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MATHCOUNTS
TOOLBOX
Facts, Formulas and Tricks

MATHCOUNTS Coaching Kit 40

I. PRIME NUMBERS from 1 through 100 (1 is not prime!)
2 3 5 7
11 13 17 19
23 29
31 37
41 43 47
53 59
61 67
71 73 79
83 89
97

II. FRACTIONS DECIMALS PERCENTS
1
/2 .5 50 %
1
/3 .3 33 .3 %
2
/3 .6 66 .6 %
1
/4 .25 25 %
3
/4 .75 75 %
1
/5 .2 20 %
2
/5 .4 40 %
3
/5 .6 60 %
4
/5 .8 80 %
1
/6 .16 16 .6 %
5
/6 .83 83 .3 %
1
/8 .125 12.5 %
3
/8 .375 37.5 %
5
/8 .625 62.5 %
7
/8 .875 87.5 %
1
/9 .1 11. 1 %
1
/10 .1 10 %
1
/11 . 09 9. 09 %
1
/12 . 083 8 .3 %
1
/16 .0625 6.25 %
1
/20 .05 5%
1
/25 .04 4%
1
/50 .02 2%

III. PERFECT SQUARES AND PERFECT CUBES
12 = 1 22 = 4 32 = 9 42 = 16 52 = 25
2 2 2
6 = 36 7 = 49 8 = 64 92 = 81 102 = 100
112 = 121 122 = 144 132 = 169 142 = 196 152 = 225
2 2 2
16 = 256 17 = 289 18 = 324 192 = 361 202 = 400
2 2 2
21 = 441 22 = 484 23 = 529 242 = 576 252 = 625

13 = 1 23 = 8 33 = 27 43 = 64 53 = 125
63 = 216 73 = 343 83 = 512 93 = 729 103 = 1000


IV. SQUARE ROOTS
1 =1 2 ≈ 1.414 3 ≈ 1.732 4 =2 5 ≈ 2.236
6 ≈ 2.449 7 ≈ 2.646 8 ≈ 2.828 9 =3 10 ≈ 3.162

V. FORMULAS

Perimeter: Volume:
Triangle p=a+b+c Cube V = s3
Square p = 4s Rectangular Prism V = lwh.
Rectangle p = 2l + 2w Cylinder V = πr2h
Circle (circumference) c = 2πr Cone V = (1/3)πr2h
c = πd Sphere V = (4/3)πr3
Pyramid V = (1/3)(area of base)h.

Area:
Rhombus A = (½)d1d2 Circle A = πr2
Square A = s2 Triangle A = (½)bh.a
Rectangle A = lw = bh Right Triangle A = (½)l1l2
Parallelogram A = bh Equilateral Triangle A = (¼) s2 3
Trapezoid A = (½)(b1 + b2)h.

Total Surface Area: Lateral Surface Area:
2
Cube T = 6s Rectangular Prism L = (2l + 2w)h
Rectangular Prism T = 2lw + 2lh + 2wh Cylinder L = 2πrh
Cylinder T = 2πr2 + 2πrh
Sphere T = 4πr2

Distance = Rate × Time
− y1 1y 6
1x 6
2
Slope of a Line with Endpoints (x1, y1) and (x2, y2): slope = m =
2 − x1
Distance Formula: distance between two points or length of segment with endpoints (x1, y1) and (x2, y2)

D= 1x 2 − x1 6 +1y
2
2 − y1 6 2

Midpoint Formula: midpoint of a line segment given two endpoints (x1, y1) and (x2, y2)
x + x y1 + y2
2
1 2
,
2
Circles:
x 12πr 6 , where x is the measure of the central angle of the arc
Length of an arc =
360
x 3πr 8 , where x is the measure of the central angle of the sector
Area of a sector =
360
2


Combinations (number of groupings when the order of the items in the groups does not matter):
N!
Number of combinations = , where N = # of total items and R = # of items being chosen
R !( N − R)!

Permutations (number of groupings when the order of the items in the groups matters):
N!
Number of permutations = , where N = # of total items and R = # of items being chosen
( N − R)!

