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
ร 250.25
ร 250.25
ร โฆ ร 250.25
= 125, so that (n-3)(n+2)....
solution:
Remember that:
๐ ๐
๐ฅ ๐ ๐
= ๐ ๐+๐
๐ ๐(๐ฅ)
= ๐ ๐
โ ๐(๐ฅ) = ๐
250.25
ร 250.25
ร 250.25
ร โฆ ร 250.25
= 125
(250.25) ๐
= 125
((52)0.25
) ๐
= 53
50.5๐
= 53
0.5๐ = 3
๐ =
3
0.5
๐ = 6
so, (๐ โ 3)(๐ + 2) = (6 โ 3)(6 + 2) = 3 ๐ฅ 8 = ๐๐
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 + โฏ is ....
Solution:
Arithmetical line formula:
๐๐ =
๐
2
(2๐ + (๐ โ 1)๐)
Logarithmic characteristic:
log(๐ ๐ฅ ๐) = log ๐ + log ๐
๐
log ๐ = log ๐ ๐
Exponent characteristic:
(๐ ๐) ๐
= ๐ ๐๐
๏ท from the problem, we found an arithmetical line,
๐1 = log 5 ; ๐2 = log 55 ; ๐3 = log 605 ; ๐4 = log 6655
๐1 = log 5 ; ๐2 = log 55 = log 5 + log 11 ; ๐3 = log 605 = ๐๐๐55 + ๐๐๐11 ; ๐4
= log 6655 = log 605 + ๐๐๐11
with initial number (a) = log 5
Difference (b) = log 11
so, the sum of first 50 digit is
๐๐ =
๐
2
(2๐ + (๐ โ 1)๐)
๐50 =
50
2
(2(log 5) + (50 โ 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 + . . . . + ๐ = ๐๐๐, then specify the value of ๐ ๐๐๐ ๐๐๐!
Solution:
1 + 2 + 3 + โฏ + ๐ =
๐
2
(๐ + 1)
๐๐๐ = 111 ร ๐ = (3 ร ๐) ร 37
๐
2
(๐ + 1) = (3 ร ๐) ร 37
๐(๐ + 1) = (6 ร ๐) ร 37
This is a sequential multiplication.
So a = 6 and n = 36
4. If a and b are arbitrary numbers, prove that
๐
๐
+
๐
๐
โฅ 2 !
Solution:
(๐ โ ๐)2
โฅ 0 โ ๐2
+ ๐2
โฅ 2๐๐ โ
๐
๐
+
๐
๐
โฅ 2
5. ๐ด = 13
โ 23
+ 33
โ 43
+ 53
โ 63
+ โฏ + 20053
, determine the value of A!
Solution :
= (13
+ 23
+ 33
+ โฏ + 20053
) โ 2(23
+ 33
+ โฏ + 20043)
= (13
+ 23
+ 33
+ โฏ + 20053
) โ 2. 23
(13
+ 23
+ 33
+ โฏ + 10023
)
= (
1
2
. 2005.2006)
2
โ 16 (
1
2
. 1002.1003)
2
= 10032
(20052
โ (4502)2
)
= 10032(20052
โ 20042)
= 10032(2005 + 2004)(2005 โ 2004)
= 10032
.4009
6. This form ๐ฅ4
โ 1 has how many factors....
Solution :
4. Faculty of Teacher Training and Education
University of Jember
2013
Note that ๐ฅ4
โ 1 = (๐ฅ2
+ 1)(๐ฅ + 1)(๐ฅ โ 1). So, the form ๐ฅ4
โ 1 has a (1+1) ร (1+1) ร (1
+1) = 8 factors.
7. If ๐๐๐๐ = (๐ฅ๐ฆ)2
, determine the value of a, b, x, and y!
Solution :
Because (๐ฅ๐ฆ)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
= ๐ด2
โ ๐ต2
where A and B are natural numbers, then determine the values of A
and B!
Solution :
13
+ 23
+ 33
+ โฏ + (๐ โ 1)3
+ ๐3
= [
1
2
๐(๐ + 1)]
2
[
1
2
(๐ โ 1)๐]
2
+ ๐3
= [
1
2
๐(๐ + 1)]
2
๐3
= [
1
2
๐(๐ + 1)]
2
โ [
1
2
(๐ โ 1)๐]
2
20043
= [
1
2
. 2004.2005]
2
โ [
1
2
. 2003.2004]
2
= (1002.2005)2
โ (1002.2003)2
๐๐, ๐ด = 1002.2005 ๐๐๐ ๐ต = 1002.2003
5. Faculty of Teacher Training and Education
University of Jember
2013
9. ๐1, ๐2, ๐3, โฆ , ๐ ๐is a different natural numbers. If 2 ๐1 + 2 ๐2 + 2 ๐3 + โฏ + 2 ๐ ๐ = 2005 then,
determine the value of ๐1 + ๐2 + ๐3 + โฆ + ๐ ๐ !
