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GMAT > Properties Of Numbers > Flashcards

Properties Of Numbers Flashcards

1
Q

Integer (Definition)

A

All numbers without a Fractional or Decimal Component inclusive of Positive numbers Negative numbers and zero.

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2
Q

Whole Numbers (Definition)

A

Set of numbers inclusive of all Non - Negative integers and Zero i.e. Positive numbers without a fractional component and Zero

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3
Q

Integer A is a Multiple of Integer B IF -

A

A/B = integer i.e A divided B equals to Z which is an Integer

Zero is Multiple of every number!
Since any number divided by zero is zero
i.e it is the only number equal to all its multiples

All numbers are multiples of one A/1 = A

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4
Q

Integer A is a Factor of Integer B IF -

A

If B/A = integer or B divided A equals to Z which is an Integer

Also, B divided by any factor of A is also equal to an integer.

Zero is not a Factor of any number!
Since division by zero is not allowed

Similarly One is a factor of all Numbers i.e B/1 =B

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5
Q

Square Root of Zero

A

Is Zero

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6
Q

The only number that is neither positive or Negative

A

Is Zero

i.e Zero is the only number that is equivalent to its opposite +0 = -0

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7
Q

Zero raised to any number

A

Is Zero

Note - the exception is Zero raised to Zero which is undefied and not tested

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8
Q

Any number raised to Zero

A

Is One

Note - the exception is Zero raised to Zero which is undefied and not tested

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9
Q

Zero divided by Any number

A

Is Zero

I.e. Zero is even since 0/2 is = 0 which is an integer

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10
Q

If (A) (B) = 1

A

Then A and B are both equal to one or minus one

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11
Q

A number that can be multiplied or divided by any number without changing its value

A

Is One it is also the only number that is neither Prime nor Composite i.e the smallest prime number is two.

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12
Q

Single-digit Primes

A

2, 3, 5, 7

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13
Q

10’s Primes

A

11, 13, 17, 19

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14
Q

20’s Primes

A

23,29

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15
Q

30’s Primes

A

31, 37

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16
Q

40’s Primes

A

41, 43, 47

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17
Q

50’s Primes

A

53, 59

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18
Q

60’s Primes

A

61, 67

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19
Q

70’s Primes

A

71, 73, 79

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20
Q

80’s Primes

A

83, 89

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21
Q

90’s Primes

A

97

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22
Q

Find the number of factors of X

A

Step 1) Prime factorize X such that X = (a^x) (b^y) (c^z)…
where a,b,c are prime numbers and x,y,z are their powers respectively.

Step 2) Add one to each of the exponents and multiply the resulting values i.e (x+1) (y+1) (z+1)…

eg- 240 = (2^4)(3^1)(5^1) i.e factors = (4+1)(1+1)(1+1) = 5x2x2=20

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23
Q

X Divides evenly into Y

A

Y/X= integer

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24
Q

X is Divisible by Y

A

X/Y = integer

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25
Q

X is a Divisor of Y

A

Y/X is an integer

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26
Q

X is a Dividend of Y

A

X/Y = integer

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27
Q

For Divisibility Questions

A

Think Prime Factorization i.e Multiple/Factor = Integer

i. e Divisibility questions are restricted to positive integers
i. e Positive multiples including the LCM of numbers exclude 0

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28
Q

If Z is Divisible by X and Z is Divisible By Y then Z is also Divisible by -

A

The LCM of X and Y because the maximum value of LCM(x,y)= to the product of X and Y if they do not share any common factors and both X and Y are factors of Z on their own.

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29
Q

X is divisible by 0 if

A

No number is divisible by zero as division by zero is not permitted

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30
Q

X is divisible by 1 if

A

All numbers are divisible by one.

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31
Q

X is divisible by 2 if

A

X ends in an even number i.e the units digit of X = 0,2,4,6,8

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32
Q

X is divisible by 3 if

A

The sum of X’s digits is divisible by 3

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33
Q

X is divisible by 4 if

A

The last two digits of X are divisible by four i.e if x=abcde then de/4 =intiger then x is divisible by four.

Also if X ends in 00 it is divisible by four as all multiples of 100 are divisible by four.