Length of a Diagonal of a Square = s 2

Length of a Diagonal of a Cube = s 3

Length of a Diagonal of a Rectangular Solid =x 2 + y 2 + z 2 , with dimensions x, y and z
N ( N − 3)
Number of Diagonals for a Convex Polygon with N Sides =
2

Sum of the Measures of the Interior Angles of a Regular Polygon with N Sides = (N − 2)180

Heron’s Formula:
For any triangle with side lengths a, b and c, Area = s( s − a )( s − b)( s − c) , where s = ½(a + b + c)

Pythagorean Theorem: (Can be used with all right triangles)
a2 + b2 = c2 , where a and b are the lengths of the legs and c is the length of the hypotenuse

Pythagorean Triples: Integer-length sides for right triangles form Pythagorean Triples – the largest
number must be on the hypotenuse. Memorizing the bold triples will also lead to other triples that are
multiples of the original.
3 4 5 5 12 13 7 24 25
6 8 10 10 24 26 8 15 17
9 12 15 15 36 39 9 40 41

Special Right Triangles:
45o – 45o – 90o 30o – 60o – 90o
hypotenuse = 2 (leg) = a 2 hypotenuse = 2(shorter leg) = 2b
hypotenuse c
leg = = longer leg = 3 (shorter leg) = b 3
2 2
longer leg hypotenuse
shorter leg = =
45°
3 2
c
a
30°
45° a c
a

60°
b

a x
Geometric Mean: = therefore, x2 = ab and x = ab
x b

360
Regular Polygon: Measure of a central angle = , where n = number of sides of the polygon
n
360
Measure of vertex angle = 180 − , where n = number of sides of the polygon
n

Ratio of Two Similar Figures: If the ratio of the measures of corresponding side lengths is A:B,
then the ratio of the perimeters is A:B, the ratio of the areas is A 2 : B 2 and the ratio of the
volumes is A 3 : B 3 .

1 61 6
Difference of Two Squares: a 2 − b 2 = a − b a + b
Example: 12 2 − 9 2 = 112 − 96112 + 96 = 3 ⋅ 21 = 63
144 − 81 = 63
Determining the Greatest Common Factor (GCF): 5 Methods
1. Prime Factorization (Factor Tree) – Collect all common factors
2. Listing all Factors
3. Multiply the two numbers and divide by the Least Common Multiple (LCM)
Example: to find the GCF of 15 and 20, multiply 15 × 20 = 300,
then divide by the LCM, 60. The GCF is 5.
4. Divide the smaller number into the larger number. If there is a remainder, divide the
remainder into the divisor until there is no remainder left. The last divisor used is the GCF.
Example: 180 385 25 180 5 25
360 175 25 5 is the GCF of 180 and 385
25 5 0

5. Single Method for finding both the GCF and LCM
Put both numbers in a lattice. On the left, put ANY divisor of the two numbers and put the
quotients below the original numbers. Repeat until the quotients have no common factors
except 1 (relatively prime). Draw a “boot” around the left-most column and the bottom
row. Multiply the vertical divisors to get the GCF. Multiply the “boot” numbers (vertical
divisors and last-row quotients) to get the LCM.

40 140 40 140 40 140 The GCF is 2×10 = 20
2 20 70 2 20 70 2 20 70 The LCM is
10 10 2 7 2×10×2×7 = 280

VI. DEFINITIONS

Real Numbers: all rational and irrational numbers

Rational Numbers: numbers that can be written as a ratio of two integers

Irrational Numbers: non-repeating, non-terminating decimals; can’t be written as a ratio of two integers
(i.e. 7 , π )


Integers: {…, -3, -2, -1, 0, 1, 2, 3, …}

Whole Numbers: {0, 1, 2, 3, …}

Natural Numbers: {1, 2, 3, 4, …}

Common Fraction: a fraction in lowest terms (Refer to “Forms of Answers” in the MATHCOUNTS
School Handbook for a complete definition.)

