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Work/Rate/Speed
Whenever a problem mentions rate, they are referring to
amount of work
time spent
.
Example 1: Suppose on a highway, the speed limit is 50 miles per hour. That means
your car should not travel more than 50 miles in one hour of time.
Speed = distance/ time
In this case, it is 50 miles
1 hour
If you continue to drive on this highway, how long would it take to reach Grandmother's
house 60 miles away?
speed * time = distance traveled
In this formula, you know all the elements except for the time. So make time = x (or any
other variable) and solve for x. In harder problems, it’s easy to lost track of what you are
looking for and assigning a variable can make it easier to solve.
50 miles
1 hour
*
x = 60 miles
Simple algebra says that x = 1.2, meaning the time needed is 1.2 hours, 1 hour and 12
minutes.
Example 2: Now say that you traveled for 45 miles on the rural highway and the rest of
the trip (15 miles) in a residential area with a speed limit of 30 mph. What was your
average speed?
First instinct may tell you to take the mean of 50 and 30, which is what an average of
several numbers usually implies. But if speed was only a matter of numbers, you could
have traveled for 30 miles on the rural highway and 30 in a residential area or 5 minutes
on the highway and the remaining time in residential and the average of 50 and 30 will
always be 40. Obviously, that is not the case.
Think of an average speed as the total distance covered in a certain time. If a car had
begun the journey alongside you, traveled at one speed throughout the entire trip, and
arrived at the same time as you did, you both would have the same average speed and the
speed that car traveled would be the average speed. This is because you covered the same
distance in the same amount of time.
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To calculate this speed, find the
total distance
time
Since average speed is what you are looking for, make x = average speed. To find the
average speed of 45 miles at 55 mph and 15 miles at 30, the distance you already know is
60 miles.
Now, rate = amount of work
time
but if you manipulate the formula, time = amount of work
rate
As you know, dividing by a fraction is the same as multiplying by a reciprocal, so we will
do just that.
45 miles * 1 hour + 15 miles * 1 hour = x
50 miles
30 miles
45 miles * 1 hour + 15 miles * 1 hour = x
50 miles
30 miles
0.9 hours + 0.5 hours = 1.4 hours
60 miles = 42 and 6/7 miles per hour = average speed of trip
1.4 hours
Example 3: Now suppose you come across one of those problems that says Susy takes
this long to do this task and Tom takes this long, how long does it take them to do it
together?
Always use the basic principle that rate = amount of work
time
Then don’t forget to identify your variable.
So let’s pretend that Susy can make a pizza in 2 hours and Tom can make 2 pizzas in 3
hours. How long does it take them to make one pizza together?
Suzy’s rate = 1 pizza
2 hours
Tom’s rate = 2 pizzas
3 hours
Just like the problem about average speed, you are trying to figure out how long it will
take to make 1 pizza using 2 different speeds. If you manipulate the original formula
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rate = amount of work
time
It is also:
rate * time = amount of work
You want to find out the amount of time for Suzy and Tom to make one pizza, so x =
total time. You also know the two rates, so:
x
1 pizza
2 pizzas
 x
 1 pizza
2hours
3hours
x*
1
2
 x * 1
2
3
x 2x

1
2 3
3x 4 x

1
6
6
7x
1
6
x = 6/7
Because x = total time to make one pizza, it is measured in hours.
It will take Susy and Tom 6/7 of an hour to make one pizza together.
Remember:


Use and manipulate the formula rate = amount of work
time
Identify the unknown variable.
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Interest
Pretend your friend had loaned you $10 and demanded that you pay her back with 6%
interest. You would give her the $10 and add 6%, which is:
10  0.06  10  $10.60
or
10  1.06  $10.60
What you just calculated is simple interest plus the original amount.
However, banks do not use simple interest. An average bank today may give you 3%
interest. This does not mean that you deposit money into an account and when you
withdraw it, you will have 3% more.
The Opulent Bank provides 3% interest compounded yearly, and Dave deposited $1000
into his account. After the first year, he will have
1000  1.03  $1030
The next year, he will have 3% interest from $1030 = $1060.90, and so on. If Dave kept
his account unchanged for five years, his sum will be
1000  1.03  1.03  1.03  1.03  1.03  $1159.27
or
1000  1.035  $1159.27
In other words,
TotalAmount  A0 (1  r )t
where: A0 = initial amount
r = rate as a decimal
t = number of years
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Dave also has an account in The Superior Bank, which provides 3% annual
interest compounded monthly. He will earn more than 3% each year. Compounding
monthly means the rate will be divided by 12 and then compounded each month. After
one month, he will have
0.03 

1000  1 
  $1002.50
12 

In one year, he will have
12
 0.03 
1000  1 
  $1030.42
12 

In 5 years, he will have
60
 0.03 
1000  1 
  $1161.62
12 

In other words to calculate annual interest that is compounded more than once a year
(quarterly, monthly, daily):
r

