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G-M > Properties of Number > Flashcards

Properties of Number Flashcards

1
Q

Zero Rule (1/4)

A
  • Zero multiplied by any number is zero
  • Zero divided by any number other than 0 is 0 (0/2=0)
  • Any number divided by zero is undefined (2/0= undefined)
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2
Q

Zero Rule (2/4)

A
  • The square root of zero is zero
  • Zero raised to any postive power is zero (0^1= 0)
  • Any number raised to the zero is equal to 1: 1^0 = 0
  • Zero is the only number thta is equal to its opposite (0=-0)
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3
Q

Zero Rule (3/4)

A
  • Zero is a mutiple of all number
  • Zero is the only number that is equal to all its multiples
  • Zero is not a factor of any number beside itself
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4
Q

Zero Rule (4/4)

A
  • Zero is an even number

- Zero can be added to or subtracted from any number without changing the value of that number

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5
Q

Properties of One (1/2)

A
  • One is a factor of all number and all numbers are multiple of 1
  • one raised to any power is one
  • multiplying or dividing any number by 1 does not changed the number
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6
Q

Properties of One (2/2)

A
  • One is an odd number
  • One is the only number with exactly 1 factor
  • One is not a prime number (the first prime number is 2)
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7
Q

Even units can be written as

A

2n

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8
Q

Odd units can be written as

A

2n-1 or 2n+1

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9
Q

odd+odd

A

even

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10
Q

odd-odd

A

even

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11
Q

even+even

A

even

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12
Q

even-even

A

even

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13
Q

odd+even

A

odd

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14
Q

odd-even

A

odd

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15
Q

even+odd

A

odd

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16
Q

even-odd

A

odd

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17
Q

even*even

A

even

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18
Q

even*odd

A

even

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19
Q

odd*even

A

even

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20
Q

odd*odd

A

odd

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21
Q

even/odd

A

even

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22
Q

odd/odd

A

odd

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23
Q

** even/even

A

can be either odd or even

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24
Q

When we add two or more positive number

A

results in a larger positive number

-> Postive+Postive = bigger positive number

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25
Q

When we add two or more (smaller) negative number

A

results in a smaller negative number

-> Negative+Negative = smaller negative number

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26
Q

Multiple numbers with the same sign

A

postive

(-5*-5)=25

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27
Q

Divide numbers with the same sign

A

divide

(-100/-5)=20

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28
Q

Divide numbers with the different sign

A

negative

(-100/5) = -20

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29
Q

Multiple numbers with the different sign

A

5*-6= -30

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30
Q

Factor of number

A

factor will divide into the number evenly without leaving a remainder
6-> factors are 2,3,6,1

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31
Q

y is a factor of x if and only if x/y

A

is an integer

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32
Q

Multiple of a number

A

a multiple of a number is the product of that number with any integer

  • > 2= 2,4,6,8….
  • > there are infinite multiple of a number
  • > x=ny
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33
Q

x is a multiple of y if and only if

A

x/y is an integer

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34
Q

Any integer is both a

A

factor and a multiple of itself

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35
Q

Prime number

A

Any integer greater than 1 that has no factors other than 1 and itself
-> so only divisible by 2 numbers

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36
Q

Prime number

A
  • 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
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37
Q

Finding total factors in a number

A

1) Find the prime factorization of the number
2) Add 1 to value of each exponent. Then multiply these results and the product will be the total number of factors for that number
-> Total factors of 240
-> 240= 2^4.3^1.5^1
Add 1 to exponent = (4+1)(1+1)(1+1)= 20

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38
Q

Total number of prime factors vs unique prime factor

A

Total number of prime factors = both unique and non unique prime factor
Unique factors = just one of each unique factor
-> prime factorization of n is xy^2z= 4 total prime factor and 3 unique factor

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39
Q

Does the number of unique prime factorization increase as as the power raised to it increase

A

no

- 5^2 has same unique factor as 5^3

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40
Q

LCM

A

smallest set of positive integers into which all of the numbers in the set will divide

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41
Q

When a set of prime factors share no prime factors

A

the LCM of that set will be the product of the number in that set
-> x and y share no prime factors then LCM of x and y is xy

