Number Properties

Question 1

Basic formula for odd/even numbers

Answer 1

Even integers can be represented by 2n, where n is an integer, and odd integers can be represented by either 2n – 1, or by 2n + 1, where n is an integer.

Question 2

Result when adding/subtracting odd/even numbers

Answer 2

When adding or subtracting two numbers, if both numbers are even, or both numbers are odd, the result will be even. Otherwise, if one is even and one odd, the result will be odd.

Question 3

Result when multiplying odd/even numbers

Answer 3

• even x any number [including multiple numbers] = even
• odd x odd = odd. with multiple numbers, if all odd, product will be odd

Question 4

Result when dividing odd/even numbers

Answer 4

• even / odd = even
• odd / odd = odd
• even / even = even or odd

Question 5

Steps to find all positive factors of a number

Answer 5

Step 1: Prime factorization of number.
Step 2: Add 1 to the value of each exponent
Step 3: Multiply the values

example: 12000

prime factorization = 2^5 * 3^1 * 5^3
Add one = (5+1)(1+2)(3+1)
Multiply=634
Factors = 48

Question 6

Unique prime factors when a number is raised to a power

Answer 6

If some number x has y unique prime factors, then x (where n is a positive integer) will have the same y unique prime factors.

Question 7

How to find least common multiple of set of numbers

Answer 7

1. Find the prime factorization of each number
2. Of repeated prime factors [i.e. same base] take only the prime factor with the largest power.
3. Minus the repeated PFs, multiply them

Question 8

LCM of two numbers that do not share prime factors

Answer 8

If two numbers X and Y do not share prime factors, the LCM is XY. If they do, it is some number less than XY.

Question 9

How to find greatest common factor of set of numbers

Answer 9

1. Find the prime factorization of each number
2. Identify repeated prime factors amongst the numbers
3. Of any repeated prime factors, take only those with smallest exponent. If no exponent, use ‘1’.
4. Multiple the numbers to find the GCF

Question 10

Relationship between LCM and GCF

Answer 10

If the LCM of x and y is p and the GCF of x and y is q, then xy = pq.

That is, xy = LCM(x, y) × GCF(x, y).

Question 11

How to find # of unique prime number in set of numbers

Answer 11

The LCM will provide the number of unique prime numbers in a set

Question 12

How to use LCM for finding coinciding rate or times

Answer 12

The LCM can be used to determine when two processes that occur at differing rates or times will coincide.

For example, let’s say that blinking light L flashes once every 32 seconds, and blinking light M flashes once every 12 seconds. If both lights flash together at 8:00:00 PM, when will be the next time the lights will flash together again?

The two lights will next flash together at the LCM of 32 seconds and 12 seconds.

Question 13

The product of any n consecutive integers will always be divisible by n!.

Answer 13

The product of any n consecutive integers will always be divisible by n!.

Also, the product of any n consecutive integers must be divisible by all of the factors of n!.

Question 14

The product of n consecutive even integers will always be divisible by 8

Answer 14

The product of n consecutive even integers will always be divisible by 8

For example, 10 × 12 = 120, and 120 is divisible by 8, because 120/8=15

Question 15

Formula for dividends, divisors, quotients, remainders

Answer 15

If an integer x is the dividend (numerator), an integer y is the divisor (denominator), Q is the integer quotient of the division, and r is the nonnegative remainder of the division, then:

x/y = q + r/y

Question 16

What are potential ranges for remainders?

Answer 16

A remainder must be a non-negative integer that is less than the divisor.

Example: If a and b are positive integers, what is the difference between the largest possible remainder and the smallest possible remainder when a is divided by b?

1) a = 9x, where x is a positive integer.

2) b=7

Statement 2 alone is sufficient. Since remainder must be less than divisor, the range of the remainder is 6-0= 6.

Question 17

What are trailing zeros?

Answer 17

In whole numbers, trailing zeros are created by (5 × 2) pairs. Each (5 × 2) pair creates one trailing zero. Thus, the number of trailing zeros of a number is the number of (5 × 2) pairs in the prime factorization of that number.

Example: ⇒ 5,200 can be expressed as 52 × 100 = 52 × 10^2 and has 2 trailing zeros

Question 18

How to find number of digits in equation? Example, how many numbers in 25^10 * 8^6

Answer 18

Step 1: Prime factorize the number
Step 2: Count the number of (5 x 2) pairs. Each pair contributes to one trailing zero.
Step 3: Collect the number of unpaired 5s or 2s, along with other nonzero prime factors (if any) and multiply them together. Count the number of digits in this product.
Step 4: : Sum the number of digits from steps 2 and 3.

Question 19

What are leading zeros?

Answer 19

In fractions, we’re going to look at what we’ll refer to as “leading zeros.” Leading zeros are the zeros that occur to the right of the decimal point but before the first nonzero number.

