1. Formulae

1.1 Geometry

Squares and Rectangles

  1. Area of Square = \(a^{2}=a \times a\)

Example:

a=4

\(a=4\)

\(A = a^{2} = a \times a = 4^{2} = 4 \times 4 = 16\)
  1. Perimeter of a Square = \(2(a + a) = 2(2a)=4a\)

Example:

a=12

\(a=12\)

\(P = 4a = 4 \times 12 = 48\)
  1. Area of a Rectangle = \(lw\)

Example:

l=4 \quad w=8

\(l=4 \quad w=8\)

\(A = lw = 4 \times 8 = 32\)
  1. Perimeter of a Rectangle = \(2(l+w)\)

Example:

l=4 \quad w=8

\(l=4 \quad w=8\)

\(P = 2(l+w)=2(4 + 8) = 2(12) = 24\)

Triangles

  1. Area of a Triangle = \(\frac{1}{2}bh\)

Example:

b=5 \quad h=12

\(b=5 \quad h=12\)

\(A = \frac{1}{2}bh = \frac{1}{2}(5)(12) = \frac{1}{2}(60) = 30\)

  1. Perimeter of a Triangle = \(P = a + b + c\)

Example:

a=3 \quad b = 3 \quad c = 3

\(a=3 \quad b = 3 \quad c = 3\)

\(P = a + b + c = 3 + 3 + 3 = 9\)
  1. The Pythagorean Theorem - Given a right triangle (a triangle that contains an angle of 90 degrees), the hypotenuse - the opposite of the 90-degree angle - can be found by the formula \(a^{2} + b^{2} = c^{2}\).

Example:

a=3 \quad b=4

\(a=3 \quad b=4\)

To find the hypotenuse \(c\), we can apply the Pythogorean Theorem as follows

\[ \begin{aligned} & a^{2} + b^{2} = c^{2} \\ & 3^{2} + 4^{2} = c^{2} = 9 + 16 \\ & c^{2} = 25 \\ & \sqrt{c^{2}} = \sqrt{25} = 5 \\ & c = 5 \end{aligned} \]

Circles

  1. Area of a Circle = \(\pi r^{2}, \qquad \text{where } r \text{ is the radius of the circle}\)

Example:

r=5

\(r=5\)

\(A = \pi r^{2} = \pi5^{2}= 78.53982\)
  1. Circumference of a Circle = \(C = 2\pi r, \qquad \text{where } r \text{ is the radius}\)

Example:

r=9

\(r=9\)

\(C=2\pi r = 2\pi9 = 56.54867\)

1.2 Arithmetic and Algebra

  1. Arithmetic mean (average) = \(\mu = \frac{1}{N}\sum_{i=1}^{N}x_{i}\)

Example:

\[ \begin{aligned} & x=\{1,5,6,7,13,21\} \quad N=6 \\ & \mu_{x} = \frac{1}{6}\sum_{i=1}^{6}x_{i}=\frac{1}{6}(1+5+6+7+13+21)=8.8\bar{3} \end{aligned} \]
  1. Combination = \(nCr=\) \({n}\choose{r}\) \(=\frac{n!}{r!(n-r)!}\)

Example:

How many possible combinations can be formed by choosing 3 people from 8 people?

\[ \begin{aligned} nCr = \binom{n}{r}=\binom{8}{3} = \frac{8!}{3!(8-3)!}=\frac{8!}{3!5!}=\frac{8\times7\times6}{3\times2\times1}=56 \end{aligned} \]
  1. Finding the Location of the Median = \(\frac{N+1}{2}\). If the resulting dividend is an integer, then that is the location of the median. If it reults in a number ending in 0.5 (5.5 for example), then the median is located between \(x_{5}\) and \(x_{6}\).