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42
Q

GCF

A

a greatest common factor or GCF of a set of positive integers is the largest number that will divide into all the number in the set

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43
Q

GCF when no common factor

A

1

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44
Q

LCM will always be

A

same or greater than GCF

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45
Q

GCF*LCF of x and y

A
  • equals to the product of the x and y
  • > lcm of x and y= 24
  • > gcf of and y= 2
  • > x and y = 24*2= 48
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46
Q

LCM of prime factor provides us with

A

unique prime factors of the set

-> thus provides all the unique factors of the product of the number in the set

47
Q

LCM can be used to determine

A

two processes that occur at differing rates or time will coincide

48
Q

When the numerator of a positive fraction is a multiple of denominator

A

an even division occurs

49
Q

2 divides into 4

A

4/2

50
Q

x/y will yield an integer of

A
  • > x is a multiple of y

- > y is factor of x

51
Q

y is a factor of x

A

x/y -> and produces an integer

52
Q

y is a divisor of x

A

x/y -> and produces an integer

53
Q

y divides into x (evenly)

A

x/y -> and produces an integer

54
Q

x is a multiple of y

A

x/y -> and produces an integer

55
Q

x is a dividend of y

A

x/y -> and produces an integer

56
Q

x is divisible by y

A

x/y -> and produces an integer

57
Q

Whenever we see large divisibility we can use

A

prime factorization

  • 3660/42= (36610)/(67) = 610/7 -> we see that not everything cancels out so it’s not a factor
  • > of prime factorization then everything needs to cancel out
58
Q

If x and y are positive integers and x/y is an integer

A

x/any factor of y is also an integer

59
Q

29Q/17 is an integer

A

Q should be divided by 17
-> whenever you find these divide evenly problem you should always try to find a prime factorization and see if it cancels or not

60
Q

If z is divisible by both x and y

A

z must also be divisible by the LCM of x and y

-> 24 and 48-> divisible by 12

61
Q

If an item went on sale for 16 percent . What is the new price of the item

A

N= (16/100)*W
W= 25N/4
-> we know that N needs to be multiplied by 4

62
Q

Number divisible by 0

A

No number is divisible by 0

63
Q

Number divisible by 1

A

All numbers are divisble by 1

64
Q

Number divisible by 2

A

Have even uni digits

65
Q

Number divisible by 3

A

a number is divisble by 3 if the sum of all the digit is divisble by 3
-> 472,071= (4+7+2+0+7+1)=21

66
Q

Number divisible by 4

A

If the last two numbers are divisble by 4 then the number is divisble by 4

  • > number that end with 00 is divisble by 4
  • > 244/4 = 44
67
Q

Number divisible by 5

A

All numbers that have last digit that end with 0 0r 5

68
Q

Number divisible by 6

A

If the number in question is an even number whose digit sum to a multiple of 3
-> 18-> even number whose sum 9 is divided by 3

69
Q

Number divisible by 8

A

-> if the number is even, divide the last three digit by 8 and if there is no remainder then it’s divisble by 8

70
Q

Number divisible by 9

A

If the sum of all digits is divisible by 9

-> 479,655 (4+7+9+6+5+5=36)

71
Q

Number divisible by 10

A

If the ones digit is zero then the number is divisible by 10

72
Q

Number divisible by 11

A

Sum of the odd numbered places digit minus the sum of the even digits is divisble by 11
-> 2915 = (9+5)-(2+1)= 11

73
Q

Number divisible by 12

A

If a number is divisble by both3 and 4, it is also divisble by 12

74
Q

Remainder (x/y)

A

Q+x/y

-> 36/5= 7+1/5

75
Q

Remainder x/y can be manipulated

A
  • > x=Qy+r
  • > Q= (x-r)/y
  • > r= x-Qy
76
Q

Expressing remainder interms of decimal

A

32/5

  • > 5 divides into 32 6 times with 2/5 left
  • > so the remainder interms of decimal is 0.4
  • > but the whole thing will be 6.4
77
Q

