1. Faculty of Teacher Training and Education
University of Jember
2013
SMART SOLUTION
By:
1. TATHMAINUL QULUB 110210101068
2. MELSI MELISSA 110210101073
1. if n complete 250.25
× 250.25
× 250.25
× … × 250.25
= 125, so that (n-3)(n+2)....
solution:
Remember that:
𝑎 𝑏
𝑥 𝑎 𝑐
= 𝑎 𝑏+𝑐
𝑎 𝑓(𝑥)
= 𝑎 𝑚
→ 𝑓(𝑥) = 𝑚
250.25
× 250.25
× 250.25
× … × 250.25
= 125
(250.25) 𝑛
= 125
((52)0.25
) 𝑛
= 53
50.5𝑛
= 53
0.5𝑛 = 3
𝑛 =
3
0.5
𝑛 = 6
so, (𝑛 − 3)(𝑛 + 2) = (6 − 3)(6 + 2) = 3 𝑥 8 = 𝟐𝟒
n factor
n factor

2. Faculty of Teacher Training and Education
University of Jember
2013
2. The summaries of 50 first line number of log5 + log55 + log605 + log 6655 + ⋯ is ....
Solution:
Arithmetical line formula:
𝑆𝑛 =
𝑛
2
(2𝑎 + (𝑛 − 1)𝑏)
Logarithmic characteristic:
log(𝑎 𝑥 𝑏) = log 𝑎 + log 𝑏
𝑎
log 𝑏 = log 𝑏 𝑎
Exponent characteristic:
(𝑎 𝑚) 𝑛
= 𝑎 𝑚𝑛
 from the problem, we found an arithmetical line,
𝑈1 = log 5 ; 𝑈2 = log 55 ; 𝑈3 = log 605 ; 𝑈4 = log 6655
𝑈1 = log 5 ; 𝑈2 = log 55 = log 5 + log 11 ; 𝑈3 = log 605 = 𝑙𝑜𝑔55 + 𝑙𝑜𝑔11 ; 𝑈4
= log 6655 = log 605 + 𝑙𝑜𝑔11
with initial number (a) = log 5
Difference (b) = log 11
so, the sum of first 50 digit is
𝑆𝑛 =
𝑛
2
(2𝑎 + (𝑛 − 1)𝑏)
𝑆50 =
50
2
(2(log 5) + (50 − 1)(log 11))
= 25(log 25 + log 1149
)
= 25 log 25 + 25 log 1149
= log 2525
+ log(1149)25
= log 2525
+ log 111225
= log(2525
111225
)

3. Faculty of Teacher Training and Education
University of Jember
2013
3. If 1 + 2 + 3 + 4 + . . . . + 𝑛 = 𝑎𝑎𝑎, then specify the value of 𝑛 𝑎𝑛𝑑 𝑎𝑎𝑎!
Solution:
1 + 2 + 3 + ⋯ + 𝑛 =
𝑛
2
(𝑛 + 1)
𝑎𝑎𝑎 = 111 × 𝑎 = (3 × 𝑎) × 37
𝑛
2
(𝑛 + 1) = (3 × 𝑎) × 37
𝑛(𝑛 + 1) = (6 × 𝑎) × 37
This is a sequential multiplication.
So a = 6 and n = 36
4. If a and b are arbitrary numbers, prove that
𝑎
𝑏
+
𝑏
𝑎
≥ 2 !
Solution:
(𝑎 − 𝑏)2
≥ 0 ↔ 𝑎2
+ 𝑏2
≥ 2𝑎𝑏 ↔
𝑎
𝑏
+
𝑏
𝑎
≥ 2
5. 𝐴 = 13
− 23
+ 33
− 43
+ 53
− 63
+ ⋯ + 20053
, determine the value of A!
Solution :
= (13
+ 23
+ 33
+ ⋯ + 20053
) − 2(23
+ 33
+ ⋯ + 20043)
= (13
+ 23
+ 33
+ ⋯ + 20053
) − 2. 23
(13
+ 23
+ 33
+ ⋯ + 10023
)
= (
1
2
. 2005.2006)
2
− 16 (
1
2
. 1002.1003)
2
= 10032
(20052
− (4502)2
)
= 10032(20052
− 20042)
= 10032(2005 + 2004)(2005 − 2004)
= 10032
.4009
6. This form 𝑥4
− 1 has how many factors....
Solution :

