1. SMART SOLUTION
by
NOVIAN NUR FATIHAH and MAULINDA FITRI SEPTIANINGRUM
In here, we will give some problem about:
1. Number Theory
2. Algebra
3. Geometry
4. Probability
1 Number Theory
1.1 Exercise
1. If A = 13 − 23 + 33 − 43 + 53 − 63 + ... + 20053 , determine the value of A!
2. Given a = 12
1 + 22
3 + 32
5 + · · · + 10022
2003 and b = 12
3 + 22
5 + 32
7 + · · · + 10022
2005
Show the nearest integer of a − b!
3. Given a, b, c, d, and e are real numbers. Ifa + b + c + d + e = 19 and
a2 + b2 + c2 + d2 + e2 = 99
4. Calculate the results of the 12 − 22 + 32 − 42 + 52 − 62 + ... + 20092 −
20102 + 20112
5. Prove for any real numbers a, b apply a2 + b2 ≥ 2ab
1.2 Answer
1. If A = 13 − 23 + 33 − 43 + 53 − 63 + ... + 20053 , determine the value of A
= (13 + 23 + 33 + ... + 20053) − 2(23 + 33 + ... + 20043)(10)
= (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)
1

2. = 10032(2005 + 2004)(2005 − 2004)
= 10032.4009
2. Answer : a − b = (12
1 + 22
3 + 32
5 + ... + 10022
2003 ) − (12
3 + 22
5 + 32
7 + ... + 10022
2005 )
= (12
1 + (22
3 − 32
3 ) + (32
5 − 22
5 ) + ... + (10022
2003 − 10012
2003 ) − 10022
2005
= 1 − 10022
2005 + (1 + 1 + 1 + ... + 1)
= 1002 − 10022
2005 = 1002(2005−1002)
2005 = 1002.1003
2005 ≈ 501
3. Given a, b, c, d, and e are real numbers. Ifa + b + c + d + e = 19 and
a2 + b2 + c2 + d2 + e2 = 99
= (19 − e)2 = (a + b + c + d)2
361 − 38e + e2 = a2 + b2 + c2 + d2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd
361 − 38e + e2 = 99 − e2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd
361−38e+e2 ≤ 99−e2 +a2 +b2 +a2 +c2 +a2 +d2 +b2 +c2 +b2 +d2 +c2 +d2
361 − 38e + e2 ≤ 99 − e2 + 99 − e2 + 99 − e2 + 99 − e2
361 − 38e + e2 ≤ 396 − 4e2
325e2 − 38e − 35 ≤ 0
4. Calculate the results of the 12 − 22 + 32 − 42 + 52 − 62 + ... + 20092 −
20102 + 20112 !
12 − 22 can be converted into (1 − 2)(1 + 2) = −1 − 2, 32 − 42 can be
converted into (3 − 4)(3 + 4) = −3 − 4, and so on.
So that the shape can be changed to : −1.−2, −3, −4, −5, −7, ..., −2009, −2010, 20112
or − (1 + 2 + 3 + 4 + ... + 2009 + 2010) + 20112
1
2 × 2010 × 2011 + 20112
2

3. 2011 − 1005 + 2011
2011 × 1006 = 2023066
5. Prove for any real numbers a, b apply a2 + b2 ≥ 2ab
= (a − b)2 ≥ 0 ↔ a2 − 2ab + b2 ≥ 0 ↔ a2 + b2 ≥ 2ab
2 Algebra
2.1 Exercise
1. Determine the value of x, y, z real numbers that satisfy the equation:
x2 + 2yz = x · · · (1)
y2 + 2zx = y · · · (2)
z2 + 2xy = z · · · (3)
2. Given a and b are real numbers that are satisfy :
a3 − 3ab2 = 44
b3 − 3a2b = 8
Determine the value a2 + b2 !
3. Given x, y, z and t are real numbers that are not zero and satisfies the
equation:
x + y + z = t · · · (1)
1
x + 1
y + 1
z = 1
t · · · (2)
x3 + y3 + z3 = 10003 · · · (3)
Determine the value of x + y + z + t !
4. Prove for any positive numbers a, b, c and d apply (a+b+c+d)
4 ≥ 4
√
abcd !
5. If α dan β roots of the equation x2−ax+b = 0 (a, b real). Make quadratic
equations who roots α3
β dan β3
α
2.2 Answer
1. If equation(1) times x, equation(2) times y and equation(3) times z then
obtained:
x3 + 2xyz = x2
3