Remainder of fraction in decimals

A
  • > we don’t really know the exact number
  • > 9.48 -> remainder in the fraction can be 48/100, can be 480/1000
  • > but there is only one reduced fraction -> for this one it will be 12/25
78
Q

9/5 = 1.8-> find the remainder in a whole number

A

.8*5= 1.8

-> we need to know what the denominator is

79
Q

When finding the remainder of number with more than one number
(121317*17) divided by 5

A

121317*17

1) Divide each by 5
- - 12/5= 2 with a reminder of 2
- - 13/5 = 2 with a reminder of 3
- - 17/7 = 3 with a reminder of 2
2) Multiply all the remainder = 232= 12
3) Divide the excess remainder by 5 and you get 2 as a remainder
- > so you remainder is 2 for the number given

80
Q

Remainder must be

A

non-negative integer that is less than divisor

81
Q

5*2 creates

A

-> a trailing zero
-> 10
-> 5^3002^298
= (52)^298*(5^2)
–> have 298 trailing zero

82
Q

Any factorials >= 5 will always have

A

zero as its unit digit

83
Q

Create a prime factor of a number (25^10*8^6)

A

1) Prime factorize the number
2) Count the number of (52) pairs
3) Collect unpaired of 5s and 2s along with nonzero prime factors. Count the number of digits in this product
4) Sum the number of digit from steps 2 and 3
–> (5^2)^10(2^3)^6= 5^202^18
= (52)^18= 18 trailing zeroes
= (5*5)=25 -> remaining, so 18+2=20

84
Q

Leading zero in the decimal

A

1) express the fraction in the form 1/x
-> If x is an integer with k digits, then 1/x will have k-1 leading zeroes unless X is a perfect power of 10, which case there will be K-2 leading zeroes
i.e:
A) 1/(5^22^11)=
x= (52)^5= 5 trailing and 2^6= 64 so 2 digits = 5+2=7-1=6

B) 1/(5^52^5)
= 5 trailing zeroes and 1 in front so 6 digits
= (52)^5 is a perfect power of 10 so we have to do k-2

85
Q

1!

A

1

86
Q

0!

A

1

87
Q

Product of set of consecutive number is

A
  • > divisible by any of the integer in the set and any factor combinations of these numbers
  • > n! must be divisible by all the integers from 1 to n inclusive. In addition, n! must be divisible by any factor combination of 1 to n inclusive
88
Q

Largest possible integer value such that
21!/3^n is an integer
-> When the denominator is prime

A

1) Divide 21 by 3^1, 3^2..3^k until you get 0
- > 21/3= 7
- > 21/3^2= 21/9=2 (ignore the remainder)
- > 21/2^3= 0 (ignore the remainder)
- > The largest number that could be a factor is of the n is 9

89
Q

Largest possible integer value such that (40!/6^n) is an interger
-> when the denominator is not a prime factor

A

(40!/6^n)
1) Break the denominator into prime number (6^n= 23)
so, (40!/2^n3^n)
2) Since there are fewer 3’s that go into 40 than 2’s, the 3’s will place the limitation
- 40/13 = 13
- 40/3^2= 4
- 40/ 3^3= 1
- 40/3^4= 0
–> 13+4+1+0= 18
-So the largest integer value that of n such that (40!/6^n) is an interger is 18

90
Q

Largest possible integer value such that (30!/4^n) is an interger
-> when the base of the divisor is a power of a prime number

A

1) Break the denominator into prime factors 4^n= (2^2)n -> can express it as 30!/2^2n
2) Divide the numerator by 2^1, 2^2…
So,
-> 30/2=15
-> 30/2^2= 7 (ignore the remainder)
-> 30/2^3= 3
-> 30/2^4=1
-> 30/2^5=0
3) Add all of them-> 15+7+3+1= 26
4) Set up an inequality
-> 2n<26
-> n<13

91
Q

Number with perfect square root end on

A 
0
1
4
5
6
9
92
Q

Number that is a perfect square never ends on

A

2
3
7
8

93
Q

If some number x (that is not 1 or 0) is a perfect square, the prime factorization of x must contain only

A

even exponents

  • > 100 is 10^2 (5^2*2^2)-> you see that both of the prime factors have exponent power
  • > 64= 2^6
94
Q