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34
Q

X is divisible by 5 if

A

The units digit of X = 0,5

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35
Q

X is divisible by 6 if

A

X is divisible by 2 and 3 i.e

1) The units digit of X = 0,2,4,6,8 i.e even
2) The sum of X’s digits is divisible by 3

36
Q

X is divisible by 7 if

A

No simple rule, check by long division

37
Q

X is divisible by 8 if

A

The last three digits of X are divisible by eight i.e if x=abcde then cde/8 =intiger then x is divisible by eight.

Also if X ends in 000 it is divisible by four as all multiples of 1000 are divisible by eight.

38
Q

X is divisible by 9 if

A

The sum of X’s digits is divisible by 9

39
Q

X is divisible by 10 if

A

X ends in a zero i.e the units digit of X = 0

40
Q

X is divisible by 11 if

A

The sum of X’s Even digits - the sum of X’s Odd digits is divisible by 11.
i.e if X=abcde then if ((a+c+e) - (b+d))/11 = Intiger then x is a multiple of 11.

41
Q

X is divisible by 12 if

A

X is divisible by 3 and 4 i.e

1) The sum of X’s digits is divisible by 3
2) The last two digits of X are divisible by four i.e if x=abcde then de/4 =intiger then x is divisible by four.

Also if X ends in 00 it is divisible by four as all multiples of 100 are divisible by four.

42
Q

X^a/X^b is an integer if

A

a is Grater than (>) or Equal (=) to b

43
Q

If X is divisible by Y then

A

X is also divisible by any factor of Y

44
Q

If Two or more processes occur at different frequencies then the time at which both processes occur/coincide will be

A

The Least common multiple of the frequency intervals of those processes (or a multiple of the same)

45
Q

The Least common multiple of a group of numbers tells us

A

The unique set of prime factors of each of the constituent numbers raised to their highest power

46
Q

(LCM (A,B)) x (GCD (A,B)) =

A

A x B

47
Q

If X/Y = integer Then
LCM (X,Y) =
GCD (X,Y) =

A

LCM (X,Y) = X since it is a multiple of Y

GCD (X,Y) = Y since it is a factor of X

48
Q

The process to find LCM

and definition

A

It is the smallest positive multiple of a given set of numbers i.e the first number that a given set of numbers will all evenly divide into.

1) Prime factorize each number and write in Prime factor -Exponent form.
2) Select the highest power of Prime factors common to the numbers.
3) Select all remaining uncommon prime factors.
4) LCM = product of the prime factors derived in steps 3 and 4

49
Q

The process to find GCD / HCF

A

It is the largest factor or greatest divisor that will divide a set of numbers. If the numbers do not share any common factors then GCD=1

1) Prime factorize each number and write in Prime factor -Exponent form.
2) Select the lowest power of the prime factors common to all the numbers.
3) GCD = product of the prime factors derived in steps 3.

50
Q

Quick LCM

A

Take the Largest number of the given set and check which one of its multiples is divisible by the other numbers in the set. The first divisible multiple will be the LCM

51
Q

Quick GCD

A

Take the Largest number of the given set and Identify which of its factors divide the other numbers in the set.

52
Q

LCM vs GCD size and Minimum / Maximum values.

A

LCM is always larger than the GCD i.e

At minimum LCM is equal to the largest number of a set of numbers I.e if the largest number is a multiple of all the other numbers in the set. At maximum, the LCM is equal to the product of all the numbers in a set if they do not contain any common factors.

GCD At Maximum is equal to the smallest number of a set of numbers if the smallest number is a factor of all the other numbers in the set. A Minimum the Gcd = 1 of the set of numbers do not contain any common factors.

53
Q

Factors of a prime number

A

Prime numbers have only 2 factors i.e 1 and the number itself as they are not divisible by any other numbers

Prime numbers are positive by definition

54
Q

Even / Even =

A

Even or Odd basically if two sets factored out of the numerator then it is odd else it is even i.e. no way to tell for sure

55
Q

ODD / Even =

A

Not an integer as an odd number can never be divided by an even number

56
Q

Even / Odd =

A

Even because an off number is not capable of factoring out two from the numerator.