TotalAmount  A0 1  
 n
where :
nt
A0 = initial amount
r = annual rate
n = times compounded per year
t = number of years
This formula assumes discrete growth (at a fixed point in time, a certain amount is
added in). However, it may be used to approximate growth and decay.
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Triangle Tips
Special Right Triangles
30°-60°-90° Triangles
30°
As you can see, in a 30°-60°-90° triangle, the ratio of the sides are
1: 3 :2. This is true in any triangle with angle measures 30°, 60°,
90°.
2x
x 3
60°
x
45°-45°-90° Triangles
The ratio of the sides of a 45°-45°-90° triangle, then is
45°
2
2
2:
2 :2.
45°
2
When a problem does not provide all the necessary elements of the Pythagorean theorem
(refer to geometry formula sheet if you don’t know what the Pythagorean Theorem is),
these two rules will become very useful.
Angle-side Relationships
C
80°
B
40°
60°
The longest side of any triangle is opposite the largest
interior angle; the shortest side is opposite the smallest
angle. In the following triangle, side A is longer than
side B, which is longer than side C, because
80>60>40.
A
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Triangle Inequality
Triangle inequality, the postulate states that the length of each side of a triangle
must be less than the sum of the lengths of the two remaining sides. So, if we think of a
triangle as the roads connecting three cities A, B, and C to each other, we know that
traveling directly from A to B will be shorter than the indirect route through C.
City A
AB
City B
AC
BC
City C
AB < AC + BC, AC < AB + BC, BC < AB + AC
To understand the concept of triangle inequality, one must consider the fact that
the shortest distance between two points on a plane is the length of the straight line
segment joining them. Since no two sides of a triangle can be collinear because all angles
in a triangle must be between 0o and 180o, exclusive, it is easy to see that why triangle
inequality must be true.
This concept is useful in that it allows us to determine whether a set of three
numbers can represent the lengths of the sides of a triangle at a glance. All one has to do
is make sure that each number is less than the sum of the other two. This can be useful in
eliminating extraneous solutions in some problems.
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Similar Triangles
What do you notice about the two triangles?
D
A
B
C
E
F
ABC and DEF are identical except that DEF is a larger version of ABC, right?
Their angles are the same and the lengths of their sides are proportional to each other.
Triangles such as these are called similar triangles.
In other words, if triangle ABC and triangle DEF are similar, then:
A = D
B = E
C = F
and
AB BC
AC


DE EF DF
What if you only knew that A = D and B = E? Would ABC and DEF still be
similar?
Yes! Because all triangles have 180°, if you know two angles, you also know the third.
One more way to identify similar triangles is called SAS (side-angle-side).
If two triangles have one common angle and the sides that make up that angle are
proportional, they are similar.
P
F
3
6
55°
55°
T
4
E
G
O
8
(Note: Unless the problem states that figures are drawn to scale, assume they’re not.)
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PET and FOG are similar by SAS.
By knowing that triangles are similar, you can figure out information about one triangle
based on information about the other.
For example:
P
3
F
7
6
55
T
50
55
4
E
G
O
8
Here is the information provided:
T = 55°
PE = 7
TE = 4
PT = 3
G = 55°
O = 50°
FG = 6
GO = 8
From this, you should be able to solve for:
a. P
b. E
c. F
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d. FO
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Shoelace
The Shoelace Algorithm is a procedure used to find the area enclosed by any set of
coordinates.
For example:
Find the area of the triangle determined by the lines: y-x=4, -x=10+2y, and y=-2x+10.
In this problem, the standard approach is to graph the lines, find the points of intersection,
and use a box plot to solve the area. Instead of using box plot, shoelace may be applied
instead.
In this case, the coordinates are: (-6,-2), (2,6), and (10,-10). (This can be solved using
a system of equations.) The area of the triangle, K, can be found by following the
procedure:
1)
 6 2 
 2 6


 10 10


 6 2 
Select a vertex and travel around the triangle ending with the
starting point. Write a matrix of the coordinates of the path
including both the starting and ending coordinates.
 6 2 
 2 6


 10 10


 6 2 
Determine the sum of all “\” products:
(-6)(6)+(2)(10)+(10)(2)
=-36+-20+-20
=-76
 6  2 
 2 6


 10 10


 6 2 
Determine the sum of all “/” products:
(-2)(2)+(6)(10)+(-10)(-6)
=-4+60+60
=116
2)
3)
4)
Determine the absolute value of the difference of the “\”
products and “/” products:
(76)  (116)
= 192
=192
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5)
Take ½ of (4)
½(192)=96= area of triangle
The formula for the area is displayed:
 6  2 
 2 6
 =1/2 |(-76)-(96)|=1/2(192)=96
K=1/2 abs 
 10 10