Ex:

.4311 has no leading zeros
.01 has one leading zero
.001 has two leading zeros

Question 20

when denominator is not a perfect power of ten: to find number of leading zeros

Answer 20

If X is an integer with k digits, and if X is not a perfect power of 10, then 1/X will have K-1 leading zeros.

A perfect power of ten is an integer where the prime factorization contains equal amounts of 5s and 2s.

Question 21

when denominator is a perfect power of ten: to find number of leading zeros

Answer 21

If X is an integer with k digits, and if X is a perfect power of 10, then 1/X will have K-2 leading zeros

Question 22

Steps for finding number of primes in a factorial

Answer 22

To determine the (largest) number of a prime number x that divides into y!, we perform the following steps:

1. Divide y by X^1, X^2, X^3.. etc until the quotient is 0.
2. Add the quotients from the previous divisions; that sum represents the number of prime number x in the prime factorization of y!. Ignore any remainders

If denominator is not a prime number but a product of primes, break the number into prime factorials and follow the above steps with the largest prime factor. [Example, if the denominator is 15^n, break into 5^n and 3^n and find the number of times 5^n goes into the factorial]

Question 23

Steps for finding number of primes in a factorial when the denominator is a power of a prime number

Answer 23

To determine the (largest) value of a non-prime number x (where x = p^k, p is a prime, and k is an integer greater than 1) that divides into y!, we perform the following steps:

example: what is the largest value of N such that 30!/4n is an integer?

1. Express x = p^k ;
   4n = (2^2)n = 2^2n
2. Apply the factorial divisibility shortcut to determine the quantity of p in y! ;
   There are 26 2s in 30!
3. Then create and simplify an inequality to determine the largest number of x that divides into y!. ;
   2n </ 26
   n </13. the largest value of n must be 13

Question 24

What must all perfect square end in?

Answer 24

0, 1, 4, 5, 6, or 9

A perfect square, other than 0 and 1, is a number such that all of its prime factors have even exponents.

Question 25

The prime factorization of a perfect cube will contain only exponents that are multiples of 3

Answer 25

A perfect cube, other than 0 or 1, is a number such that all of its prime factors have exponents that are divisible by 3. 2^3 is a perfect cube because its exponent is divisible by 3.

Question 26

How to find out whether fraction is a terminating decimal or not

Answer 26

The decimal equivalent of a fraction will terminate if and only if the denominator of the reduced fraction has a prime factorization that contains only 2s or 5s, or both.

If the prime factorization of the reduced fraction’s denominator contains anything other than 2s or 5s, the decimal equivalent will not terminate.

Question 27

Patterns in unit digits

Answer 27

0: All powers of 0 end in 0.
1: all powers of 1 end in 1.
2. The units digits of positive powers of 2 will follow the four-number pattern 2-4-8-6.
3. The units digits of powers of 3 will follow the four-number pattern 3-9-7-1.
4. The units digits of powers of 4 follow a two-number pattern: 4-6. All positive odd powers of 4 end in 4, and all positive even powers of 4 end in 6.
5. All positive integer powers of 5 end in 5.
6. All positive integer powers of 6 end in 6.
7. The units digits of positive powers of 7 will follow the four-number pattern 7-9-3-1.
8. The units digits of positive powers of 8 will follow the four-number pattern 8-4-2-6.
9. The units digits of powers of 9 follow a two-number pattern: 9-1. All positive odd powers of 9 end in 9, and all positive even powers of 9 end in 1.

The units-digit pattern of the powers of any integer greater than 9 has the same units-digit pattern as the powers of its units digit.

Question 28

Remainders after division by 10^n

Answer 28

When a whole number is divided by 10, the remainder will be the units digit of the dividend (numerator).

For example, when 153 is divided by 10, the result is 15.3, which is 15 3/10 [remainder 3]

and so on with 100, last two digits of dividend, 1000, last 3 digits etc

Question 29

Remainders after division by 5

Answer 29

When integers with the same units digit are divided by 5, the remainder is constant.

Question 30

Difference between combinator and permutation

Answer 30

Combinator, order does not matter

Permutation, order/sequence does matter

Question 31

Formulas for basic combinations

Answer 31

1. n!/k!(n-k)!

N= total n in pool
K = number of objects we will actually choose

2. step 1: fill number of ‘boxes’/’objects’ with total n.
   step 2: keep going until you fill the last box.
   step 3. divide by number of boxes factorial.

example: pool of 7, how many different combinations of 3?

765/321

Question 32

Fundamental counting principle

Answer 32

If there are m ways to perform task 1 and n ways to perform task 2 and the tasks are independent, then there are m × n ways to perform both of the tasks together.

