1st Assignment

SMA 3043 – ELEMENTARY NUMBER THEORY

ASSIGNMENT 1

Prepared by:

LECTURER : DR. NOR’ASHIQIN MOHD IDRUS

NAME MATRIX NUMBER

SITI SHUHADA BINTI MOHD YUSOFF D20111048895

HALEEDA BINTI ROSDI D20111048896

MASITAH HUDA BINTI HUSSIN D20111048897

QUESTION 1

1. a) For all 𝒏 ≥ 𝟏, prove: 𝒂𝒏 − 𝟏 = 𝒂 − 𝟏 𝒂𝒏−𝟏 + 𝒂𝒏−𝟐 + 𝒂𝒏−𝟑 +⋯+ 𝒂 + 𝟏 . [Hint: 𝒂𝒏+𝟏 − 𝟏 = 𝒂 + 𝟏 𝒂𝒏 − 𝟏 − 𝒂 𝒂𝒏−𝟏 − 𝟏 .] Solution: Given:

𝒂𝒏 − 𝟏 = 𝒂 − 𝟏 𝒂𝒏−𝟏 + 𝒂𝒏−𝟐 + 𝒂𝒏−𝟑 +⋯+ 𝒂 + 𝟏

Want to show:

𝒂𝒏+𝟏 − 𝟏 = 𝒂 + 𝟏 𝒂𝒏 − 𝟏 − 𝒂 𝒂𝒏−𝟏 − 𝟏

Let 𝑺 = 𝒌 𝒌 ∈ 𝒁+ such that

i. 𝟏 ∈ 𝑺

𝐋𝐇𝐒:

𝒂𝒌 − 𝟏 = 𝒂𝟏 − 𝟏 = 𝒂 − 𝟏

𝐑𝐇𝐒:

𝒂 − 𝟏 𝒂 − 𝟏 𝒂𝒌−𝟏 + 𝒂𝒌−𝟐 + 𝒂𝒌−𝟑 +⋯+ 𝒂 + 𝟏

= 𝒂𝒌 + 𝒂𝒌−𝟏 + 𝒂𝒌−𝟐 + 𝒂𝒌−𝟑 +⋯+ 𝒂𝟐 − 𝒂𝒌−𝟏 − 𝒂𝒌−𝟐 − 𝒂𝒌−𝟑 −

…− 𝒂 − 𝟏 = 𝒂𝒌 − 𝟏

= 𝒂𝟏 − 𝟏

= 𝒂 − 𝟏

∴ 𝐋𝐇𝐒 = 𝐑𝐇𝐒

ii. 𝒌 + 𝟏 ∈ 𝑺, given that 𝒌 ∈ 𝑺.

𝐋𝐇𝐒 ∶

𝒂𝒌+𝟏 − 𝟏 = 𝒂 − 𝟏 𝒂𝒌−𝟏+𝟏 + 𝒂𝒌−𝟐+𝟏 + 𝒂𝒌−𝟑+𝟏 +⋯+ 𝒂 + 𝟏

= 𝒂 − 𝟏 𝒂𝒌 + 𝒂𝒌−𝟏 + 𝒂𝒌−𝟐 +⋯+ 𝒂 + 𝟏

= 𝒂𝒌+𝟏 + 𝒂𝒌 + 𝒂𝒌−𝟏 +⋯+ 𝒂𝟐 + 𝒂 − 𝒂𝒌 − 𝒂𝒌−𝟏 − 𝒂𝒌−𝟐−⋯− 𝒂 − 𝟏

= 𝒂𝒌+𝟏 − 𝟏

𝐑𝐇𝐒:

𝒂 + 𝟏 𝒂𝒌 − 𝟏 − 𝒂 𝒂𝒌−𝟏 − 𝟏 = 𝒂𝒌+𝟏 − 𝒂 + 𝒂𝒌 − 𝟏 − 𝒂𝒌 + 𝒂

= 𝒂𝒌+𝟏 − 𝟏


b) Verify that for all 𝒏 ≥ 𝟏,

𝟐 • 𝟔 • 𝟏𝟎 • 𝟏𝟒… 𝟒𝒏 − 𝟐 =𝟐𝒏 !