 6 2 
Note: Do you see the “shoelaces”?
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Circles
1. A central angle is equal in degrees to its intercepted arc. A central angle is one in
which the point at which the rays meet is at the center of the circle (shown below).
If arc AB = 80º, then
Angle AOB = 80º
O
A
80º
B
2. An inscribed angle is equal in degree to one-half its intercepted arc. An inscribed
angle is one in which the point at which the rays meet is on the circumference of the
circle (shown below).
If arc AC = 120º, then
Angle ABC = 60º
A
B
120º
C
3. An interior angle, formed by two chords intersecting in a circle, is equal in degrees to
one half the sum of its intercepted arcs.
If arc AD = 40º and arc CB = 80º,
Then angle CEB = 60º = angle AED
C
A
E
40
Angle AED = ½(AD + BC)
=1/2(40 + 80) = 60
80
D
B
4. An exterior angle, formed outside of a circle by two secants, a secant and a tangent,
or two tangents, is equal in degrees to one-half the difference if its intercepted arcs.
C
If arc AD = 120º and
B
70
arc BD = 70º, then angle ACD = 25º
A
Angle ACD = ½(AD-BD)
=1/2(120°-70°)=25°
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D
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5. Two tangent segments drawn to a circle from the same external point are congruent
(same length).
If AC and CE are tangent
to circle O at B and D,
Then CB = CD.
B
A
90
C
O
90
D
6.
E
Angle O and angle C are supplementary angles when OB is tangent to AC and OD is
tangent to CE.
A
B
90
O
C
90
D
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Interior Angles
Polygons are closed plane figures bounded by three or more line segments.
Polygons cannot be composed of any curves since true curves cannot be constructed
using straight-line segments. Circles and ellipses, therefore, are not polygons.
Polygons are named according to their number of line segments, or sides. Examples of
polygons:
name
number of sides
triangle
3
quadrilateral
4
pentagon
5
hexagon
6
7
heptagon
octagon
8
nonagon
9
decagon
10
11-gon
11
dodecagon
12
…
…
n-gon
n
When we refer to the angles of a polygon, we are in fact referring to interior
angles. An important property of polygons is that all polygons with the same number of
sides have the same sum of interior angles. For example, adding the three interior angles
of any triangle (three-sided polygon) always produces 180°, adding the four angles of any
quadrilateral always produces 360°, and adding the five angles of any pentagon always
produces 540°. In general, adding the interior angles of any n-sided polygon always
produces 180(n-2) degrees.
Why does this pattern occur?
Every time we add another side to a polygon, a triangle is added on to the
polygon. Since the angles of a triangle total 180 degrees, 180 degrees are added to the
interior angle sum of the polygon.
In summary, every three-sided polygon has an interior angle sum of 180°. For
every side more than 3, 180 degrees is added to the sum. Thus, the interior angle sum of
a polygon with n sides is 180°(n-2).
E
F
G
D
C
B
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Shown here is an example. Each triangle has
180 degrees and the number of triangles that can
be drawn is (n-2) because from any vertex, you
can draw diagonals to every other vertex except
the two adjacent ones.
A
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Diagonals of a Polygon
The number of diagonals in a polygon can be found by using the formula
n (n-3)/2 where n is the number of sides of the polygon. For example, say you had a
pentagon. Then, n=5 so when you plug it into the formula, you get 5*2/2 = 5. According
to the formula, a pentagon should have 5 diagonals. As shown below, a pentagon really
does have 5 diagonals.
Now the reason for the formula is because when you draw a diagonal, you are
merely choosing 2 points and connecting them. So, in essence, drawing all the diagonals
and sides consists of taking all the vertices and connecting them in every way.
This is the same as n choose 2 (explained in the combinations and permutations section).
The formula for n choose 2 is n (n-1)/2 which looks similar to our formula above but
when you do n choose 2, you are including the sides so in order to get the number of
diagonals, you have to subtract n and you have n (n-1)/2 –n which when multiplied out
becomes
(n2 – n –2n)/2 which is equal to (n2-3n)/2 which is the formula for the number of
diagonals of a polygon.
Alternative explanation: Given an n-gon. From each vertex you can draw (n-3)
diagonals (excluding itself and its two adjacent vertices). This would yield n(n-3)
diagonals. However, each diagonal has been counted twice, once from each endpoint.
Therefore, divide by 2 to get the result: n(n-3)/2, same as above.
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Completing the Square
( a + b )2 = a2 + 2ab + b2
The method of completing the square is a technique used in a variety of problems to
change the appearance of quadratic expressions.
The basic idea is to make a perfect binomial square show up by manipulating the
original expression.
The method is based on the simple observation that, while x2+10x is not a perfect square,
x2 + 10x + 25 is. "Completing the square" means putting in the missing 25 that is needed
to form the perfect square.
Two obvious concerns are "How do you know you need 25?" and "How can you just add
25 any place you want?"
Here are the steps used:
Start with a quadratic expression like x2+10x
Square half of the linear coefficient 10. That would be 52=25.
Simultaneously add and subtract 25 and factor the perfect square.
So x2 + 10x = x2 + 10x + 25 - 25 = (x+5) 2-25.
Note that the constant in the binomial turns out to be half of the linear coefficient.
That's the main idea. Of course there are some variations to encounter in different types
of problems.
Here are three more examples of completing the square. Make sure you understand the
steps involved. Look especially at Example 3 in which the leading coefficient is not one.
Example 1
x2 + 8x
x2 + 8x + 16 – 16
(x + 4) 2 - 16
Example 2
x2 - 3x + 1
x2 - 3x + 9/4 – 9/4 + 1
(x – 3/2) 2 - 5/4 by
since (8/2) 2 = 42 = 16
by factoring x2 + 8x + 16
since (-3/2) 2 = 9/4
factoring x2 - 3x + 9/4
Example 3
3x2 + 5x + 6
3(x2 + (5/3)x) + 6
3(x2 + (5/3)x + 25/26 – 25/26) + 6
3(x2 + (5/3)x + 25/26) – 25/12 + 6
3(x + 5/6) 2 + 47/12
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factor leading coefficient 3 from first 2 terms
since (5/6) 2 = 25/26
separating –25/26 from the parentheses
by factoring x2 + (5/3)x + 25/36
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Factoring
Factoring has confounded mathematicians since the inception of number theory.
Specifically, they have endlessly tried to determine an effective method for factoring
large numbers into prime components. However, it may seem unclear what practical
value such factorization would produce. Quite simply, many number theoretical theorems
depend upon the ability to factor numbers into primes. Without such information, they
would become merely intellectual exercises. Both of the following theorems, popular in
mathematical contests, require prime factorization. Keep this in mind.
r
r
r
THEOREM: Let n be any positive integer. Also, let n  p1 1 p2 2  pk k , where
p1 , p 2 ,, p k are the different prime factors of n. Then, d(n), the number of divisors in n,
is determined by
d (n)  (r1  1)( r2  1) (rk  1)
Most mathematical theorems are written in this format. Its conciseness helps for
reference. See if you can understand it. In plain language, this theorem says that the
number of divisors in some number can be determined by factoring it into its prime
components, adding one to their powers, and multiplying them together.
Example: Find the number of divisors of 24 by hand and through the theorem.
Solution: By hand, 24 has the following divisors: 1, 2, 3, 4, 6, 8, 12, 24, making 8. By the
theorem, since 24  2 3  3 1, (3  1)(1  1)  4  2  8 .
The justification for this rests in the multiplicative nature of d(n); that is, d(ab) = d(a)
d(b).
Another theorem is closely related to this.
r
r
r
THEOREM: Let n be any positive integer. Also, let n  p1 1 p2 2  pk k , where
p1 , p 2 ,, p k are the different prime factors of n. Then, (n), the sum of all the positive
divisors of n, is determined by
 p r1 1  1  p 2 r2 1  1   p1 rk 1  1 

