3. This is also the area of the
rectangle. Since the diagonal of the rectangle has
length 8, we have: xxxx xxx√3 and xx� + yy� = 8�. We
must solve these equations for xxx xx. The second
equation yields yy� = 64 − xx�. Substituting into the
�irst one and squaring, we get:
xx�(64 − xx�) = 432, ∴ xx� − 64xx� + 432 = 0.
Treating this as a quadratic equation in xx�, we get:
�� � �� � √��� � � a ���
� � �� � √����
� � �� � �√��.
So the sides of the rectangle are �32 + 4√37 and
�32 − 4√37.
Solution to problem III-1-M.5 Find the value of
� 2014�
2012 × 2013�−� 2012�
2013 × 2014� .
Let aa aaaaa. The expression within the �irst ‘⌊ ⌋’
then equals:
(aa aaa�
(aa aaaaa = aa� + 3aa� + 3aa aa
aa� − aa
= aa aaa
8
aa aa − 1
aa,
∴ � (aa aaa�
(aa aaaaa � = aa + 4x
since 1 > �
��� > �
� . Similarly, the expression
within the second ‘⌊ ⌋’ equals:
(aa aaa�
aaaaa aaa = aa� − 3aa� − 3aa aa
aa� + aa
= aa aaa
8
aa aa − 1
aa,
∴ � (aa aaa�
aaaaa aaa � = aa − 4x
since 1 > �
��� > �
� . Therefore the given quantity
equals (aa aaaaaaa aaaaa.
3
Solution to problem III-1-M.4 In a circle with
radius 4 units, a rectangle and an equilateral
triangle are inscribed. If their areas are equal, find
the dimensions of the rectangle.
Let the side of the equilateral triangle be a, and
let the rectangle have dimensions x, y (see Figure
2). The radius of the circle is 4 units. The height of
the equilateral triangle is
Solution to problem III-1-M.4 In a circle with
radius 4 units, a rectangle and an equilateral triangle
are inscri�ed. If their areas are equal, �ind the
dimensions of the rectangle. Let the side of the
equilateral triangle be aa, and let the rectangle have
dimensions xxx xx (see Figure 2). The radius of the
circle is 4 units. The height of the equilateral triangle
is aa a √3/2, and since the radius of the circle is 2/3
of the height, we get:
4 =
2
3
× aa a
√3
2 , ∴ aa aa√3.
Hence the area of the triangle is
√�
� aa� = √�
� × 48 = 12√3. This is also the area of the
rectangle. Since the diagonal of the rectangle has
length 8, we have: xxxx xxx√3 and xx� + yy� = 8�. We
must solve these equations for xxx xx. The second
equation yields yy� = 64 − xx�. Substituting into the
�irst one and squaring, we get:
xx�(64 − xx�) = 432, ∴ xx� − 64xx� + 432 = 0.
Treating this as a quadratic equation in xx�, we get:
�� � �� � √��� � � a ���
� � �� � √����
� � �� � �√��.
So the sides of the rectangle are �32 + 4√37 and
�32 − 4√37.
Solution to problem III-1-M.5 Find the value of
� 2014�
2012 × 2013�−� 2012�
2013 × 2014� .
Let aa aaaaa. The expression within the �irst ‘⌊ ⌋’
then equals:
(aa aaa�
(aa aaaaa = aa� + 3aa� + 3aa aa
aa� − aa
= aa aaa
8
aa aa − 1
aa,
∴ � (aa aaa�
(aa aaaaa � = aa + 4x
since 1 > �
��� > �
� . Similarly, the expression
within the second ‘⌊ ⌋’ equals:
(aa aaa�
aaaaa aaa = aa� − 3aa� − 3aa aa
aa� + aa
= aa aaa
8
aa aa − 1
aa,
∴ � (aa aaa�
aaaaa aaa � = aa − 4x
since 1 > �
��� > �
� . Therefore the given quantity
equals (aa aaaaaaa aaaaa.
3
, and since the
radius of the circle is 2/3 of the height, we get
Solution to problem III-1-M.4 In a circle with
radius 4 units, a rectangle and an equilateral triangle
are inscri�ed. If their areas are equal, �ind the
dimensions of the rectangle. Let the side of the
equilateral triangle be aa, and let the rectangle have
dimensions xxx xx (see Figure 2). The radius of the
circle is 4 units. The height of the equilateral triangle
is aa a √3/2, and since the radius of the circle is 2/3
of the height, we get:
4 =
2
3
× aa a
√3
2 , ∴ aa aa√3.
Hence the area of the triangle is
√�
� aa� = √�
� × 48 = 12√3. This is also the area of the
rectangle. Since the diagonal of the rectangle has
length 8, we have: xxxx xxx√3 and xx� + yy� = 8�. We
must solve these equations for xxx xx. The second
equation yields yy� = 64 − xx�. Substituting into the
�irst one and squaring, we get:
xx�(64 − xx�) = 432, ∴ xx� − 64xx� + 432 = 0.
Treating this as a quadratic equation in xx�, we get:
�� � �� � √��� � � a ���
� � �� � √����
� � �� � �√��.
So the sides of the rectangle are �32 + 4√37 and
�32 − 4√37.
Solution to problem III-1-M.5 Find the value of
� 2014�
2012 × 2013�−� 2012�
2013 × 2014� .
Let aa aaaaa. The expression within the �irst ‘⌊ ⌋’
then equals:
(aa aaa�
(aa aaaaa = aa� + 3aa� + 3aa aa
aa� − aa
= aa aaa
8
aa aa − 1
aa,
∴ � (aa aaa�
(aa aaaaa � = aa + 4x
since 1 > �
��� > �
� . Similarly, the expression
within the second ‘⌊ ⌋’ equals:
(aa aaa�
aaaaa aaa = aa� − 3aa� − 3aa aa
aa� + aa
= aa aaa
8
aa aa − 1
aa,
∴ � (aa aaa�
aaaaa aaa � = aa − 4x
since 1 > �
��� > �
� . Therefore the given quantity
equals (aa aaaaaaa aaaaa.
3