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68 At Right Angles | Vol. 3, No. 1, March 2014

problem corner

Problems for the

Middle School

Problem Editor : R. Athmaraman

Problems for Solution Problems for Solution

Problem III-1-M.1

Show that the following number is a perfect

square for every positive integer nnnn:

111111 ... 111111 �������������

�� digits

− 222 ... 222 �������

� digits

.

For example, 11 − 2 = 9 and 1111 − 22 = 1089

are perfect squares.

Problem III-1-M.2

On a digital clock, the display reads 6 ∶ 38. What

will the clock display twenty-eight digit changes

later?

Problem III-1-M.3

�he �igure shows a hall AAAAAAAAAAAAAAAAAAAAAAAA with right angles

at its corners. Its area is 2520 sq units, and

AAAAAAAA = AAAAAAAA, AAAAAAAA AAA units, AAAAAAAA = 60 units. A point PPPP

is located on AAAAAAAA such that line AAAAAAAA divides the hall

into two parts with equal area. Find the length AAAAAAAA.

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, �ind the dimensions of the rectangle.

Problem III-1-M.5

Find the value of the following (no calculators!):

� 2014�

2012 × 2013� − � 2012�

2013 × 2014� .

Here the symbol ⌊ ⌋ has the following meaning: if xxxx

is any real number, ⌊xxxx⌋ is the largest integer not

greater than xxxx. For example, ⌊3.2⌋ = 3, and

⌊−1.7⌋ = −2. It is called the “greatest integer

function”.

� �

� �

� � �

��

��

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Vol. 3, No. 1, March 2014 | At Right Angles 69

SOLUTIONS OF PROBLEMS IN ISSUE-II-3 Solutions of Problems in Issue - II - 3

Solution to problem II-3-M.1

Find the value of the following (no calculators!):

(2013� − 2019) × (2013� + 4023) × 2014

2010 × 2012 × 2015 × 2016 .

Let aaaa aa 2013; then 2019 aa aaaa + 6, 4023 aa 2aaaa − 3 ,

etc., so the given expression equals:

�aaaa� − aaaa aaa aaaaa� + 2aaaa aaa aaaaa aaa

(aaaa − 3)(aaaa − 1)(aaaa + 2)(aaaa + 3)

aa (aaaa − 3)(aaaa + 2)(aaaa + 3)(aaaa − 1)(aaaa + 1)

(aaaa − 3)(aaaa − 1)(aaaa + 2)(aaaa + 3) aa aaaa + 1.

So the expression simpli�ies to 2014.

Solution to problem II-3-M.2

�an you �ind a pair of perfect squares that differ by

2014?

The answer is No. For suppose that

aaaa� − bbbb� aa 2014 where aaaaaa bbbb are integers. Then

(aaaa − bbbbbbaaaa + bbbbb aa 2014. Since 2014 is even, at

least one of the quantities aaaa − bbbb, aaaa + bbbb is an even

number. But aaaa − bbbb and aaaa + bbbb have the same parity

(they are both odd or both even), so if one of them

is even, then so is the other one. This means that

the product (aaaa − bbbbbbaaaa + bbbbb is a multiple of 4.

However, 2014 is not a multiple of 4. Hence the

given representation is not possible.

Solution to problem II-3-M.3

From a two-digit number nnnn we subtract the number

obtained by reversing its digits. The answer is a

perfect cube. What could nnnn be?

Let nnnn aa aaaaaa aaaaa where aaaaaa bbbb are digits. On

subtracting its reversal, 10bbbb + aaaa, we get the

number xxxx aa aaaaaa aaaaaa aa a�(aaaa − bbbbb. For xxxx to be a

cube, aaaa − bbbb would have to be 3 times a cube (this

would lead to xxxx aa a� × a cube). Since aaaa and bbbb are

digits, the absolute value of aaaa − bbbb cannot exceed 9.