4. Faculty of Teacher Training and Education
University of Jember
2013
Note that 𝑥4
− 1 = (𝑥2
+ 1)(𝑥 + 1)(𝑥 − 1). So, the form 𝑥4
− 1 has a (1+1) × (1+1) × (1
+1) = 8 factors.
7. If 𝑎𝑎𝑏𝑏 = (𝑥𝑦)2
, determine the value of a, b, x, and y!
Solution :
Because (𝑥𝑦)2
is the square of the digits unit numbers 0, 1, 4, 5, 6 or 9.
Mean b = 00, 11, 44, 55, 66 or 99
Quadratic number when divided by 4 the remaining 0 (for even) or 1 (for odd)
Numbers is divisible by 4 if the last 2 digits is divisible by 4, so b = 44
aabb = aa44 = 11 x a04 then a = 7
aabb = 7744 = 882
So a = 7, b = 4, x = 8, and y = 8
8. If 20043
= 𝐴2
− 𝐵2
where A and B are natural numbers, then determine the values of A
and B!
Solution :
13
+ 23
+ 33
+ ⋯ + (𝑛 − 1)3
+ 𝑛3
= [
1
2
𝑛(𝑛 + 1)]
2
[
1
2
(𝑛 − 1)𝑛]
2
+ 𝑛3
= [
1
2
𝑛(𝑛 + 1)]
2
𝑛3
= [
1
2
𝑛(𝑛 + 1)]
2
− [
1
2
(𝑛 − 1)𝑛]
2
20043
= [
1
2
. 2004.2005]
2
− [
1
2
. 2003.2004]
2
= (1002.2005)2
− (1002.2003)2
𝑆𝑜, 𝐴 = 1002.2005 𝑎𝑛𝑑 𝐵 = 1002.2003

5. Faculty of Teacher Training and Education
University of Jember
2013
9. 𝑎1, 𝑎2, 𝑎3, … , 𝑎 𝑛is a different natural numbers. If 2 𝑎1 + 2 𝑎2 + 2 𝑎3 + ⋯ + 2 𝑎 𝑛 = 2005 then,
determine the value of 𝑎1 + 𝑎2 + 𝑎3 + … + 𝑎 𝑛 !
Solution :
2005 = 111110101012
2005 = 210
+ 29
+ 28
+ 27
+ 26
+ 0 + 24
+ 0 + 22
+ 0 + 20
𝑎1 + 𝑎2 + 𝑎3 + … + 𝑎 𝑛 = 10 + 9 + 8 + 7 + 6 + 4 + 2 + 0 = 46
10. Determinethe value of𝑥, 𝑦, 𝑧real numbersthatsatisfythe equation:
𝑥2
+ 2𝑦𝑧 = 𝑥 … (1)
𝑦2
+ 2𝑧𝑥 = 𝑦 … (2)
𝑧2
+ 2𝑥𝑦 = 𝑧 … (3)
Answer:
If equation (1) times 𝑥, equation (2) times 𝑦 and equation (3) times 𝑧 then obtained:
𝑥3
+ 2𝑥𝑦𝑧 = 𝑥2
𝑦3
+ 2𝑥𝑦𝑧 = 𝑦2
𝑧3
+ 2𝑥𝑦𝑧 = 𝑧2
by eliminating2𝑥𝑦𝑧 then obtained 𝑥 = 𝑦 = 𝑧
𝑥2
+ 2𝑦𝑧 = 𝑥 → 𝑥2
+ 2𝑥. 𝑥 = 𝑥 → 𝑥 =
1
3
= 𝑦 = 𝑧
11. Equation of a circle centered at (-3.2) and intersect to the line 3𝑥 − 4𝑦 − 8 = 0 is
Solution:
(𝑥 − 𝑎)2
+ (𝑦 − 𝑏)2
= 𝑟2
𝑤ℎ𝑒𝑛 𝑟 = |
𝑎𝑥 + 𝑏𝑦 + 𝑐
√𝑎2 + 𝑏2
|
= |
3(−3)+4(2)−8
√32+42
|
= 5
𝑠𝑜, (𝑥 + 3)2
+ (𝑦 − 2)2
= 25