4. y3 + 2xyz = y2
z3 + 2xyz = z2
by eliminating 2xyz then obtained x = y = z
x2 + 2yz = x −→ x2 + 2x.x = x −→ x = 1
3 = y = z
2. a3 −3ab2 = 44 −→ (a3 −3ab2)2 = 442 ←→ a6 −6a4b2 +9a2b4 = 1936...(1)
b3 − 3a2b = 8 −→ (b3 − 3a2b)2 = 82 ←→ b6 − 6a2b4 + 9a4b2 = 64...(2)
Equation (1) addition by (2)
a6 + 3a4b2 + 3a2b4 + b6 = 2000 (a2 + b2)3 = 2000 ←→ a2 + b2 = 3
√
2000 =
10 3
√
2
3. Determine the value of x + y + z + t !
1
x + 1
y + 1
z = (xy+xz+yz)
xyz = 1
t ←→ xy + xz + yz = xyz
t
(x + y + z)3 = x3 + y3 + z3 + 3(x + y + z)(xy + xz + yz) − 3xyz
t3 = x3 + y3 + z3 + 3txyz
t − 3xyz
x3 + y3 + z3 = t3 = 10003 −→ t = 1000
x + y + z + t = t + t = 2t = 2000
4. (a+b+c+d)
4 = (
(a+b)
2
+
(c+d)
2
2 ≥
√
ab+
√
cd
2 ≥
√
ab
√
cd =
√
abcd
5. x2 − ax + b = 0
α + β = a dan αβ = b
α3
β + β3
α = α4+β4
αβ = (α2+β2)2−2α2β2
αβ
=(α+β)2−2αβ
2
−2(αβ)2
αβ
=(α2−ab)2
b
=a4−4a2b+2b2
b
α3
β .β3
α = (αβ)2= b2
Quadratic equations who roots α3
β dan β3
α is:
4

5. (x − α3
β )(x − β3
α ) = 0
x2 − (α3
β + β3
α )x + α3
β .β3
α = 0
x2 − a4−4a2b+2b2
b x + b2 = 0
bx2 − (a4 − 4a2b + 2b2)x + b3 = 0
3 Geometry
3.1 Exercise
1. The area bounded by the curve y = x2 − 4x + 3 and y = 3 − x are ...
2. The area bounded by the curve y = x2 + 3x + 4 and y = 1 − x are ...
3. Turn the volume of the object that occurs when the area bounded by the
curve y = x2 and y = 4x − 3 rotated 360o around the X axis is ...
4. A triangle has base length 3 cm shorter than the height and breadth of
less than 27cm2. If height of triangle t cm so...
5. Given a triangle ABC with sides AB, BC and CA respectively is 5 cm, 6
cm and 4 cm. What is sin2(< BAC)
3.2 Answer
1. y1 = y2
⇒ x2 − 4x + 3 = 3 − x
⇔ x2 − 3x = 0
So, D = b2 − 4ac = 9
L = D
√
D
6a2 = 9
√
9
6.12 = 27
6 = 9
2
So, L = 9
2
2. y1 = y2
⇒ x2 + 3x + 4 = 1 − x
⇔ x2 + 4x + 3 = 0
5