Hence if aaaa − bbbb is 3 times a cube, it must be that

aaaa − bbbb aa 3 or −3. Therefore nnnn is one of the

following: 14, 41, 25, 52, 36, 63, 47, 74, 58, 85, 69,

96. For each of these, xxxx aa xx27 aa axx3)�.

Solution to problem II-3-M.4

To a certain two-digit number mmmm we add the

number obtained by reversing its digits. The

answer is a perfect square. What could mmmm be?

Let mmmm aa aaaaaa aaaaa where aaaaaa bbbb are digits. On adding

its reversal, 10bbbb + aaaa, we get the number

yyyy aa aaaaaaa aaaaaa. For yyyy to be a square, aaaa + bbbb would

have to be 11 times a square (this would lead to

yyyy aa aa� × a square). Since aaaa and bbbb are digits, the

sum aaaa + bbbb cannot exceed 18. Hence if aaaa + bbbb is 11

times a square, it must be that aaaa + bbbb aa 11.

Therefore mmmm is one of the following: 29, 92, 38, 83,

47, 74, 56, 65. For each of these, yyyy aa aaa aa aa�.

Solution to problem II-3-M.5

The rectangle shown has been divided into equal

squares. The squares along the perimeter are

green. Note that the number of red squares is

greater than the number of green squares. What

should be the dimensions of the rectangle if the

number of red squares equals the number of green

squares?

Let the dimensions of the rectangle be aaaa × bbbb, with

aaaa aa bbbb. The inner rectangle (shaded green) has

dimensions (aaaa − 2) × (bbbb bbb. For the areas of the

red and green regions to be the same, the area of

the green region must be half the area of the large

rectangle, so we must have:

aaaabbbb aa 2(aaaa − 2)(bbbb bbbaa

∴ aaaabbbb aa 2(aaaabbbb bbaaaa − 2bbbb bbbaa

∴ aaaabbbb bbaaaa − 4bbbb bb aa 0.

�e must therefore �ind pairs (aaaaaa bbbbb of positive

integers that satisfy the equation

aaaabbbb bbaaaa − 4bbbb bb aa 0. The way we do this is based

on factorization. (It is a fairly standard

procedure.) It draws on the observation that

aaaabbbb bbaaaa − 4bbbb bb is `almost' equal to the product

(aaaa − 4)(bbbb bbb, but not quite: we get 16 in place of

8. This prompts the following:

aaaabbbb bbaaaa − 4bbbb bb aa 0aa

∴ aaaabbbb bbaaaa − 4bbbb bbb aa 8aa

∴ (aaaa − 4)(bbbb bbb aa 8.

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70 At Right Angles | Vol. 3, No. 1, March 2014

Hence are a pair of positive integers

whose product is 8. In what ways can 8 be

expressed as a product of two integers? The only

ways are: 8 × 1 and 4 × 2. Hence

or (4, 2), and therefore,

or (8, 6). So the dimensions of the

large rectangle are either 12 × 5 or 8 × 6. Observe

that these satisfy the stated conditions:

12 × 5 = 60, (12 − 2) × (5 − 2) = 30,

30 = half of 60, 8 × 6 = 48, (8 − 2) × (6 − 2) = 24,

24 = half of 48.

3

Clues Across :

1. Half of 16 A

3. The first digit is followed by its successor

and then by its predecessor

5. The middle digit is the sum of the end digits

6. Area of a square of side 74

9. Digits in arithmetic progression

10. The square root of 417316

12. Two complete rotations and two degrees

13. Two centuries, two decades and two years

Clues Down :

1. 16A divided by 4D

3. One less than a positive multiple of 10

4. A dozen more than 19D

6. 14 D written in reverse

7. Twice the difference between 15D and 17D

8. Two and a half times 20A

9. A score of unlucky numbers

10. 3 D times the cube of 3

11. 9 times the second 3 digit prime.

14. A perfect square between 30 and 40

15. Square root of 12 A written in reverse

16. One day short of 10 weeks

17. 2 score and 2

19. One tenth of 9D

1 2 3 4

11

17 18 19

20

15

5 6

9 10

13

8

12

16

14 