 

 (n)   1
 p  1   p  1 
p

1
1
2
k


 

Do any of the terms in this product look familiar? They are all reminiscent of the sum of
a geometric series. This is no coincidence, as shown in the following example.
Example: Find the sum of the divisors of 24 by hand and through the theorem.
Solution: 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60.
Through the theorem,
 2 4  1  32  1 

  15  4  60 . Interestingly, though, observe that 1 + 2 + 3 + 4
 (24)  
 2  1  3  1 
+ 6 + 8 + 12 + 24 can also be written as
 2 4  1  32  1 

 .
1  2  3  2 2  2  3  2 3  2 2  3  2 3  3  (1  2  2 2  2 3 )(1  3)  
 2  1  3  1 
Of course, this is not a proof by any means, but it should strengthen your belief
that the theorem holds true.
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Greatest Common Factor
The greatest common divisor, abbreviated GCD, of two positive integers is the
largest number that divides both of them (GCD is at times depicted as GCF, the
greatest common factor). For example, the GCD of 8 and 12 is 4. Notationally, the
GCD of two numbers is expressed with parentheses, such as (8, 12) = 4. Do not confuse
this with ordered pairs. For small numbers, finding the GCD does not pose very large of a
problem. Expressing the two numbers in their prime factors and choosing the common
factors suffices to determine the GCD. For example, to find the GCD of 16 and 1000, one
expresses 16 as 24 and 1000 = 103 = (5*2)3 = 53*23. The number 23 is common to both
numbers, so 23 = 8 is the GCD.
Example: Find the GCD of 1800 and 168.
Solution: 1800 = 18*100 = 6*3*52*22 = 3*2*3*52*22 = 23*32*52.
168 = 2*84 = 22*42 = 23*21 = 23*3*7. 23*3 = 24 is common to both numbers, so 24 is
the GCD.
With large numbers, finding the GCD becomes considerably difficult using the prime
factor method. However, a method called the Euclidian algorithm circumvents
determining the factors. The application is not very hard, but the proof requires
knowledge of mathematical induction and some experience. If desired, any number
theory book should explain all the steps of the proof. The Euclidian algorithm involves
dividing numbers, finding their remainders, taking the result, and reapplying the method.
Though hard to explain, an example clearly illustrates this process.
Example: Find the GCD of 12345 and 54321.
Solution: First, divide the larger (the dividend) by the smaller (the divisor) and look at the
remainder. Here, 54321 = 4*12345 + 4941. Then, express the divisor in terms of the
remainder. Here, 12345 = 2*4941 + 2463. Repeat the process.
4941 = 2*2463 + 15
2463 = 164*15 + 3
15 = 5*3 + 0.
When the remainder becomes 0, the GCD is the last quotient, 3 in this case. When two
numbers have a GCD of 1, they are called relatively prime. The Euclidian algorithm is
the quintessential algorithm, a series of steps that, when performed a finite number of
times, yields an answer. Usually, algorithms are used heavily in computer programs,
where calculations can be accomplished very quickly.
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Related to the GCD is the LCM, or the least common multiple. You may know it as the
least common denominator when adding fractions. Typically, the LCM can be calculated
by factoring both numbers into primes. The notation for LCM is [a, b]. Take all the
factors that are either in a or b, and multiply them together. For example, to determine [6,
8], express 6 as 2*3 and 8 as 23. Then, [6, 8] = 23*3 = 8*3 = 24. this is difficult with
large numbers. However, a relationship exists between GCDs and LCMs: (a, b)*[a, b] =
ab. Convince yourself that this is true. (a, b) contains the factors common to both a and b,
ab
whereas [a, b] contains the rest. Written in another form, [a, b] 
.
( a, b)
Example: Calculate the LCM of 3542 and 6783.
Solution: (3542, 6783) = 7, so [3542, 6783] = 3542*6783 / 7 = 3432198.
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Geometric Series
A geometric sequence is a sequence in which each term differs from the previous
by a constant ratio. That is, for a sequence a0r = a1, a1r = a2, … a n+1 = anr.
Explicitly, an = a0rn. (Convince yourself that this works)
Example: John has 1000 dollars. He places it in a bank that returns 5% interest yearly.
How much money will John have in 5 years? (to the nearest cent)
a0 = 1000, the initial principal. The goal is to find a5, which equals a0r5 = 1000r5. The
interest is 5%, which means the money gains 0.05 of its value each year. The amount at
the end of each year is computed if the amount each year is multiplied by 1.05. This is
because 5% is equal to 5/100 = 0.05 so you’re adding the original amount (1) plus the
interest 0.05 all multiplied by the original amount. Thus, r = 1.05, and a5 = 1000(1.05)5 =
1276.28…
A geometric series is the summation of a geometric sequence, such as a0 + a1 + a2 … = a0
+ a0r + a0r2 + … Through clever symbol manipulation, a general formula can be
determined for such a series.
Let Sn = a0 + a0r + a0r2 + … + aorn. rSn = a0r + a0r2 + … + a0r n+1
Subtract rSn from Sn.
Sn = a0 + a0r + a0r2 + … + a0rn
-rSn =
-a0r – a0r2 -…….- a0rn – a0r n+1
Sn-rSn= a0
-a0r n+1
Sn-rSn = a0 – a0r n+1
Sn(1-r) = a0 (1-r n+1)
 1  r n 1 