Equation of a Line:
A
Standard form: Ax + By = C with slope = −
B
Slope-intercept form: y = mx + b with slope = m and y-intercept = b

Regular Polygon: a convex polygon with all equal sides and all equal angles

1 1
Negative Exponents: x − n = and −n
= xn
xn x

Systems of Equations: x + y = 10 8 + y = 10 (8, 2) is the solution
x−y = 6 y = 2 of the system
2 x = 16
x = 8

Mean = Arithmetic Mean = Average

Mode = the number(s) occurring the most often; there may be more than one

Median = the middle number when written from least to greatest
If there is an even number of terms, the median is the average of the two middle terms.

Range = the difference between the greatest and least values

Measurements:
1 mile = 5280 feet
1 square foot = 144 square inches
1 square yard = 9 square feet
1 cubic yard = 27 cubic feet

VII. PATTERNS
Divisibility Rules:
Number is divisible by 2: last digit is 0,2,4,6 or 8
3: sum of digits is divisible by 3
4: two-digit number formed by the last two digits is divisible by 4
5: last digit is 0 or 5
6: number is divisible by both 2 and 3
8: three-digit number formed by the last 3 digits is divisible by 8
9: sum of digits is divisible by 9
10: last digit is 0

Sum of the First N Odd Natural Numbers = N 2
Sum of the First N Even Natural Numbers = N 2 + N = N(N + 1)
N
Sum of an Arithmetic Sequence of Integers: × (first term + last term), where N = amount of
2
numbers/terms in the sequence

Find the digit in the units place of a particular power of a particular integer
Find the pattern of units digits: 71 ends in 7
72 ends in 9
(pattern repeats 73 ends in 3
every 4 exponents) 74 ends in 1
75 ends in 7
Divide 4 into the given exponent and compare the remainder with the first four exponents.
(a remainder of 0 matches with the exponent of 4)
Example: What is the units digit of 722?
22 ÷ 4 = 5 r. 2, so the units digit of 722 is the same as the units digit of 72, which is 9.

VIII. FACTORIALS (“n!” is read “n factorial”)
n! = (n)×(n −1)×(n − 2)×…×(2)×(1) Example: 5! = 5 × 4 × 3 × 2 × 1 = 120
0! = 1
1! = 1
2! = 2 Notice 6! 6 × 5 × 4 × 3 × 2 ×1
= = 30
3! = 6 4! 4 × 3 × 2 ×1
4! = 24
5! = 120
6! = 720
7! = 5040

IX. PASCAL’S TRIANGLE
Pascal’s Triangle Used for Probability:
Remember that the first row is row zero (0). Row 4 is 1 4 6 4 1. This can be used to
determine the different outcomes when flipping four coins.
1 4 6 4 1
way to get ways to get ways to get ways to get way to get
4 heads 0 tails 3 heads 1 tail 2 heads 2 tails 1 head 3 tails 0 heads 4 tails

For the Expansion of (a + b)n , use numbers in Pascal’s Triangle as coefficients.
1 (a + b)0 = 1
1 1 (a + b)1 = a + b
1 2 1 (a + b)2 = a2 + 2ab + b2
1 3 3 1 (a + b)3 = a3 + 3a2b + 3ab2 + b3
1 4 6 4 1 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
1 5 10 10 5 1 (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5


For 2n, add all the numbers in the nth row. (Remember the triangle starts with row 0.)
1 20 = 1
1 1 21 = 1 + 1 = 2
1 2 1 22 = 1 + 2 + 1 = 4
1 3 3 1 23 = 1 + 3 + 3 + 1 = 8
1 4 6 4 1 24 = 1 + 4 + 6 + 4 + 1 = 16
1 5 10 10 5 1 25 = 1 + 5 + 10 + 10 + 5 + 1 = 32

X. SQUARING A NUMBER WITH A UNITS DIGIT OF 5

(n5)2 = n × (n + 1) 2 5 , where n represents the block of digits before the units digit of 5
Examples:
(35)2 = 3×(3+1) 2 5 (125)2 = 12×(12+1) 2 5
= 3×(4) 2 5 = 12×(13) 2 5
=1225 =15625
= 1,225 = 15,625

XI. BASES
%DVH GHFLPDO ² RQO XVHV GLJLWV ²
Base 2 = binary – only uses digits 0 – 1
%DVH RFWDO ² RQO XVHV GLJLWV ²
Base 16 = hexadecimal – only uses digits 0 – 9, A – F (where A=10, B=11, …, F=15)

Changing from Base 10 to Another Base:
What is the base 2 representation of 125 (or “125 base 10” or “12510”)?
We know 125 = 1(102) + 2(101) + 5(100) = 100 + 20 + 5, but what is it equal to in base 2?
12510 = ?(2n) + ?(2n-1) + … + ?(20)
The largest power of 2 in 125 is 64 = 26, so we now know our base 2 number will be:
?(26) + ?(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20) and it will have 7 digits of 1’s and/or 0’s.