Example:

\[ \begin{aligned} & x \text{ is a series of 72 integers. Where is the median?} \\ & \frac{72}{2} = 36 \end{aligned} \]

The median is located at \(x_{36}\)

  1. The Distance or Length of a Line Segment on an Cartesian Coordinate System (xy-coordinate plane) can be calculated using the Euclidean Distance Formula = \(\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}\)

  2. The sum of a consecutive series of integers is \(A = \{x_{1}, x_{2}, x_{3}, \ldots, x_{n}\} = \frac{N(N+1)}{2}\) where \(N\) is the number of terms in the series.

Example:

\[ \begin{aligned} & A = \{1,2,3,4,5,6,7,8,9 \} \quad N=9 \\ & \sum A = \frac{N(N+1)}{2} = \frac{9(9+1)}{2} = \frac{9(10)}{2} = 45 \end{aligned} \]
  1. The sum of all even integers in a consecutive series is \(N(N+1)\) where \(N = \frac{\text{First even integer + Last even integer}}{2} - 1\)

Example:

\[ \begin{aligned} & A = \{1,2,3,4,5,6,7,8, \ldots, 21\} \qquad N = \frac{2 + 20}{2} - 1=10 \\ & \sum A= N(N+1) = 10(11) = 110 \end{aligned} \]
  1. If \(x\) is a repeating decimal (e.g. 0.333333), you can convert any repeating decimal by:
  2. Let \(x\) represent the repeating decimal. Multiply \(x\) by \(10\).
  3. Subtract by \(x\).
  4. Solve for \(x\).

Example:

\[ \begin{aligned} & x = 0.777 \\ & 10x = 7.777 \\ & 10x - x = 7.777 - 0.777 = 7 \\ & 9x = 7 \\ & x = \frac{7}{9} \end{aligned} \]
  1. An expression is quadratic if it is in the form \(ax^{2} + bx + c\). Setting the form equal to zero gives you a quadratic equation \(ax^{2} + bx + c = 0\). To find the roots of a quadratic equation you can either use the quadratic forumula or simply “guess” using trial and error. In general, the trial and error method is usually the quickest if you can find divisors of \(c\) that sum to \(b\). However, if you can’t, then you’ll have to use the quadratic formula to find the roots. The quadratic formula is given by

\[ \begin{aligned} \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \end{aligned} \]

Example:

Find the roots of \(x^{2} + 2x - 15\)

For the trial and error method, the divisors of \(15\) are \(\{1, 3, 5, 15\}\). Which of the divisors, when added together, sum to \(2\) AND result in a product of \(c=-15\)? The answer is \(5\) and \(-3\), since \(5 - 3 = 2\) and \(5 \times -3 = -15\). Therefore, the roots are \(x = 5\) and \(x = -3\). Factoring the equation yields \((x + 5)(x - 3)\). To check your solution, multiply the expression using the acronym FOIL. FOIL is the order in which you multiply the expression, and it stands for first, outer, inner, and last. That is, multiply each term in both paranthesis in that order. Continuing with our example, you would multiply in the following order

\[ \begin{aligned} & 1. \quad x \times x = x^{2} \\ & 2. \quad x \times -3 = -3x \\ & 3. \quad 5 \times x = 5x \\ & 4. \quad 5 \times -3 = -15 \end{aligned} \]

This yields \(x^{2} -3x + 5x - 15 = x^{2} + 2x - 15\).

2. Rules and Properties

2.1 Geometry

Special Triangles

  1. Right Triangles with Whole Number Solutions (Pythagorean Triples)

    • 3-4-5
    • 5-12-13
    • 8-15-17

  2. 30-60-90 Triangle

Other

  1. Angles on a straight line always sum to 180 degrees.
  1. The radius of a circle inscribed in a square is equal to the side of the square.

Example:

If the side of the square is 10, then the diameter of the circle is also 10.
  1. A tangent line to a circle forms a \(90^{\circ}\) angle where the radius intersects the tangent line, as shown below

2.2 Arithmetic, Algebra, and Probability & Statistics

Consecutive Series of Integers

Consecutive integers are a series of integers that follow one another and have a difference of 1 between every two numbers. \(\{1,2,3,4,5,\ldots, x_{n}\}\), \(\{2,4,6,8,10, \ldots, x_{n}\}\) are two examples of consecutive integers.