S n  a 0 
 1 r 
Admittedly, this formula appears awkward. Nonetheless, practice should help you
memorize it. This formula works even for r < 0.
Example: A runner starts by running 3 miles on Sunday. He increases his distance by
10% every day. How many miles will he have run by the end of Saturday?
a0 = 3 and r = 1.1, clearly, n = 7, the number of days he has run.
 1  1.18 
  34.307
S 7  3
1

1
.
1


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Infinite Geometric Series
An infinite geometric series is one in which there are an infinite number of terms.
Basically, it is a geometric series (explained above) that doesn’t end.
Obviously, if |r| > 1, calculating our infinite geometric series will be a hopeless endeavor.
However, if |r| < 1, should the series still be meaningless if the terms keep decreasing?
Examining the formula for summation provides the answer.
 1  r n 1 

S n  a 0 
 1 r 
As n grows, r n+1 shrinks, if |r| < 1. (Convince yourself). Indeed rn+1 goes towards 0, even
if it never reaches there! Committing a technical symbolic error (don’t tell your teacher),
we can write
1  r 
S   a 0 
 1 r

 1  0  a0
  a 0 

1 r  1 r

Weird, but true.
Example: The Greek philosopher Zenos once stated the following dilemma.
Say a person is running a marathon. Once she reached the halfway point, she would have
to reach the halfway point of the remaining distance, and then the halfway point of the
remainder, and so forth. How is it possible she will reach the end?
Solution: Although there are an infinite number of points, the total distance is finite. Let
the entire length of the route equal 1 unit. She runs ½ unit initially, so let a0 = ½. The
length to each point is halved each time, resulting in the infinite geometric series ½ + ½
(1/2) + ½ (1/2)2 + … Using the formula previously derived, such a sum equals
 1 
 1 
1 / 2
  1 / 2
  1/ 2  2  1
1  1/ 2 
1/ 2 
just as said earlier.
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Permutation and Combination
Permutation and combination are methods that are used to find how many ways
something can be done. The formal definitions are:
A permutation: an arrangement of a group of objects in a definite order.
A combination: a group of objects in which the order or arrangement is not
considered.
Permutations of Objects that are all Different
The number of permutations of 7 things taken 4 at a time is represented by 7P4.
Following the same form, the number of permutations of n things taken r at a time is
represented by nPr. Suppose you had 4 digits and you wished to make as many 4 digit
numbers as possible; how many could you make? Well, since there are 4 possibilities
for the first digit, 3 for the second, etc. the answer is 4*3*2*1 = 24 = 4P4. Assuming that
the digits were 1, 2, 3, and 4, this is how it would be shown diagrammatically:
3
4
2
4
2
3
2
1
3
4
1
2
3
4
3
4
1
4
1
3
1
3
2
4
2
4
1
4
1
2
2
3
1
3
1
2
1
4
2
3
As you can see, there are 24 different numbers. Following the reasoning above, we can
conclude that
P = n(n-1)(n-2) … (n-r+1)
there are always r factors.
n r
Along the same line:
P = n(n-1) … 2*1 = n!
n n
This applies only when each of the objects being chosen are different and each object can
only occupy one space.
So, for example, if you had 4 terms and you wanted to pick 2 with order as a factor, it
would be:
4P2 = 4*3 = 12
2
1
3
4
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Other Permutations
If each object could be used multiple times, then, in the above example with the
4-digit numbers, the solution would be 4*4*4*4 = 256 because then each of the slots can
have each of the 4 digits filling it.
There is also a different twist to the problem if some of the objects are the same.
For example, if you wanted to find the total number of permutations of the word Book, it
wouldn’t just be 4P4 because if you switched two of the o’s around, the permutation
would still be the same. What the answer actually becomes is
4
p4
2!
which is because there are 2 o’s.
The number of permutations of n things taken n at a time with r things alike, s
things alike, and so on is:
N!
(r! s!...)
Another aspect of permutations is circular permutations. A circular permutation
is a way in which a number of different things can be arranged about a circle or any other
closed curve. The difference between linear permutations and circular permutations is
that when it is arranged around a closed curve, the starting place doesn’t matter so if you
have n objects and the number of linear permutations taking n at a time would be n! but
the number of circular permutations would be n! / n because then the first place no longer
applies. For example, suppose you had 4 people sitting around a circular table.
A
D
B
D
C
C
A
C
B
D
B
B
A
C
A
D
All 4 of the above are the same permutation because with respect to the table, everyone is
in the same relative position. That is the reason you divide by n when dealing with
circular permutations.
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Combinations
As stated previously, a combination is a group of objects in which order does not
matter. To make it clearer, the different permutations ABC, CBA, and BCA are all the
same combination.
C = the number of combinations of n things taken r at a time.
n r
Because for each combination of r objects there are r! permutations, r!nCr = nPr. Then,
since nPr = n(n-1)(n-2) … (n-r+1), by substituting, we can arrive at
n
Cr = n(n-1)(n-2) … (n-r+1) / r!
