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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
∴ 𝐋𝐇𝐒 = 𝐑𝐇𝐒
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
∴ 𝐋𝐇𝐒 = 𝐑𝐇𝐒
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 !
∴ 𝐋𝐇𝐒 = 𝐑𝐇𝐒
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,
𝒂𝒏 =𝒏 𝒏+𝟏
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
∴ 𝐋𝐇𝐒 = 𝐑𝐇𝐒
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.
𝒂𝒏 =𝒏 𝒏+𝟏
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+
So, this proved that it is triangular number when 𝒏 ∈ 𝒁+.
𝟒𝟗𝒏 + 𝟔 = 𝟒𝟗𝒏 𝒏+𝟏
𝟐+ 𝟔
= 𝟒𝟗𝒏𝟐+𝒏
𝟐+ 𝟔
=𝟒𝟗𝒏𝟐+𝟒𝟗𝒏+𝟏𝟐
𝟐
=𝟕𝒏+𝟑 𝟕𝒏+𝟒
𝟐 , 𝒏 ∈ 𝒁+
So, this proved that it is triangular number when 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.
• Case 2 where t=62
10 calves at RM120, 22 lambs at RM50
and 68 chicken at RM25.
• Case 3 where t=63
15 calves at RM120, 3 lambs at RM50
and 82 chicken at RM25.
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