  1. If you have a series of consecutive integers, the mean is equal to the median.

Example:

\[ \begin{aligned} & X = \{4, 8, 12, 16, 20\} \\ & \bar{X} = \frac{4+8+12+16+20}{5} = 12 \\ & \text{Median = } 12 \end{aligned} \]
  1. Any number \(x\) raised to a negative power is the reciprocal of \(x\).

Example:

\[ \begin{aligned} &x = \frac{2^{-2}}{4^{-2}} \\ & = \frac{4^{2}}{2^{2}} = \frac{16}{4} = 4 \end{aligned} \]
  1. Given a Gaussian (Normal Distribution) with mean \(\mu\) and standard deviation \(\sigma\),
    • 68% of the data lie between \(\mu \pm \sigma\) of the distribution.
    • 95% of the data lie between \(\mu \pm 2\sigma\) of the distribution.
    • 99.7% of the data lie between \(\mu \pm 3\sigma\) of the distribution.
  2. Arithmetic Rules
    • Even + Even = Even
    • Even + Odd = Odd
    • Odd + Odd = Even

  3. Given a line \(m\), the slope of a perpendicular line \(k\) is the negative reciprocal of line \(m\).

Example:

\[ \begin{aligned} & \text{Line } k \text{ given by the following equation: }y = 2x + 5 \qquad \text{The slope of the line is 2} \\ & \text{Line } m \text{ is perpendicular to line } k \\ & \text{The slope of the perpendicular line } k = -\frac{1}{2} \end{aligned} \]
  1. If a line passes two points on an xy-coordinate plane, its slope can be obtained by the formula

\[ \begin{aligned} \frac{∆y}{∆x}=\frac{(y_{2}-y_{1})}{(x_{2}-x_{1})} \end{aligned} \]

Example:

Line \(k\) passes through \((1,1)\) and \((5,2)\). What is the slope of line \(k\)?

\[ \begin{aligned} \text{Slope = } \frac{∆y}{∆x} = \frac{2 - 1}{5 - 1} = \frac{1}{4} \end{aligned} \]
  1. 0 is a multiple of every integer except itself.

Example:

How many multiples of 2 or 3 are there in \([-11, 11]\)?

Answer: 15

Proof:

x <- c(-11:11)
x
##  [1] -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5
## [18]   6   7   8   9  10  11
multiples <- vector()
for (i in x){
  if (i %% 2 == 0 | i %% 3 == 0){
    multiples <- append(multiples, i)
  }
}
multiples
##  [1] -10  -9  -8  -6  -4  -3  -2   0   2   3   4   6   8   9  10
length(multiples)
## [1] 15
  1. If \(n\) is a positive integer, and \(\tau(n)\) represents the number of positive divisors of \(n\), then the number of positive divisors can be found using the following theorem:

Let \(\sigma\) denote the sum of divisors function, which can be defined by setting \(\sigma(n)\) equal to the sum of all the positive divisors of \(n\).

Let \(\tau\) denote the number of divisors function, which can be defined by setting \(\tau(n)\) equal to the number of positive divsiors of \(n\).

Both functions can be expressed as

\[ \begin{aligned} \sigma(n) = \sum_{d|n}d \qquad \tau(n) = \sum_{d|n}1 \end{aligned} \]

Both \(\sigma(n)\) and \(\tau(n)\) are multiplicative functions. That is,

\[ \begin{aligned} \sigma(mn)=\sigma(m)\sigma(n) \qquad \tau(mn)=\tau(m)\tau(n) \qquad (1) \end{aligned} \] where \(m\) and \(n\) are relatively prime positive integers.