Another way to look at combinations is that nCr is the number of ways to take r objects
from a group of n. If you take r, then you leave n-r. So what remains is the number of
ways to take n-r objects from a group of n or nCn-r.
For example, if you wanted to choose 2 objects from 5, 5C2 = (5*4) / 2! = 10. You can
also see that 5C3 = (5*4*3) / 3! = 10. They are the same, but is that coincidence?
Actually, if you write out the formulas for nCr and nCn-r you get:
Cr = n(n-1)(n-2) … (n-r+1) / r!
C = n(n-1)(n-2) … (r+1) / (n-r)!
n n-r
n
Now to prove that those are equal, multiply the nCr equation by (n-r)! / (n-r)! to result in
n! / (r!(n-r)!). Multiply the nCn-r equation by r! / r! to get the same result, which proves
that nCr = nC n-r and it also gives us the formula for nCr.
n
Cr = n! / (r!(n-r)!)
The total number of combinations of n things means to find the total of the number of
combinations taking things 0 at a time, 1 at a time, 2 at a time … n at a time. In notation,
this means
Total number of combinations = nC0 + nC1 … nCn.
n
It just so happens that the total number of combinations of n objects is 2 . This can be
proven but it may be a bit complicated so for now it is just a handy fact to keep in mind.
It is related as well to Pascal’s Triangle, an extremely deep topic with numerous
interesting properties.
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Probability
Probability is the ratio of the number of favorable events to the total number of
possible events. Probability is often referred to as chance, the chance of an event
occurring. As such, the probability of an event can never be less than 0 because there can
never be a negative number of events. Also, the probability can never be greater than 1
because the number of favorable events cannot exceed the total number of events.
Example 1: Suppose a bag contains 7 white balls and 4 blue balls. What is the
probability that, drawing one ball at random, you’ll get a blue ball?
Solution: Well, following the definition, you consider the number of favorable events. In
this case, you see that there are 4 blue balls, so the number of favorable is 4. Then, you
see that there is a total of 11 balls. So, the probability is 4 out of 11 or 4/11.
Example 2: Assume you have a bucket containing 5 slips of red paper and 6 slips of
green paper. What is the probability that, drawing 2 slips of paper at random, you will
get one red and one green?
Solution 2: There are 2 methods of approaching a problem of this type. First, you can
view it as a single probability. Then, you would say that there are a total of 11 choose 2
possible events (choose, meaning combination, is explained in the section of the guide
titled permutation and combination). So, you do (11*10)/(2*1) = 55. Then, you say that
there are 6 ways of drawing a green and 5 ways of drawing a red so there are 6*5 = 30
favorable cases. So, the answer you arrive at is 30/55 or 6/11.
The second method is to consider the problem as 2 separate probabilities. What is
the probability of drawing a green on the first draw? Well, following example 1, you find
that it is 6/11. After that, there are 10 slips of paper left, 5 of which are red so the
probability of drawing a red on the second draw would be 5/10 or ½. If
You multiply because you want A to occur and then B to occur in that order. 6/11*1/2 =
3/11. Then, because you can also draw the red slip first and the green second, and that
also has a probability of 3/11, you add the 2 probabilities 3/11 + 3/11 and the final result
is 6/11, same as the first method.
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With probability, there are numerous different twists and turns that questions can take.
For example, in example 2, instead of asking what the probability of drawing a red and a
green in 2 draws is, what if you now had 3 draws? Or, what if in example 3, instead of
coins, you were rolling die. Then, the question might be, what is the probability of
obtaining a sum greater than 18? The key is to think the problem through logically so
you can determine exactly what data is necessary and what to do about it.
Odds are another aspect of probability that ought to be kept in mind. Odds are the
number of favorable cases versus the number of unfavorable cases or, the probability of
the occurrence of the favorable event versus the probability of the occurrence of the
unfavorable event. For example, if you had 7 blue balls and 5 red ones, the odds of
drawing a blue ball are 7 to 5. In reality, it is (7/12)/(5/12) and the 12’s just cancel out.
You could also say that the odds of not drawing a red ball is 7 to 5.
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Divisibility Rules
Divisibility rules come in handy when you want to factor something quickly.
There isn’t really much explanation that is required to go along with them so I will
merely state them and give examples. However, for the extremely curious, these rules
can be proven by Number Theory so feel free to look into that. Some will be extremely
obvious and everyone will know them but others will be obscure. So everyone should be
able to learn something.
Divisibility Rule for 2 = any even number is divisible by 2.
Divisibility Rule for 3 = if the sum of all the digits in the number is divisible by 3, then
the number is divisible by 3.
Divisibility Rule for 4 = if the 2-digit number that is made up of the last 2 digits of the
number are divisible by 4, then the number is divisible by 4.