𝒏!

Solution:

Given:

𝟐 • 𝟔 • 𝟏𝟎 • 𝟏𝟒… 𝟒𝒏 − 𝟐 =𝟐𝒏 !

𝒏!

Want to show:

𝟐 • 𝟔 • 𝟏𝟎 • 𝟏𝟒… 𝟒𝒏 − 𝟐 • 𝟒𝒏 + 𝟐 =𝟐𝒏 + 𝟐 !

𝒏 + 𝟏 !

Let 𝑺 = 𝒌 𝒌 ∈ 𝒁+ such that

i. 𝟏 ∈ 𝑺

𝐋𝐇𝐒 = 𝟐 • 𝟔 • 𝟏𝟎 • 𝟏𝟒… 𝟒𝒌 − 𝟐

= 𝟐 • 𝟔 • 𝟏𝟎 • 𝟏𝟒… 𝟒(𝟏) − 𝟐

= 𝟐

𝐑𝐇𝐒 =𝟐𝒌 !

𝒌!

=𝟐(𝟏) !

𝟏!

= 𝟐


ii. 𝒌 + 𝟏 ∈ 𝑺, given that 𝒌 ∈ 𝑺.

𝐋𝐇𝐒 ∶

𝟐 • 𝟔 • 𝟏𝟎 • 𝟏𝟒… 𝟒𝒌 − 𝟐 • 𝟒𝒌 + 𝟐

= 𝟐𝒌 !

𝒌!• 𝟒𝒌 + 𝟐

=𝟐𝒌 𝟐𝒌−𝟏 !

𝒌 𝒌−𝟏 !• 𝟒𝒌 + 𝟐

=𝟐 𝟐𝒌−𝟏 !•𝟐 𝟐𝒌+𝟏

𝒌−𝟏 !

=𝟒 𝟐𝒌+𝟏 𝟐𝒌−𝟏 !

𝒌−𝟏 !

𝐑𝐇𝐒:

𝟐𝒌+𝟐 !

𝒌+𝟏 ! =

𝟐𝒌+𝟐 𝟐𝒌+𝟏 𝟐𝒌 𝟐𝒌−𝟏 !

𝒌+𝟏 𝒌 𝒌−𝟏 !

=𝟐 𝒌+𝟏 𝟐𝒌+𝟏 𝟐 𝒌 𝟐𝒌−𝟏 !

𝒌+𝟏 𝒌 𝒌−𝟏 !

=𝟒 𝟐𝒌+𝟏 𝟐𝒌−𝟏 !

𝒌−𝟏 !


QUESTION 2

For all , derive each of the identities below:

a)

[Hint: Let a and b be some integers so that a+b=0.]

b)

[Hint: i) After expanding by the Binomial Theorem,let b=1

ii) Note also that ]

1n

0)1(...210

n

nnnnn

12...3

32

21

nnn

nn

nnn

1)1( nbn

.1

)1(1

k

nk

k

nn

ANSWER 2(a)

n

nnnn

n

nn

nn

k

n

balet

n

nnnn

n

nnnn

nnn

kknn

k

n

1...210

0

11...112

111

110

)1(1

110

1,1

01...210

22

1100

0

Shown

ANSWER 2(b)

1)1(

12

1,)1(

2...3

32

21

0

1

1

1

k

nk

k

nnn

bbn

nn

nn

nnn

n

k

n

n

n

n

nn

nn

nn

nn

n

nn

nn

nnnn

bak

nba

nnn

n

nnn

k

knn

k

n

1...

2

1

1

1

0

1

111

...

112

1

111

111

0

1-n)11(

1,blet

Theorem Binomial usingby

1

221

1110011

0

n

nn

nnnnso

n

nn

nnnRHS

n

n

nn

nnnn

n

nn

nnnn

n

n

n

n

...3

32

21

12,

...3

32

21

1

2LHS

...3

32

21

1)2(

1...

