This document provides examples of solutions to problems involving number theory, algebra, geometry, and probability. It contains the following types of problems: - Number theory word problems involving finding missing terms in sequences - Solving quadratic equations and determining equations with transformed roots - Calculating areas and perimeters of geometric shapes like octagons and hexagons - Finding equations of lines transformed by parallel shifts - Probability problems involving counting outcomes, finding probabilities of events, and calculating combinations and permutations The document aims to demonstrate smart solutions to a variety of mathematical problems across different topics through clear explanations and step-by-step working.

1. ENGLISH MEDIUM OF INSTRUCTION
Teacher Training and Education Science Faculty
Jember University
By: Tedy Dwi Ariyanto (110210101048)
Rizka Nurul Kurnia (110210101075)
SMART SOLUTION
0.1 Number Theory
1. 70, 10, 80, 7, 90, 4, 100, A. What is A?
Solution: Type 1 = 70, 80, 90, 100 (difference 10)
Type 2 = 10, 7, 4, A (difference −3)
So A = 1
2. 999999 ∗ 111112
= ...
Solution: 111112
= 123454321
999999 ∗ 123454321 = (1000000 − 1) ∗ 123454321 = 123454197545679
3. 1
2
, 1
6
, 1
12
, 1
20
, X, 1
42
, 1
56
, 1
72
. What is X?
Soluion: X = 1
30
1
2∗1
, 1
3∗2
, 1
4∗3
, 1
5∗4
, 1
6∗5
, 1
7∗6
, 1
8∗7
, 1
9∗8
4. 25, 15, 83, 73, 51, Y. What is Y?
Solution: Y = 41
Type = 25, 15 (difference −10), 83, 73 (difference −10) and 51, 41 (differ-
ence −10 too)
5. 1
1+ 1
1+ 1
1+ 1
1+ 1
1+1
= ...
Solution: Count from the bottom
1 + 1 = 2, so 1
2
, 1
2
+ 1 = 3
2
2
3
, 2
3
+ 1 = 5
3
3
5
, 3
5
+ 1 = 8
5
5
8
, 5
8
+ 1 = 13
8
8
13
, the result is 8
13
1

2. 0.2 Algebra
1. If the quadratic equation x2
+ 2x + 3 = 0 has the roots P and Q, then the
new equation with quadratic roots (P − 2) and (Q − 2) is . . .
Solution: a(x + k)2
+ b(x + k) + c = 0
(x + 2)2
+ 2(x + 2) + 3 = 0
x2
+ 6x + 11 = 0
2. If the quadratic equation x2
− 3x − 43 = 0 has the roots X and Y, then the
new equation with quadratic roots (X + 3) and (Y + 3) is . . .
Solution: a(x − k)2
+ b(x − k) + c = 0
(x − 3)2
− 3(x − 3) − 4 = 0
x2
− 9x + 11 = 0
3. The quadratic equation that has roots 2 times the roots of a quadratic
equation x2
+ 8x + 10 = 0 is . . .
Solution: ax2
+ nbx + n2
c = 0
x2
+ 2 ∗ 8x + 22
∗ 10 = 0
x2
+ 16x + 40 = 0
4. Equation 3x3
+ (m + 2)x2
− 16x − 12 = 0 has roots x = 2. Then the sum
of squares the third root of the equation is . . .
Solution: x2
1 + x2
2 + x2
3 = b2−2ac
a2
f(2) = 3(2)3
+ (m + 2)(22
) − 16(2) − 12 = 0 then 4m − 12 = 0, m = 3
(m+2)2−2∗3∗(−16)
32 = 52+96
9
= 121
9
the result is 121
9
5. The quadratic equation 2x2
+ 3x + 5 = 0 has the roots A and B, then the
new equation with quadratic roots 1
A
and 1
B
is . . .
Solution: ax2
+ bx + c = 0 become cx2
+ bx + a = 0
5x2
+ 3x + 2 = 0
0.3 Geometry
1. What area of octagon with out radius 8 cm . . .
Solution: area of n-gon = n1
2
r2
sin(360
n
) = 81
2
82
sin(360
8
) = 128
√
2 cm2
2. What roving of hexagon with out radius 11 cm . . .
Solution: roving of n-gon = nr 2(1 − cos(360
n
) = 6 ∗ 11 2(1 − cos(360
6
) =
66 cm
3. Find the equation of a new line if the line 3x + 2y = 6 shifted to right as
far as 3 unit . . .
2

3. Solution: ax + by = c shifted to right as far as n unit: ax + by = c + na
3x + 2y = 6 + 3 ∗ 3
3x + 2y = 15
4. Find the equation of a new line if the line 2x + 3y = 9 shifted to left as far
as 5 unit . . .
Solution: ax + by = c shifted to left as far as n unit: ax + by = c - na
2x + 3y = 9 − 5 ∗ 2
2x + 3y = −1
5. Find the equation of a new line if the line 5x + 9y = 2 shifted to up as far
as 4 unit . . .
Solution: ax + by = c shifted to up as far as n unit: ax + by = c + nb
5x + 9y = 2 + 4 ∗ 9
5x + 9y = 38
0.4 Probability
1. In a box there are 6 red balls, 5 blue balls and 4 green balls. If the first de-
cision taken a red ball and not returned in box, then take again the second
taking, then chances are both red ball is....
Solution: 1C6
1C15
∗ 1C5
1C14
=
6
15
∗ 5
14
= 1
7
2. In a box there are 5 red balls, 4 blue balls and 6 green balls, will be taken
3 balls randomly. What is the probability of taken at least 1 red ball and
1 green ball?
Solution: 1 red ball, 1 blue balls and 1 green ball = 1C5∗1C4∗1C6
3C15
= 5∗4∗6
455
=
120
455
= 24
91
1 red ball and 2 green balls = 1C5∗2C6
3C15
= 5∗15
455
= 75
455
= 15
91
2 red balls and 1 green ball = 2C5∗1C6
3C15
= 10∗6
455
= 60
455
= 12
91
24
91
+ 15
91
+ 12
91
= 51
91
3. 2 dices thrown once simultaneously. Probability of appearance the sum is
5 or 10 on the dice is . . .
Solution: The possibility of the occurrence of an event = {(1, 1), (1, 2),
(1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1),
(3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5),
(6, 6)}
The sum of dice 5 = {(1, 4), (2, 3), (3, 2), (4, 1)} so P = 4
36
The sum of dice 10 = {(4, 6), (5, 5), (6, 4)} so P = 3
36
So the sum 5 or 10 on the dice is 4
36
+ 3
36
= 7
36
3

4. 4. From 5 candidates to the board of the class will be selected 2 peoples as
chairman and vice chairman of the class. How many pairs of candidates
that might happen?
Solution: 2P5 = 5!
(5−2)!
= 20
5. There are 5 different balls, each ball will be put in a box. How many chances
ball will be put in a box?
Solution: 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = 120
4