Lemma: Let \(p\) be prime and \(a\) be a positive integer. Then

\[ \begin{aligned} \sigma(p^{a}) = 1 + p + p^{2} + \ldots + p^{a} = \frac{p^{a+1}-1}{p-1} \qquad (2) \end{aligned} \]

and

\[ \begin{aligned} \tau(p^{a})=a+1 \qquad (3) \end{aligned} \]

Proof. The divisors of \(p^{a}\) are \(1, p, p^{2}, \ldots, p^{a-1}, p^{a}\). Therefore, \(p^{a}\) has exactly \(a + 1\) divisors, so that \(\tau(p^{a}) = a + 1\). Note that \(\sigma(p^{a}) = 1 + p + p^{2} + \ldots + p^{a-1} + p^{a} = \frac{p^{a-1}-1}{p-1}\) (this is the sum of terms in a geometric progression).

Theorem: Let the positive integer \(n\) have a prime factorization \(n=p_{1}^{a_{1}} \times p_{2}^{a_{2}} \times \ldots \times p_{s}^{a_{s}}\). Then we have

\[ \begin{aligned} \sigma(n)= \frac{p_{1}^{a_{1}+1}-1}{p_{1}-1} \times \frac{p_{2}^{a_{2}+1}-1}{p_{2}-1} \times \ldots \times \frac{p_{s}^{a_{s}+1}-1}{p_{s}-1} = \prod_{j=1}^{s}\frac{p_{j}^{a_{j+1}}-1}{p_{j}-1} \qquad (4) \end{aligned} \] and

\[ \begin{aligned} \tau(n)=(a_{1}+1)(a_{2}+1) \ldots (a_{s}+ 1) = \prod_{j=1}^{s}(a_{j}+1) \qquad (5) \end{aligned} \]

Proof. Since \(\sigma\) and \(\tau\) are both multiplicative, it follows that

\[ \begin{aligned} \sigma(n) = \sigma(p_{1}^{a_{1}} \times p_{2}^{a_{2}} \times \ldots \times p_{s}^{a_{s}}) = \sigma(p_{1}^{a_{1}}) \times \sigma(p_{2}^{a_{2}}) \times \ldots \times \sigma(p_{s}^{a_{s}}), \end{aligned} \]

and

\[ \begin{aligned} \tau(n) = \tau(p_{1}^{a_{1}} \times p_{2}^{a_{2}} \times \ldots \times p_{s}^{a_{s}}) = \tau(p_{1}^{a_{1}}) \times \tau(p_{2}^{a_{2}}) \times \ldots \times \tau(p_{s}^{a_{s}}) \end{aligned} \]

Inserting the values \(\sigma(p_{i}^{a_{i}})\) and \(\tau(p_{i}^{a_{i}})\) found in \((2)\) and \((3)\), we can obtain the desired formulas.

Example:

Calculate \(\sigma(200)\) and \(\tau(200)\).

Using \((4)\) and \((5)\), we get

\[ \begin{aligned} \sigma(200) = \sigma(2^{3}5^{2}) = \frac{2^{4}-1}{2-1} \times \frac{5^{3}-1}{5-1} = 15 \times 31 = 465 \end{aligned} \]

and

\[ \begin{aligned} \tau(200) = \tau(2^{3}5^{2}) = (3 + 1)(2 + 1) = 4 \times 3 = 12 \end{aligned} \]

Therefore, if \(n\) is a positive integer and \(\tau(n)\) is the number of positive divisors of \(n\), then \(n=200\) has \(12\) positive divisors. We can use R to check if this is correct.

library(numbers)
d <- divisors(200)
d
##  [1]   1   2   4   5   8  10  20  25  40  50 100 200
length(d)
## [1] 12

It checks out.

  1. The only even prime number is 2.

  2. The midpoint of a line segment on an xy-coordinate plane is \(\sqrt{\frac{(x_{2}+x_{1})}{2}, \frac{(y_{2}+y_{1})}{2}}\)

  3. 0 is an even number.

  4. If \(n\) is an integer divided by another integer \(x\), the remainder \(r\) is always less than the divisor \(x\). That is, \(r < x\).