1212

1111

1010)11(

1

1

1

1

QUESTION 3

3. The ancient Greek called a number triangular if it is the sum of consecutive integers, beginning with 1. Prove the following facts concerning triangular numbers:

a) A number is triangular if and only if it is of

the form 𝒏 𝒏+𝟏

𝟐 for some 𝒏 ≥ 𝟏.

Solution: i. If a number 𝒂𝒏 is triangular, then 𝒂𝒏 can be in the form of

𝒏 𝒏+𝟏

𝟐 for some 𝒏 ≥ 𝟏.

𝒂𝒏 is triangular, then we assume 𝒂𝒏 as 𝟏, 𝟑, 𝟔, 𝟏𝟎, 𝟏𝟓, 𝟐𝟏,… ,𝒏 𝒏+𝟏

𝟐.

We know that, 𝒂𝟏 = 𝟏 𝒂𝟐 = 𝟑 = 𝒂𝟏 + 𝟐 = 𝟏 + 𝟐 𝒂𝟑 = 𝟔 = 𝒂𝟐 + 𝟑 = 𝟏 + 𝟐 + 𝟑 𝒂𝟒 = 𝟏𝟎 = 𝒂𝟑 + 𝟒 = 𝟏 + 𝟐 + 𝟑 + 𝟒 . . . 𝒂𝒏 = 𝟏 + 𝟐 + 𝟑 +⋯+ 𝒏

Therefore, from previous slide, we can assume that triangular number, 𝒂𝒏 as:

𝒂𝒏 = 𝟏 + 𝟐 + 𝟑 +⋯+ 𝒏 − 𝟏 + 𝒏

= 𝒊𝒏𝒊=𝟏

By the other way,

𝒂𝒏 = 𝒏 + 𝒏 − 𝟏 +⋯+ 𝟑 + 𝟐 + 𝟏

= 𝒏 − 𝒊 + 𝟏𝒏𝒊=𝟏

𝒂𝒏 + 𝒂𝒏 = 𝒊𝒏𝒊=𝟏 + 𝒏 − 𝒊 + 𝟏𝒏

𝒊=𝟏

𝟐𝒂𝒏 = 𝒏 + 𝟏𝒏𝒊=𝟏

𝒏 + 𝟏𝒏

𝒊=𝟏 = 𝒏 + 𝟏 + 𝒏 + 𝟏 +⋯+ 𝒏 + 𝟏

= 𝒏 + 𝟏 will be added 𝒏 times

= 𝒏 𝒏 + 𝟏

𝟐𝒂𝒏 = 𝒏 𝒏 + 𝟏

𝒂𝒏 =𝒏 𝒏+𝟏

𝟐

ii. If 𝒂𝒏 can be in the form of 𝒏 𝒏+𝟏

𝟐 for some 𝒏 ≥ 𝟏, then 𝒂𝒏 is

triangular number. By mathematical induction,


𝟐

= 𝟏 + 𝟐 + 𝟑 +⋯+ 𝒏 − 𝟏 + 𝒏 When 𝒏 = 𝟏,

𝐋𝐇𝐒 =𝒏 𝒏+𝟏

𝟐 𝐑𝐇𝐒 = 𝟏

=𝟏 𝟏+𝟏

𝟐

= 𝟏 ∴ 𝐋𝐇𝐒 = 𝐑𝐇𝐒

We want to show that 𝒏 = 𝒌 + 𝟏 is true, given that it is true for 𝒏 = 𝒌.