Since there is one 64, we have: 1(26) + ?(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20)
We now have 125 – 64 = 61 left over, which is one 32 = 25 and 29 left over, so we have:
1(26) + 1(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20)
In the left-over 29, there is one 16 = 24, with 13 left over, so we have:
1(26) + 1(25) + 1 (24) + ?(23) + ?(22) + ?(21)+ ?(20)
1(26) + 1(25) + 1(24) + 1(23) + ?(22) + ?(21)+ ?(20)
1(26) + 1(25) + 1(24) + 1(23) + 1(22) + ?(21)+ ?(20)
In the left-over 1, there is no 2 = 21, so we still have 1 left over, and our expression is:
1(26) + 1(25) + 1(24) + 1(23) + 1(22) + 0(21)+ ?(20)
The left-over 1 is one 20, so we finally have:
1(26) + 1(25) + 1(24) + 1(23) + 1(22) + 0(21)+ 1(20) = 11111012


Now try What is the base 3 representation of 105?
The largest power of 3 in 105 is 81 = 34, so we now know our base 3 number will be:
?(34) + ?(33) + ?(32) + ?(31)+ ?(30) and will have 5 digits of 2’s, 1’s, and/or 0’s.
Since there is one 81, we have: 1(34) + ?(33) + ?(32) + ?(31)+ ?(30)
In the left-over 105 – 81 = 24, there is no 27 = 33, so we still have 24 and the expression:
1(34) + 0(33) + ?(32) + ?(31)+ ?(30)
In the left-over 24, there are two 9’s (or 32’s), with 6 left over, so we have:
1(34) + 0(33) + 2(32) + ?(31)+ ?(30)
In the left-over 6, there are two 3’s (or 31’s), with 0 left over, so we have:
1(34) + 0(33) + 2(32) + 2(31)+ ?(30)
Since there is nothing left over, we have no 1’s (or 30’s), so our final expression is:
1(34) + 0(33) + 2(32) + 2(31)+ 0(30) = 102203

The following is another fun algorithm for converting base 10 numbers to other bases:

12510 = ?2 10510 = ?3 12510 = ?16

1 r.1 1 r.0 7 r.13(D)
2 3 r.1 3 3 r.2 16 125
2 7 r.1 3 11 r.2
2 15 r.1 3 35 r.0 12510 = 7D16
2 31 r.0 3 105
2 62 r.1
Start
here 2 125 10510 = 102203

12510 = 11111012 xyzn = (x × n2) + (y × n1) + (z × n0)

Notice: Everything in bold shows the first division operation. The first remainder will be the last digit
in the base n representation, and the quotient is then divided again by the desired base. The process is
repeated until a quotient is reached that is less than the desired base. At that time, the final quotient
and remainders are read downward.

XII. FACTORS
Determining the Number of Factors of a Number: First find the prime factorization (include the 1 if a
factor is to the first power). Increase each exponent by 1 and multiply these new numbers together.
Example: How many factors does 300 have?
The prime factorization of 300 is 22 × 31 × 52 . Increase each of the exponents by 1 and multiply
these new values: (2+1) × (1+1) × (2+1) = 3 × 2 × 3 = 18. So 300 has 18 factors.

Finding the Sum of the Factors of a Number:
Example: What is the sum of the factors of 10,500?
(From the prime factorization 22 × 31 × 53 × 71, we know 10,500 has 3 × 2 × 4 × 2 = 48 factors.)
The sum of these 48 factors can be calculated from the prime factorization, too:
(20 + 21 + 22)(30 + 31)(50 + 51 + 52 + 53)(70 + 71) = 7 × 4 × 156 × 8 = 34,944.

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