Divisibility Rule for 5 = if the units digit of the number is 5 or 0, then the number is
divisible by 5.
Divisibility Rule for 6 = just test both 2 and 3 and if the number is divisible by both, then
it is divisible by 6.
Divisibility Rule for 7 = this one is pretty complicated. Starting from the units digit and
moving left, you group the digits into groups of 3. Then, you take all the odd
groups and subtract all the even groups or vice versa and then if the resulting
number is divisible by 7, then the original number is divisible by 7. If the number
does not have a multiple of 3 number of digits, that’s ok, just pretend there are
zeroes added on to the left of the number.
Divisibility Rule for 8 = if the 3-digit number that is made up of the last 3 digits of the
number are divisible by 8, then the number is divisible by 8.
Divisibility Rule for 9 = if the sum of the digits is divisible by 9, then the number is
divisible by 9.
Divisibility Rule for 10 = if the units digit of the number is 0, then the number is divisible
by 10.
Divisibility Rule for 11 = if the sum of the even placed digits minus the sum of the odd
placed digits is a multiple of 11 or vice versa, then the number is divisible by 11.
Divisibility Rule for 12 = if it is divisible by 3 and 4, then it is divisible by 12.
Divisibility Rule for 13 = same as the divisibility rule for 7 except you divide the result of
the odd groups minus the even groups (or vice versa) by 13 instead of 7.
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Examples:
Example of 3: Let’s say your number was 137568. So, following the rule, you add the
digits 1+3+7+5+6+8 = 30. Since 30 is divisible by 3, you know that 137568 is as well. 9
follows the same process.
Example of 4: To test the number 107184, you take the last two digits so you have the
number 84. Since 84 is divisible by 4, 107184 is as well. The divisibility rule for 8
follows the same process except with the last 3 digits.
Example of 7: If your number is 1723575, what you do is you split it into groups of 3 so
what you have is 1 | 723 | 575. Now, it is the odd groups minus the even groups or vice
versa so you can do 723-575-1 = 147. Since 147 is a multiple of 7 (147 / 7 = 21), you
know 1723575 is a multiple of 7. 13 follows the same process.
Example of 11: Assume the number in question is 93159. It’s the odd placed digits
minus the even placed digits or vice versa so you can do 9 - 3 + 1 – 5 + 9 = 11. Since 11
is a multiple of 11, we know that 93159 is a multiple of 11.
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Possible Paths
The following is a useful method to employ when encountering a problem of this
specific type. I think the best way to teach this method is by example so here we go.
Problem: How many different paths are there from point A to point B, provided that you
can only move downwards?
A
B
Solution: Begin at point A and give it a value of 1. Then, for every following
intersection, label it the sum of the values above it that connect to it. For example:
3
2
2
1
C
D
C’s value would be 2 + 3 + 2 = 7 because those are the lines that are above it and connect
to it. Likewise, D’s value would be 2 + 1 = 3. You begin with point A and just use this
method throughout the entire diagram and your result, the value at point B, is the number
of possible paths from point A to point B.
A
1
1
1
1
1
2
1
4
1
4
5
1
10
7
15
5
7
15
27
30
72
1
27
72
B 174
So, there are a total of 174 possible paths from A to B =)
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Checkerboards Squares/Rectangles
How many squares can be formed from:
The answer is 1.
Well, how many squares can be formed from:
How many of you answered 5?
Well, the correct answer is 6.
There are 4
small ones
4
There is one
big one
+
1
There is one
diagonal one
+
1
=
6
Well, now what if there are 16 dots?
How many squares are there in this?
Well, let’s see, shall we?
There are 9
small squares
There are 4 diagonal
squares that are 2 by 2
There are 4 squares that
are 2 by 2 (only 1 shown)
There is 1
square 3 by 3
There are 2 other different diagonal
squares (only 1 shown for simplicity)
So, 9 + 4 + 1 + 4 + 2 = 20. There are 20 different squares in 16 dots.
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Ok, so the sequence is 1, 6, 20… Does anyone see a pattern?
Let’s take a look at Pascal’s triangle.
1
1
1
1
1
1
1
1
7
2
3
4
5
6
1
3
6
10
15
21
1
4
1
10
20
30
1
5
15
30
1
6
21
1
7
1
If you take a look at the circles, you’ll note that the first circle is around 1, the second is
around 1 and 5, and the last circle is around 5 and 15. 1 + 5 = 6 and 5 + 15 = 20,
fascinating, isn’t it?
So, in order to find the number of squares, all you have to do is find that diagonal in
Pascal’s triangle. If you had 25 dots, then there would be 15 + 30 =45 squares and so
forth.
This sequence can also be represented by the formula:
n
9 2 35
x  x  18
2
2
where n is the number of squares and x is the number of dots that comprise one side of
the largest square.
Another point to note is that there is another pattern if you disregard the diagonal squares
(you only count the squares whose sides would be drawn if there was a cartesian grid).
Then, for 4 dots, there would still be only 1, but for 9 dots, instead of 6, there would only
be 5 and for 16 dots, instead of 20, there would only be 14. Now, look at the diagonal in
Pascal’s triangle right above the one we were just using and you’ll see that