𝒌 𝒌+𝟏

𝟐= 𝟏 + 𝟐 + 𝟑 +⋯+ 𝒌 − 𝟏 + 𝒌

𝒏 = 𝒌 + 𝟏 𝒌 + 𝟏 𝒌 + 𝟐

𝟐= 𝟏 + 𝟐 + 𝟑 +⋯+ 𝒌 + 𝒌 + 𝟏

𝑳𝑯𝑺 =𝒌𝟐+𝟑𝒌+𝟐

𝟐 , 𝑹𝑯𝑺 =

𝒌 𝒌+𝟏

𝟐+ 𝒌 + 𝟏

=𝒌 𝒌+𝟏 +𝟐 𝒌+𝟏

𝟐

=𝒌𝟐+𝟑𝒌+𝟐

𝟐


b) The integer 𝒏 is a triangular number if and only if 𝟖𝒏 + 𝟏 is a perfect square.

Solution: Let 𝒏 = 𝒂𝒏 i. If the integer 𝒂𝒏 is triangular, then 𝟖𝒂𝒏 + 𝟏 is a perfect square.


𝟐, 𝟖𝒂𝒏 + 𝟏 =?

𝒂 is a perfect square if 𝒂 = 𝒃. 𝒂 = 𝒃𝟐

𝟖𝒂𝒏 + 𝟏 = 𝒃𝟐

𝟖𝒂𝒏 + 𝟏 = 𝒃

𝒂𝒏triangular → 𝟖𝒂𝒏 + 𝟏 = 𝒃𝟐

𝒂𝒏 triangular → 𝒂𝒏 = 𝒌 𝒌+𝟏

𝟐

Suppose 𝒂𝒏 triangular, then 𝒂𝒏 = 𝒌 𝒌+𝟏

𝟐.

So, 𝟖𝒂𝒏 + 𝟏 = 𝟖𝒌 𝒌+𝟏

𝟐+ 𝟏

= 𝟒𝒌 𝒌 + 𝟏 + 𝟏

= 𝟒𝒌𝟐 + 𝟒𝒌 + 𝟏

= 𝟐𝒌 + 𝟏 𝟐 , 𝒌 ∈ 𝒁+

ii. If 𝟖𝒂𝒏 + 𝟏 is a perfect square, then integer 𝒂𝒏 is triangular.

𝟖𝒂𝒏 + 𝟏 = 𝒃

𝟐 𝟖𝒂𝒏 + 𝟏 = 𝟐𝒌 + 𝟏 𝟐 𝟖𝒂𝒏 + 𝟏 = 𝟒𝒌

𝟐 + 𝟒𝒌 + 𝟏 𝟖𝒏 = 𝟒𝒌𝟐 + 𝟒𝒌

𝒏 =𝒌𝟐+𝒌

𝟐

𝒏 =𝒌 𝒌+𝟏

𝟐 , 𝒌 ∈ 𝒁+

c) The sum of any two consecutive triangular numbers is a perfect square.

Solution: By using 𝒌𝒕𝒉 term: 𝒌 − 𝟏 𝒕𝒉 term and 𝒌𝒕𝒉 term.

𝑻 𝒌 − 𝟏 + 𝑻 𝒌 =𝒏 − 𝟏 𝒏 + 𝟏 − 𝟏

𝟐+𝒏 𝒏 + 𝟏

𝟐

=𝒏𝟐−𝒏+𝒏𝟐+𝒏

𝟐

=𝟐𝒏𝟐

𝟐

= 𝒏𝟐

d) If 𝒏 is a triangular number, then so are 𝟗𝒏 + 𝟏, 𝟐𝟓𝒏 + 𝟑 and 𝟒𝟗𝒏 + 𝟔.

Solution:

𝟗𝒏 + 𝟏 = 𝟗𝒏 𝒏+𝟏

𝟐+ 𝟏

= 𝟗𝒏𝟐+𝒏

𝟐+ 𝟏

=𝟗𝒏𝟐+𝟗𝒏+𝟐

𝟐

=𝟑𝒏+𝟏 𝟑𝒏+𝟐

𝟐 , 𝒏 ∈ 𝒁+

So, this proved that it is triangular number when 𝒏 ∈ 𝒁+.