57
Q

Even + or - Even =

A

Even i.e The result will be even if both numbers are either even or odd

58
Q

Even + or - Odd =

A

Odd i.e The result will be odd if one of the numbers is even and the other is odd i.e both numbers are not either even or odd

59
Q

Odd + or - Odd =

A

Even i.e The result will be even if both numbers are either even or odd

60
Q

Odd x Odd =

A

Odd i.e The result will be even if at least one of the numbers is even i.e 2 becomes a factor of the number

61
Q

Even x Even =

A

Even i.e The result will be even if at least one of the numbers is even i.e 2 becomes a factor of the number

62
Q

Even x Odd =

A

Even i.e The result will be even if at least one of the numbers is even i.e 2 becomes a factor of the number

63
Q

Formula for Division

A

(Dividend / Divisor) = Quotient + (Remainder / Divisor)

i. e X/Y = Q + R/Y i.e X= Qy+R
i. e Manipualting the equation we can express any integer X as the product of a Quotient (Q) and Divisor (Y) plus some remainder (R)

the Quotient (Q) can be equal to zero leaving us with only a remainder i.e 7/9=0+7

The remainder is the numerator of the un simplified fractional component of the division i.e we do not factor out common terms from the remainder over divisor fraction

64
Q

Converting a fractional remainder to its decimal form

A

Similar to a mixed number we perform a long division on the fractional component of the number and add it to the Quotient i.e 25/4 = 6+1/4 = 6+0.25 = 6.25 (1/4=0.25)

65
Q

Converting a decimal remainder to its fractional form

A

We Cannot determine the exact remainder from its fractional component without knowing the original numbers as there are multiple fractional equivalents for a decimal value.

However, we can determine the most reduced form of the fractional component from which the actual remainder must be a multiple of this reduced form

i.e. Consider 9.48 i.e R = 48 in 48/100
However R = 24 in 24/50
and R = 12 in 12/25 which is the most reduced form i.e in all cases R will be a multiple of 12

66
Q

Converting a decimal Remainder to its integer form

A

If we know the value of the denominator Y in X/Y= AB.C then the integer form of the remainder C = YxC

i.e for 9/5 = 1.8 we can multiple the decimal component of the Quotient(1.8) i.e 0.8 with the denominator or divisor 5 to get the integer version of the remainder i.e 0.8x5=4

67
Q

Multiplication of Remainders

A

We can multiply the numerator of the fractional component of remainders however incase the numerator exceeds the denominator we need to perform a division to correct for any excess.

Eg. Remainder for (17x13x12)/5 = 2x3x2= 12/5 = 2 2/5
i.e R=2 since (17/5= 3 2/5)(13/5= 3 3/5)(12/5= 2 2/5)

68
Q

Adding or Subtracting Remainders

A

Remainders can be added or subtracted however we need to correct for any excess i.e resulting remainders that are less that are greater than the divisor in case of addition (divide the remainder by divisor to get the true remainder) or negative remainders in the case of subtraction (add the divisor to the negative remainder to get the true remainder)

eg- 17/5 - 13/5 = 3 2/5 - 2 3/5 = 2-3 = -1 = -1 + 5 = 4
i.e the true remainder = 4 not -1.

69
Q

Remainder definition and range

A

A Remainder is a non-negative integer that is less than the Divisor. If the Divisor = n then the remainder ranges from 0 to n-1

70
Q

Trailing Zeros

A

Each pair of (5x2) factors that a number possesses form one trailing zero. This is because 5x2=10 i.e each time 10 is multiplied by the number another zero is added to the right of its last non zero digit. Similarly, the units digit of any factorial greater or equal to 5 is Zero.

71
Q

Determine the number of digits in an integer from its factors.

A

1) Prime factorize the number and write it in integer exponent form.
2) Collect all pairs of (5x2)s each one constitutes to a single trailing zero.
3) Collect all unpaired 5s, 2s, and any other factors. Multiply these factors together and count the resulting digits.
4) Sum up the number of trailing zeros determined in step 2 and the number of digits derived in step 3 to determine to total number of digits.

Note if the integer is a perfect power of 10 i.e it consists only of (5x2) pairs and no factors are leftover in step 3 then number of digits = number of tailing zeros + 1
i.e digits in 10^8 = 8 trailing zeros plus one i.e nine digits.

72
Q

Definition of Leading zeros and determination process.

A

The number of zeros that occur to the right of the Decimal point prior to the first non-zero digit are called Leading zeros.