6. So, D = b2 − 4ac = 4
L = D
√
D
6a2 = 4
√
4
6.1 = 8
6 = 4
3
So, L = 4
3
3. Volume benda putar
V = π
b
a y2
1 − y2
2dx
= π
3
1 (4x − 3)2 − (x2)2dx
= π
3
1 (−x4 + 16x2 − 24x + 9)dx
= [−1
5 x5 + 16
3 x3 − 12x2 + 9x]3
1
= (−1
5 (3)5 + 16
3 (3)3 − 12(3)2 + 9(3)) − (−1
5 (1)5 + 16
3 (1)1 − 12(1)2 + 9(1))
= (−243
5 + 144 − 108 + 27) − (−1
5 + 16
3 − 12 + 9)
= (216
15 ) − (32
15 )
= 184
15 = 124
5 satuan volume
4. Suppose TD = x
x
9 = x+8
15
15x = 9x + 72
6x = 72, so x = 12
TA = x + 8 = 12 + 8 = 20
5. a2 = b2 + c2 − 2bccosA
62 = 42 + 52 − 2.4.5.CosA
36 = 16 + 25 − 40cosA
40cosA = 5
cosA = 1
8
sinA = 1 − cosA = 1 − 1
64 = 63
64
6

7. 4 Probability
4.1 Exercise
1. Suppose there are 5 cards where each card is given a different number
is 2, 3, 4, 5, 6. The cards are then lined up from left to right so that the
randomly shaped line. What is the probability that the number of cards
lined up from left to right and placed on the i − th place will be greater
or equal to i for each i with 1 ≤ i ≤ 5?
2. N circle drawn on a flat surface so that there are six point where the six
points are contained in at least three circles. What is the smallest N that
satisfies these conditions?
3. Given four dice equilibrium and different, each of which is octagonal ir-
regularly edged 1, 2, 3, ..., 8. Four dice collide (thrown) together one time.
Probability of occurrence of two dice with the same eye for emerging are...
4. Five students A, B, C, D, E are in one group in a relay race. If A is not
able to run the first and D could not run Last reviewed, then the number
of possible arrangements is ...
5. In a pouch there are 2 white balls and six balls red. Taken one ball at
random and ball took its color is noted. After which the ball is returned
to the pockets and later retrieved again one ball at random. Opportunity
took two balls of different colors is...
4.2 Answer
1. 2
15
The simplest arrangement is 2, 3, 4, 5, 6. To satisfy the conditions of
the respective matter each number 2, 3, 4, and 5 can only be shifted to
the right one step. This way there are as many as 24 = 16 . As for
the figures probably do not need to be shifted to the left we noticed, be-
cause if we shift the numbers to the left then there must be a number
that should be shifted to the right so that already in the first calculation
above. Therefore, the greater the probability 2
15 .
2. 5
If we draw three circles on a flat, then the maximum would be formed
6 cut point. Because through any three points that are not in line can
be formed a circle passing through the three points, then by making two
circles each through 3 points A, B and C, D, E : 5 F will form a circle
in which there are 6 points respectively - each found in 3 circles, as what
was requested.
7

8. 3. Number of occurrences of all numbers different dice is 8 × 7 × 6 × 5
Opportunitiestherearethesamenumber
1 − 8×7×6×5
84 = 151
256
So, chances there are the same number are 151
256
4. There are 2 case :
-If D as the primary runner
Ways of choosing 2nd runners there 4 , 3rd runner there are 4, 4th runner
there 2, and 5th runner there 1.
The number of ways is = 4 × 3 × 2 × 1 = 24
-If D is not a primary runner
Ways of choosing 1st runner there 3 Ways of choosing 5th runner
there 3 Ways of choosing 2nd runners there 3 and 3rd runner there
are 2, and 4th runner there 1.
The number of ways is = 3 × 3 × 3 × 2 × 1 × ×3 = 54
Many ways to compose runners is = 24+54=78
So, Many ways to compose runners is 78
5. P(2 different color balls)
=P(white dan red)+P(red dan white)
= 2
8 .6
8 + 6
8 .2
8
= 24
64 = 3
8
8