𝟐𝟓𝒏 + 𝟑 = 𝟐𝟓𝒏 𝒏+𝟏

𝟐+ 𝟑

= 𝟐𝟓𝒏𝟐+𝒏

𝟐+ 𝟑

=𝟐𝟓𝒏𝟐+𝟐𝟓𝒏+𝟔

𝟐

=𝟓𝒏+𝟐 𝟓𝒏+𝟑

𝟐 , 𝒏 ∈ 𝒁+


𝟒𝟗𝒏 + 𝟔 = 𝟒𝟗𝒏 𝒏+𝟏

𝟐+ 𝟔

= 𝟒𝟗𝒏𝟐+𝒏

𝟐+ 𝟔

=𝟒𝟗𝒏𝟐+𝟒𝟗𝒏+𝟏𝟐

𝟐

=𝟕𝒏+𝟑 𝟕𝒏+𝟒

𝟐 , 𝒏 ∈ 𝒁+


QUESTION 4 4. Use the Division Algorithm to establish the following:

a) The square of any integer is either of the form 𝟑𝒌 or 𝟑𝒌 + 𝟏.

Solution:

Let 𝒃 = 𝟑; 𝒓 = 𝟎, 𝟏, 𝟐; 𝒂 = 𝒃𝒒 + 𝒓;

𝒂𝟏 = 𝒃𝒒 + 𝒓 = 𝟑𝒒 + 𝟎

𝒂𝟐 = 𝒃𝒒 + 𝒓 = 𝟑𝒒 + 𝟏

𝒂𝟑 = 𝒃𝒒 + 𝒓 = 𝟑𝒒 + 𝟐

Value of 𝒓 must be in the range 𝟎 ≤ 𝒓 < 𝟑.

For 𝒂𝟏, 𝟑𝒒 + 𝟎 𝟐 = 𝟗𝒒𝟐

= 𝟑 𝟑𝒒𝟐

= 𝟑𝒌 ; when 𝒌 = 𝟑𝒒𝟐 For 𝒂𝟐, 𝟑𝒒 + 𝟏 𝟐 = 𝟗𝒒𝟐 + 𝟔𝒒 + 𝟏

= 𝟑 𝟑𝒒𝟐 + 𝟐𝒒 + 𝟏

= 𝟑𝒌 + 𝟏 ; when 𝒌 = 𝟑𝒒𝟐 + 𝟐𝒒 For 𝒂𝟑, 𝟑𝒒 + 𝟐 𝟐 = 𝟗𝒒𝟐 + 𝟏𝟐𝒒 + 𝟒 = 𝟗𝒒𝟐 + 𝟏𝟐𝒒 + 𝟑 + 𝟏

= 𝟑 𝟑𝒒𝟐 + 𝟒𝒒 + 𝟏 + 𝟏

= 𝟑𝒌 + 𝟏 ; when 𝒌 = 𝟑𝒒𝟐 + 𝟒𝒒 + 𝟏 ∴ The square of any integers is either of the form 𝟑𝒌 or 𝟑𝒌 + 𝟏.

b) The cube of any integer has one of the forms: 𝟗𝒌, 𝟗𝒌 + 𝟏 and 𝟗𝒌 + 𝟖.

Solution: Let 𝒃 = 𝟑; 𝒓 = 𝟎, 𝟏, 𝟐; 𝒂 = 𝒃𝒒 + 𝒓; 𝒂𝟏 = 𝒃𝒒 + 𝒓 = 𝟑𝒒 + 𝟎 𝒂𝟐 = 𝒃𝒒 + 𝒓 = 𝟑𝒒 + 𝟏 𝒂𝟑 = 𝒃𝒒 + 𝒓 = 𝟑𝒒 + 𝟐 Value of 𝒓 must not exceed 2 since 𝒃 = 𝟑.