Question 33

When you see ‘or’ in combinator problems

Answer 33

‘or’ is a mutually exclusive event. you would add the probabilities rather than multiply.

also if you see ‘at least x amount of combinators’ start with the lowest and fill to the top. add the different combos

Question 34

Combinators: some items can be chosen, others cant

Answer 34

When some number of the items in a set must be chosen while another number of items in the set cannot be chosen, we can do the following:

1. Mentally eliminate from the larger group those items that are excluded from the subgroup. Note that restricting items from a subgroup does not change the number of spots available in that subgroup.
2. Visualize those items that must be in the subgroup as already placed in the subgroup.
3. We then can determine both how many items remain in the larger group and how many spots remain to be filled within the subgroup.

Question 35

Combinator formula where subjects cannot be together in same group

Answer 35

formula;

The Total Number of 3-Person Club Configurations That Can be Made = The Number of 3-Person Club Configurations in which Both People are together in the club + The Number of 3-Person Club Configurations in which neither People are together in the club

Question 36

Permutation formula

Answer 36

1. nPk = n!/(n-k)!

n= total number of objects
k=number of objects that are to be chosen

2. step 1: fill number of ‘boxes’/’objects’ with total n.
   step 2: keep going until you fill the last box.
   step 3. do not divide by anything

Question 37

permutation formula for distinct permutations with repeating/identical items

Answer 37

n!/(r1!)(r2!)…(rn!)

r represents the number of times an item repeats

Question 38

Circle permutation formula

Answer 38

The Circular Permutation Formula states that K items can be arranged in a circle in (K – 1)! ways.

Question 39

If items must be together in a permutation

Answer 39

When x items must be together in an arrangement of y items, treat those x items together as one unit. Find permutation, then times that by x!

(y-x+1!)(x!)

Question 40

Direct variation equation [ratios]

Answer 40

When we have a direct variation relationship between two variables x and y, they can be related by the equation y = kx, where k is a positive constant.

Question 41

Inverse variation equation [ratios]

Answer 41

When we have an inverse variation relationship between two variables x and y, they can be related by the equation y=k/x

Question 42

When ratio varies directly AND inversely

Answer 42

A single variable can vary directly with one variable and inversely with another variable.

For example, if y varies directly with x and inversely with z, then the three variables are related by the equation y=kx/z, where k is a constant.

Question 43

Joint variation equations [ratios]

Answer 43

”y varies directly jointly with x and z” → y = kxz

“y varies inversely jointly with x and z” → y = k/xz

Question 44

Two common ways to determine ratio multiplier

Answer 44

There are two common scenarios that will allow us to determine the ratio multiplier:
1) We are given the quantity of one of the items in the ratio.
2) We are given the quantity of all of the items in the ratio.

Question 45

Converting to squared units, cubed units

Answer 45

what is 81 inches squared in feet squared?

1. Manipulate the unit conversion by raising both sides of the equation to the second power

e.g.

(1 foot ^)2 = (12 inches)^ 2

1 foot ^ 2 = 144 inches ^ 2

2. convert squared units

81 inches squared = 1 foot squared/144 inches squared

answer = 9/16 feet squared

Question 46

What does percent mean?

Answer 46

percent = per 100

divide by 100

Question 47

Percent less than

Answer 47

In “Percent Less Than” problems, x% less than y is equivalent to (100 – x)% of y.

final value = initial value * (1 - ‘percent less than’/100)

Question 48

Percent greater than

Answer 48

In “Percent Greater Than” problems, x% greater than y is equivalent to (100 + x)% of y.

final value = initial value * (1 + ‘percent greater than’/100)

examples:

3 percent greater than z = 103 percent of z

25 percent greater than z = 125 percent of z

Question 49

Mark up an item by x percent

Answer 49

multiply the value of the item by (1 + x/100)

Question 50

mark down an item by x percent

Answer 50

multiply the value of the item by (1 - x/100)

Question 51

if there are multiple percent changes to a value in a word problem, how should that be solved

Answer 51

if variables:

by multiplying the percent changes in one equation rather than setting up several

if numbers:
set up several equations

Question 52

percent change formula #1

Answer 52

We can use this Percent Change Formula to determine any percent change, regardless of whether it’s a percent increase, decrease, or combination of the two.

percent change = (final value - initial value/initial value) x 100

Question 53

percent change formula #2

Answer 53

We can use the Percent Greater Than formula, discussed earlier in the chapter, to account for percent increases only. Replace “Percent Greater Than” with some variable x and solve for it.

final value = initial value x (1 + percent greater than/100)

Question 54

percent change formula #3

Answer 54

We can use the Percent Less Than formula, discussed earlier in the chapter, to account for percent decreases only. Replace “Percent Less Than” with some variable x and solve for it.

final value = initial value x (1 - percent less than/100)

Question 55

Quick solution for if question asks ‘If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?’

Answer 55

Sequence of 4 numbers [number that is divisible]: 4
Numbers we have [numbers in n n+1 sequence]: 2

Answer 1-2/4 = 1/2