The first nine non-cube perfect cubes are

A
  • 0
  • 1
  • 8
  • 27
  • 64
  • 125
  • 216
  • 343
  • 512
95
Q

A perfect cube, other than 0 or 1, is a number such that all of its prime factors

A

have exponents that are divisible by 3

-> perfect cube of 64 is 4^3

96
Q

Terminating decimals

A

is one that has a finite number of non-zero digits to the right of the decimal

97
Q

The decimal will be terminating if

A

the denominator of the reduced fraction has a prime factorization that contains only 2s and 5s

  • > 1/4= 0.25, 1/20= 0.10 (termianting)
  • > 1/6= 0.1666666 (non-terminating)
98
Q

When we divide, is there a pattern in the remainder

A

-> yes there is

99
Q

Unit digit of the power

A

forms a pattern (when both the base and the exponent are positive)

100
Q

Power of 0

A

0

-> 0^1=0, 0^2=0….0

101
Q

Power of 1

A

1

-> 1^1, 1^2=1…1^n=1..

102
Q

Power of 2

A 
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=128
2^8=256
2^9=512
2^10=1024
-> all the base will follow the pattern of 2-4-8-6
103
Q

Power of 3

A 
3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
3^6=729
3^7=2187
3^8=6561
3^9=19,683
3^10=59,049
-> all the base will follow the pattern of 3-9-7-1
104
Q

Power of 4

A 
4^1= 4
4^2= 16
4^3= 64
4^4= 256
4^5= 1024
4^6= 4096
4^7= 16,384
4^8= 65,536
4^9= 262,144
4^10= 1,048,576
-> all the base will follow the pattern 
\: odd power end with 4 
\: even power end with 6
105
Q

Power of 5

A 
5^ 1=5
5^2= 25
5^3= 125
5^4 = 625
5^5 = 3125
-> all the positive integer powers of 5 ends in 5
106
Q

Power of 6

A 
6^1 = 6
6^2 = 36
6^3 = 216
6^4 = 1296
6^ 5= 7776
-> all positive integer power of 6 end in 6
107
Q

Power of 7

A 
7^1=7
7^2 =49
7^3 = 343
7^4 =2,401
7^5 =16,807
7^ 6=117,649
7^ 7=823,543
7^8 =5,764,801
7^9 =40,353,607
7^10 =282,475,249
-> all the base will follow the pattern 7-9-3-1
108
Q

Power of 8

A 
8^1= 8
8^2 = 64
8^3 = 512
8^4= 4096
8^5 = 32,768
8^6 = 262,144
8^7 = 2,091,152
8^8 = 16,777,216
8^9 = 134,217,728
8^10 = 1,073,741,824
-> all the base will follow the pattern 8-4-2-6
109
Q

Power of 9

A 
9^1 = 9
9^2= 81
9^3 = 729
9^4 = 6561
9^5 = 59,049
9^6 = 531,441
9^7 = 4,782,969
9^8 = 43,046,721
9^9 = 387,420,489
9^10 = 3,486,784,401
-> all the base will follow the pattern 9-1
110
Q

All the number will follow some kind of pattern in their

A

base

111
Q

When a whole number is divided by 10

A

the remainder unit digit of the dividend (numerator)
i.e x= 255 and y= 10, what is x^2 divided by y
= unit digit of 255^2=5
so, 5/10= you have a remainder of 5
-> similar pattern follow for 100 and so on

112
Q

When digit with same unit digits are divided by 5, the remainder is

A

constant

  • > 9/5 = 1 4/5, 19/5= 3 4/5, 59/5= 11 4/5
  • > 7/5= 1 2/5, 17/5 = 3 2/5, 57/5= 11 2/5
113
Q

3 ways that consecutive number appear on GMAT

A

1) A set of consecutive integers
2) A set of consecutive multiples of a given number such as (2,4,6,8) or (10,20,30,40)
3) A set of consecutive numbers with a given remainder when divided by some integers (1, 6, 11, 16, 21) 0r (3,7,11,15,19)

114
Q

Two consecutive number will never share the same

A

prime factors

-> so GCF of two consecutive integers is 1

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