For 𝒂𝟏, 𝟑𝒒 + 𝟎 𝟑 = 𝟐𝟕𝒒𝟑

= 𝟗 𝟑𝒒𝟑 = 𝟗𝒌 For 𝒂𝟐, 𝟑𝒒 + 𝟏 𝟑 = 𝟐𝟕𝒒𝟑 + 𝟐𝟕𝒒𝟐 + 𝟗𝒒 + 𝟏

= 𝟗 𝟑𝒒𝟐 + 𝟑𝒒 + 𝒒 + 𝟏 = 𝟗𝒌 + 𝟏 For 𝒂𝟑, 𝟑𝒒 + 𝟐 𝟑 = 𝟐𝟕𝒒𝟑 + 𝟓𝟒𝒒𝟐 + 𝟑𝟔𝒒 + 𝟖

= 𝟗 𝟑𝒒𝟑 + 𝟔𝒒𝟐 + 𝟒𝒒 + 𝟖 = 𝟗𝒌 + 𝟖 ∴ The cube of any integer has one of the forms: 𝟗𝒌, 𝟗𝒌 + 𝟏, 𝟗𝒌 + 𝟖

QUESTION 5

For , use the mathematical induction to establish each of the following divisibility.

1n

55372|24 b)

)757()75(5)75:[hint

75|8 a)

2221)2(k

2n

nn

k

ANSWER 5(a)

true.isk n given that trueis 1kn Show

875 Given that

32|8

3275

1nlet

1nfor )7(5|8

2k

2(1)

2n

lwherel

trueis 75|8,

,8

)375(8

)3(8785

)15(7)75(5

7)7(5)7(555

75575

true.is 75|8 WTS

)1(2

2

2

222

2222

221)2(k

1)2(k

k

k

k

k

so

mm

l

l

ANSWER 5(b)

5515714

245553772

)1.....(5245)3(7)2(

2455)3(7)2( i.e ,55)3(7)2(|24 Given that

55)3(7)2(|24 WTS

1

)1(24245)5)(3()7)(2(

1nlet

,2455)3((2)7

Def.By

)55372(|24

11

2

kk

kk

kk

kkkk

kk

n

nn

ll

l

l

. k is true n given that is true, k n Show that

ll

ss

lqplqp

lqp

lqp

lqp

l

qpqplet

kallforoddeachareandthatObserve

l

l

l

withsubs

kk

kk

kk

kk

kkkk

kkkk

24

,,)1(24

24)1(24

24)222(12

24)1212(12

24)57(12

,125127;

; 57

24)315(5)214(7

245)3(7)2(5)15(7)14(

]245)3(7)2[(515714

)1(

QUESTION 6

Given integers a and b, prove the following:

a) There exist integers x and y for which c=ax+by if and only if gcd(a,b)|c.

b) If there exist integers x and y for which ax+by=gcd(a,b), then gcd(x,y)=1.

ANSWER 6(a)

• If gcd(a,b)|c then

Let gcd(a,b)=d

By theorem,

d|a , d|b, so d|(ax+by) then d|c

where

conversely suppose gcd(a,b)|c

WTS, c=ax+by

gcd(a,b)=d

byaxc

byaxba ),gcd(

dbyaxba ),gcd(

kdkc ,

By the Euclidean Algorithm, there are integers p and q such that

multiply by k,

we have,

bqapd

bqkapkdk

byaxc

qkypkxletqkbpka

bqkapk

kbqap

dkc

,)()(

)(

assume

WTS

since gcd(a,b)=d (by theorem)

byaxc

cd |

byaxdba ),gcd(

cd

cba

lbyaxc

ldlc

cd

|

|),gcd(

)(

,

|

ANSWER 6(b) • Suppose

d

ydqxdpbyaxd

qdbdpa

qp

bdad

yqxp

yxbabyax

yx

by sideboth dividing

)()(then

and

such that and integers are thereso

.| and|then db)gcd(a,let

1 WTS

1),gcd(then ),gcd(such that

,

shown. isit then,

1 yqxp

q

dqy

d

dpx

d

d

QUESTION 7

Use the Euclidean Algorithm to obtain integers x and y (for (a)), x, y and z (for (b)) satisfying the following:

a) gcd(1769, 2378) = 1769x + 2378y

b) gcd(198, 288, 512) = 198x + 288y + 512z

ANSWER

• a) gcd(1769, 2378) = 1769x + 2378y

)2378,1769gcd(29

029258

29589551

55160921769

609176917692378

29 and 39 have weso,

)29(2378)39(1769

)29(2378)39(1769

)29(1769)29(2378)10(1769

)29)(17692378()10(1769

)29(609)10(1769

)9(609)10)(60921769(

)9(609)10(551

)551(9)609(9551

]551609[9551

)58(955129

yx

• b) gcd(198, 288, 512) = 198x + 288y + 512z

• First assume that

Then,find

18)288,198gcd(

0)5(1890

18)2(90198

90)1(198288

d

)288,198gcd(d

)512,18gcd()512,gcd( d

)512,18gcd(2

0)4(28

2)2(818

8)28(18512

)2(512)114(288)171(198

)2(512)114(288)171(198

)2(512)114(198)114(288)57(198

)2(512)114)(198288()57(198

)2(512)114(90)57(198

)2(512)57)](2(90198[

)2(512)57(18

)2(512)57(18

)56(18)2(512)1(18

)2)](28(18512[)1(18

)2(8182

• So,we have

2

114

171

z

y

x

QUESTION 8

Determine all solution in the positive integers of the following Diophantine equation.

9062154 yx

ANSWER 8

30239063,

0)3(39

3)1(912

9)1(1221

12)2(2154

)21,54gcd(

9062154

wheredhence

dfind

yx

1510,604

3025,2

)5(21)2(54

)5(21)2(54

)1(21)]4(21)2(54[

)1(21)2)](2(2154[

)1(21)2(12

)1(12)1(2112

)1)](1(1221[12

)1(9123

|

yx

bymultiplyyx

cd

8486

8.8328.86

18

1510

7

604

018151007604

00

1815107604

3

541510

3

21604

integers positive find

00

tt

tt

tt

tt

yx

tt

tytx

-86 -84

Overlapping at . Hence, there are an integer solution when

8486 t

8486 t

o

o

QUESTION 9

• A farmer purchased 100 head of livestock for a total cost of RM4000. Prices were as follows: calves, RM120 each; lamb, RM50 each; chicken RM25 each. If the farmer obtained at least one animal of each type, how many of each did he buy?

ANSWER

• Let:

chicken ofnumber

lambs ofnumber

calves ofnumber

z

y

x

)3...(100

)2...(100

)1...(40002550120

yxz

zyx

zyx

15002595

40002525250050120

4000)100(2550120

(1) into (3) substitute

yx

yxyx

yxyx

5)25,95(gcd

0)4(520

5)1(2025

20)3(2595

)25,95( gcd

for

)1200(25)300(951500

300 with everythingmultiply *

)4(25)1(95

)1(95)4(25

)3(25)1(9525

)1)](3(2595[25

)1(20255

60

3005

05300

5300

)5

25(300

)(0

t

t

t

tfor

t

t

td

bxx

63

2.63

120019

0191200

191200

)5

95(1200

)(0

t

t

t

t

tfor

t

t

td

ayy

58

1.57

80014

014800

14800

)191200()5300(100

100

t

t

t

t

tfor

t

tt

yxz

There are overlapping of t when 60 < 𝑡 ≤ 63.

Hence, t = 61, t=62 and t = 63.

• When t=61, 𝑥 = 5, 𝑦 = 41 and 𝑧 = 54

• When t=62, 𝑥 = 10, 𝑦 = 22 and 𝑧 = 68

• When t=63, 𝑥 = 15, 𝑦 = 3 and 𝑧 = 82

57 60 63

o o

o

t

• The only integral values of t to satisfy three inequalities are t=61,t=62 and t=63.

• Thus, there are three possible purchases:

• Case 1 where t=61

5 calves at RM120, 41 lambs at RM50

and 54 chicken at RM25.







Documents

1st Assignment