1 = 1, 1 + 4 = 5, and 4 + 10 = 14 so if you disregard the diagonal squares, you could find
the number of squares by applying the same procedure to the diagonal above.
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Rectangles:
There exists an interesting method for finding the number of rectangles given
vertical and horizontal lines (rectangles whose sides are formed by the lines, no diagonal
ones). For example, say you had 5 vertical lines and 4 horizontal lines.
The method involves using combination. To find the number of rectangles, all you have
to do is:
(Number of vertical lines choose 2) multiplied by (Number of horizontal lines choose 2)
In this case, it would be:
(5 choose 2) x (4 choose 2) = 10 x 6 = 60 rectangles.
If you’d like to test this for correctness, I’d advise choosing a simple diagram,
applying this method, and then counting to check.
The reason this method works is because if you think about it, a rectangle is
composed of 2 vertical lines and 2 horizontal lines. So, by doing (number of vertical
lines choose 2), you’re getting all the different combinations of the 2 vertical lines that
could possibly form the vertical sides of the rectangle. You do the same with the
horizontal sides and then when you multiply them, that gives you the total number of
possibilities of different rectangles.
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Factors of Factorials Technique
The Factors of Factorials Technique is another handy trick that comes in useful at times
on math competitions. It can be shown by example better than anything else so that is the
method I’ll use to explain it.
Example: If 7n is a factor of 400!, what is the largest n?
Well, what you would do is “scan”.
(1) You divide 400 by 7 (rounded down) to get the number of numbers in 400! that are
divisible by 7 and that is 57.
(2) Then, you take the quotient, 57, and divide by 7 (rounded down) to get the number of
numbers in 400! That are divisible by 49 since those are the ones with at least 72 as
factors. 57 / 7 = 8.
(3) Then, take the quotient, 8, and divide that by 7 rounded down to get 1 and that is the
number of numbers in 400! that contain 73.
(4) So, to find n, all you have to do is add up the numbers, 57+8+1 = 66. There are 66
sevens in 400!.
Now, if you were trying to find the number of 10s in 400!, or any other number that is
composite, you would follow a slightly different process.
(1) Since 10 is 5*2, first you must decide which factor there will be less of. In this case,
it is fairly obvious to know that there will be more factors of 2 than 5 in 400!. So, all
you need to do is scan for 5’s.
(2) Following the same procedure as the previous problem, we get (400 / 5 = 80), (80 / 5
= 16), and (16 / 5 = 3 ) => 80 + 16 + 3 = 99. Since you know that for every 5 there is
at least a 2, you know that there are 99 tens in 400!
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Interesting Formulas
For all formulas given below: an = nth term
r = ratio
d = common difference
n = number of terms
An arithmetic sequence is a sequence of numbers in which there is a constant difference
between each term and its following term. Meaning that a n+1 = an + d
A geometric sequence is one in which each number differs from the following number by
a common ratio. Meaning that a n+1 = an * r
(1)
Nth term of an arithmetic sequence: an = a1 + d(n-1)
(2)
Sum of first n terms of an arithmetic sequence: sum = n (2a1 + d(n-1))/2
(3)
Using the formula given in (2), we can figure out that if you want to add 1 + 2 +
3… n, the formula is: sum = n(n+1)/2. This is because in a sequence like 1 + 2 +
3… the difference is 1 and the first term is 1 so your 2a1 + d(n-1) term simplifies
to 2 + n – 1 = n + 1.
(4)
Nth term of a geometric sequence: an = a1 * r n-1
(5)
Sum of first n terms of a geometric sequence: sum = a1(1-rn)/(1-r)
(6)
As n nears infinity, if r > 1, then the sum goes to infinity as well. However, if r <
1, then the formula for the sum of the terms of an infinite geometric series is:
a1/(1-r) = sum
(7)
There is a special formula for the sum of a sequence consisting of 12 + 22 + 32 …
and that is: n(n+1)(2n+1)/6 = sum of first n terms.
(8)
For 13 + 23 + 33 … the formula is (the square of the formula for 1 + 2 + 3…):
(
(9)
n(n  1) 2
)  sum
2
The formula for the distance from a point (c , d) to a line ex + fy + g = 0 where e,
f, and g are constants that define the line is:
ec  fd  g
e2  f 2
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JHMC Guide Table of Contents
Title:
Page Number(s):
Author(s):
Work/Rate/Speed
1-3
Cindy Xi
Interest
4-5
Cindy Xi
Triangle Tips
6–9
Lucy Lu and Cindy Xi
Shoelace
10 – 11
Cindy Xi
Circles
12
Lucy Lu
Interior Angles
13
Lucy Lu
Diagonals of a Polygon
14
Denny Tu
Completing the Square
15 – 16
Lucy Lu
Factoring
18
Eric Syu
Greatest Common Factor
19
Eric Syu
Geometric Series
20 – 21
Eric Syu
Combination & Permutation
22 – 24
Denny Tu
Probability
25 – 26
Denny Tu
Divisibility
27 – 28
Denny Tu
Possible Paths
29
Denny Tu
Checkerboards
30 – 32
Denny Tu
Factors of Factorials
33
Denny Tu
Interesting Formulas
34
Denny Tu
Authors: Lucy Lu, Tao Luo, Eric Syu, Denny Tu, and Cindy Xi
Inspired by Cindy Xi
Editors: Mr. Ron Vavrinek and Denny Tu
Produced by Denny Tu
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Please send any comments,
suggestions, or questions to
[email protected] or Denny Tu at
IMSA. Thank You again, best of
luck with the competition! =)
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