Example:

Let \(n\) be a positive integer. Set \(n = 13\)

Which is greater?

A: The remainder when \(n\) is divided by 7 or B: 7

A: \(\frac{13}{7} = 1 \frac{6}{7}, \quad r= 6\)

B: \(7\)

7 > 6

\(\therefore\) B > A
  1. Order of Operations: PEMDAS (Please Excuse My Dear Aunt Sally). When given an expression, the order of operations is as follows:
    1. Calculate anything in parentheses.
    2. Calculate any exponent(s).
    3. Divide and multiply from left to right.
    4. Add and subtract from left to right.

Example:

\(24 \times (6 + 2)^{2} = 24 \times (8)^{2} = 24 \times 64 = 1,536\)

\(24 \div 4 \times 4 = 6 \times 4 = 24\)

  1. When dividing by a fraction simply take the reciprocal of the denominator and multiply by the numerator.

Example:

\[ \begin{aligned} & 1. \quad \frac{2}{\frac{1}{2}} = 2 \times \frac{2}{1} = 4 \\ & 2. \quad \frac{\frac{1}{5}}{\frac{2}{5}} = \frac{1}{5} \times \frac{5}{2} = \frac{5}{10} = \frac{1}{2} \end{aligned} \]

3. Miscellaneous Strategies

  1. When given an inequality (e.g. \(7 < x \leq 11\)), do not assume that \(x\) is an integer unless explicitly stated.

  2. If a question asks, “Which of the following must be true?” Any choice that doesn’t satisfy the expression or equation can automatically be ruled out. However, the opposite does not apply. Even if an answer satsifies the problem, it does not necessarily mean that it must be true.

  3. When employing a “trial and error” strategy (i.e., substituting in different numbers for a particular variable), \(100\) and \(2\) are good numbers to use in place of a variable. \(100\) is especially nice when dealing with percentages. Thus, whenever you’re dealing with an unkown percentage, expressing the uknown percentage in terms of \(\frac{x}{100}\) makes it very easy to calculate.

Example:

Big Toe usually deposits the same amount of money each month into a vacay fund for her annual trip to Las Vegas. However, this year, she decided not to make any contributions during November and December. In order to make the same annual contribution that she had originally planned, by what percent should Big Toe increase her monthly deposits from January through October?

Solution: We don’t know how much money Big Toe contributes monthly to her vacation fund, and consequently don’t know how much her annual savings are. However, we do know that she makes the deposits the same amount every month. Knowing this, we can let \(x\) represent her monthly contributions. You can set \(x\) to whatever number you like, but let’s set \(x=\$100\). If she contributes \(\$100\) every month, her annual savings is \(12x = 12(\$100) = \$1,200\). However, because of her mental deficiencies and imprudent decision-making, she decides not to contribute anything during November and December, and only decides to deposite money 10 months out of the year (from January to October). Since we want to know how much she needs to increase her monthly contributions to ensure that she still saves \(\$1,200\), we simply set \(10x = \$1,200\), solve for \(x\), and find the \(\%\) increase she needs to make to her monthly contributions.

\[ \begin{aligned} & 10x = 1200 \\ & x = \frac{1200}{10} = 120 \\ & \frac{120-100}{100} = \frac{20}{100} = \frac{1}{5} = 0.2 \times 100 = 20\% \end{aligned} \] We can check our answer by \(10x = 10(\$120) = \$1,200\). And there you go. Big Toe has enough money to go ham in Vegas. Note that because the only variable in this problem is her monthly contributions, it doesn’t matter what you set \(x\) equal to. If you set \(x=\$50\), you will get the same answer.

\[ \begin{aligned} & 12x = 12(50) = 600 \qquad \text{(Her regular annual savings)} \\ & 10x = 600 = \frac{600}{10} = 60 \\ & \frac{60-50}{50} = \frac{10}{50} = \frac{1}{5} = 0.2 \times 100 = 20\% \end{aligned} \]