Solution :
2005 = 111110101012
2005 = 210
+ 29
+ 28
+ 27
+ 26
+ 0 + 24
+ 0 + 22
+ 0 + 20
๐1 + ๐2 + ๐3 + โฆ + ๐ ๐ = 10 + 9 + 8 + 7 + 6 + 4 + 2 + 0 = 46
10. Determinethe value of๐ฅ, ๐ฆ, ๐งreal numbersthatsatisfythe equation:
๐ฅ2
+ 2๐ฆ๐ง = ๐ฅ โฆ (1)
๐ฆ2
+ 2๐ง๐ฅ = ๐ฆ โฆ (2)
๐ง2
+ 2๐ฅ๐ฆ = ๐ง โฆ (3)
Answer:
If equation (1) times ๐ฅ, equation (2) times ๐ฆ and equation (3) times ๐ง then obtained:
๐ฅ3
+ 2๐ฅ๐ฆ๐ง = ๐ฅ2
๐ฆ3
+ 2๐ฅ๐ฆ๐ง = ๐ฆ2
๐ง3
+ 2๐ฅ๐ฆ๐ง = ๐ง2
by eliminating2๐ฅ๐ฆ๐ง then obtained ๐ฅ = ๐ฆ = ๐ง
๐ฅ2
+ 2๐ฆ๐ง = ๐ฅ โ ๐ฅ2
+ 2๐ฅ. ๐ฅ = ๐ฅ โ ๐ฅ =
1
3
= ๐ฆ = ๐ง
11. Equation of a circle centered at (-3.2) and intersect to the line 3๐ฅ โ 4๐ฆ โ 8 = 0 is
Solution:
(๐ฅ โ ๐)2
+ (๐ฆ โ ๐)2
= ๐2
๐คโ๐๐ ๐ = |
๐๐ฅ + ๐๐ฆ + ๐
โ๐2 + ๐2
|
= |
3(โ3)+4(2)โ8
โ32+42
|
= 5
๐ ๐, (๐ฅ + 3)2
+ (๐ฆ โ 2)2
= 25
6. Faculty of Teacher Training and Education
University of Jember
2013
12. Circle ๐ฟ โฎ (๐ฅ โ 4)2
+ (๐ฆ + 2)2
= 9 intersect the line ๐ฅ = 4. Equation of the tangent line at
the point of intersection circle and the line ๐ฅ = 4 is
Solution:
๐ฆ = ยฑ๐ + ๐
๐ฆ = ยฑ3 โ 2 โ ๐ฆ1 = 1, ๐ฆ2 = โ5
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 โ a = (3,4, -5) โ |x|= 5โ2
Y = AC = c โ a = (1, -2, 2) โ |y|= 3
so, cos ๐ =
๐ฅ๐ฆ
|๐ฅ||๐ฆ|
=
โ15
15โ2
= โ
1
2
โ2
14. If OA(1,2), OB(4,2) and ๐ = โ (OA,OB), so that tan ๐=
Solution:
|a|= โ5 dan |b| = โ20 = 2โ5
cos ๐ =
๐๐
|๐||๐|
โ ๐๐๐ ๐ =
8
10
=
4
5
, ๐ ๐ ๐ฆ = 3
so, tan ๐ =
4
3
15. If the line 3x+2y=6 is translated to the matrix (3 -2), then the transformation is
Solution:
๐๐ฅ + ๐๐ฆ = ๐ โ ๐(๐ ๐): ๐๐ฅ + ๐๐ฆ = ๐ + ๐๐ + ๐๐
3๐ฅ + 2๐ฆ = 6 โ ๐(3 โ 2): 3๐ฅ + 2๐ฆ = 6 + 9 โ 4 โ 3๐ฅ + 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
ร 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
ร 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 :
๐(๐๐๐, ๐๐๐) =
๐๐๐ ๐๐๐๐ โ 1
๐๐๐ ๐๐๐๐ + ๐๐๐ข๐ ๐๐๐๐ + ๐๐๐๐๐ ๐๐๐๐ โ 1
=
6โ1
6+5+4โ1
=
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
= ๐4
+ 4๐3
๐ผ + 6๐2
๐ผ2
+ 4๐๐ผ3
+ ๐ผ4
๐๐, ๐กโ๐ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐๐ 3๐ผ ๐๐๐ 1๐ =
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:
๐บ๐๐๐๐ โถ ๐ต๐๐ฆ๐ = (๐ฅฬ โ ๐ฅฬ ๐๐๐ฆ) โถ (๐ฅฬ ๐๐๐๐ โ ๐ฅฬ )
= (7.0 โ 6.4) โถ (7.4 โ 7.0)
= 0.6 โถ 0.4
= 3 โถ 2