6. Faculty of Teacher Training and Education
University of Jember
2013
12. Circle 𝐿 ⋮ (𝑥 − 4)2
+ (𝑦 + 2)2
= 9 intersect the line 𝑥 = 4. Equation of the tangent line at
the point of intersection circle and the line 𝑥 = 4 is
Solution:
𝑦 = ±𝑟 + 𝑏
𝑦 = ±3 − 2 → 𝑦1 = 1, 𝑦2 = −5
13. Given some point A(1,-1,-2), B(4,3,-7) and C(2,-3,0). The cosines angle between AB and AC
is
Solution:
Let x = AB = b – a = (3,4, -5) → |x|= 5√2
Y = AC = c – a = (1, -2, 2) → |y|= 3
so, cos 𝜃 =
𝑥𝑦
|𝑥||𝑦|
=
−15
15√2
= −
1
2
√2
14. If OA(1,2), OB(4,2) and 𝜃 = ∠(OA,OB), so that tan 𝜃=
Solution:
|a|= √5 dan |b| = √20 = 2√5
cos 𝜃 =
𝑎𝑏
|𝑎||𝑏|
→ 𝑐𝑜𝑠𝜃 =
8
10
=
4
5
, 𝑠𝑜 𝑦 = 3
so, tan 𝜃 =
4
3
15. If the line 3x+2y=6 is translated to the matrix (3 -2), then the transformation is
Solution:
𝑎𝑥 + 𝑏𝑦 = 𝑐 → 𝑇(𝑝 𝑞): 𝑎𝑥 + 𝑏𝑦 = 𝑐 + 𝑎𝑝 + 𝑏𝑞
3𝑥 + 2𝑦 = 6 → 𝑇(3 − 2): 3𝑥 + 2𝑦 = 6 + 9 − 4 → 3𝑥 + 2 = 11
16. From experiment of throwing two dices for 900 times, found the expectation frequency
totaled 5 of the dice
Solution:
P (probability of the dice numbered 5) =
4
36
=
1
9
F (the expectation frequency) = P x the number of experiments

7. Faculty of Teacher Training and Education
University of Jember
2013
=
1
9
× 900
= 100
17. from experiment of throwing a dice for 60 times, found the expectation frequency of the
dice factor of 6
Solution:
P (factor of 6) =
4
6
=
2
3
F (the expectation frequency) = P x the number of experiments
=
2
3
× 60
= 40
18. In a bag there are 6 red balls, 5 blue balls and 4 green balls. If the first decision taken a red
ball and did not return to the bag. probability for second decision red ball taken from the
bag.
Solution :
𝑝(𝑟𝑒𝑑, 𝑟𝑒𝑑) =
𝑟𝑒𝑑 𝑏𝑎𝑙𝑙 − 1
𝑟𝑒𝑑 𝑏𝑎𝑙𝑙 + 𝑏𝑙𝑢𝑒 𝑏𝑎𝑙𝑙 + 𝑔𝑟𝑒𝑒𝑛 𝑏𝑎𝑙𝑙 − 1
=
6−1
6+5+4−1
=
5
14
19. On experiment of throwing 4 coins, the probability of appearing 3 image faces and 1
numbering face is.
Solution:
(number face + image face)4
= (N + I)4
= 𝑁4
+ 4𝑁3
𝐼 + 6𝑁2
𝐼2
+ 4𝑁𝐼3
+ 𝐼4
𝑆𝑜, 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 3𝐼 𝑎𝑛𝑑 1𝑁 =
4
16
=
1
4

8. Faculty of Teacher Training and Education
University of Jember
2013
20. average score of some boy students is 6.4 and average score of some girl students is 7.4. if
the average score of whole of them is 7.0. so the amount ratio of the boys and the girl
student is
Solution:
𝐺𝑖𝑟𝑙𝑠 ∶ 𝐵𝑜𝑦𝑠 = (𝑥̅ − 𝑥̅ 𝑏𝑜𝑦) ∶ (𝑥̅ 𝑔𝑖𝑟𝑙 − 𝑥̅)
= (7.0 − 6.4) ∶ (7.4 − 7.0)
= 0.6 ∶ 0.4
= 3 ∶ 2