1) Take the denominator of the fraction and determine its number of digits using the trailing zero method.
2) If X is an integer with K digits then 1/X will have K-1 leading zeros unless X is a perfect power of 10 in which case it will have K-2 leading zeros.

Note if the integer is a perfect power of 10 i.e it consists only of (5x2) pairs and no factors are leftover in step 3 then number of digits = number of tailing zeros + 1
i.e digits in 10^8 = 8 trailing zeros plus one i.e nine digits.

73
Q

The product of “n” consecutive integers is always divisible by-

A

n! i.e all integers from one to n.

eg1. 7x8x9 is divisible by 3! i.e 504/6 = 84
eg2. 3x4x5x6 is divisible by 4! i.e 360/24 =15

i.e the largest integer that X consecutive integers are divisible by is X! or the product of the integers from one to X.

74
Q

N Factorial is divisible by -

A

All integers from one to n inclusive and it must also be divisible by any factor combinations of one to n inclusive.

75
Q

Determine the largest number/exponential power of Prime Factor “X” that is present in Y! factorial

A

1) Divide Y by X followed by each consecutive power of X until X raised to some power divides Y Zero times leaving only a remainder.
2) During each division keep track of the Quotients i.e number of times that each of the powers of X divides Y and ignore any remainders.
3) Add up all the Quotient values from Step 2 which equals to the largest exponential power of the prime factor X present in Y!

eg- There are 99 5’s in 400! i.e
400/5 = 80, 400/25=16, 400/125=3 + ignored
remainders and 400/625= 0 plus only remainder hence we stop here i.e 80+16+3=99

76
Q

Determine the largest number/exponential power of a composite number “Z” that is present in Y!

A

For composite numbers break the number (Z) down into its prime factors and Identify the largest prime factor let’s call it X.

1) Divide Y by X followed by each consecutive power of X until X raised to some power divides Y Zero times leaving only a remainder.
2) During each division keep track of the Quotients i.e number of times that each of the powers of X divides Y and ignore any remainders.
3) Add up all the Quotient values from Step 2 which equals to the largest exponential power of the composite number Z present in Y!

eg There are 21 15’s in 90! i.e 15 = 3x5 hence
90/5=18, 90/25= 3 & 90/125= 0 + only remainder i.e 18+3=21 i.e there 21 15’s in 90.

77
Q

Determine the largest number/exponential power of a composite number “Z” that is present in Y! such that Z is an exponential power of prime factor X

A

Express composite Z as X raised to n where X is a prime factor and n is the exponential power of X present in and w is the exponential power of Z present in Y! i.e Z=X^nw

1) Divide Y by X followed by each consecutive power of X until X raised to some power divides Y Zero times leaving only a remainder.
2) During each division keep track of the Quotients i.e number of times that each of the powers of X divides Y and ignore any remainders.
3) Add up all the Quotient values from Step 2 which equals to the largest exponential power of X in Y! let this power be equal to v

4) Now form inequality X^nw < or = X^v
i. e nw < or = v therefore w < or = v/n.

eg- there are 32 8s in 100! i.e 8=2^3 and
100/2=50, 100/4=25, 100/8=12+r, 100/16=6+r, 100/32=3+r, 100/64=1+r, 100/128=0+r.
therefore there are 50+25+12+6+3+1=97 2s in 100!

i. e 2^3n < or = 2^97
i. e 3n < or = 97
i. e n < or = 97/3 ll N < or = 32 (ignore the remainder)

78
Q

Properties of Perfect squares (4)

exceptional numbers -

A

1) The square root of a perfect square will yield an integer.
2) A perfect square will never end in 2,3,7,8 i.e the squares of the first 9 digits do not possess those numbers
3) A perfect square will never end in an odd number of trailing zeros i.e it has to have an even pair of (5x2)
4) Prime factorizations of a perfect square will only yield factors with even exponents.

0 and 1 are the only perfect squares exempt from the above rules.

79
Q

Properties of Perfect cubes (2)

exceptional numbers -

A

1) The cube root of a perfect Cube will yield an integer.
2) Prime factorizations of perfect cubes will yield only factors with exponents that have powers that are multiples of 3.
0 and 1 are the only perfect Cubes exempt from the above rules.

80
Q

The Decimal equivalent of a fraction will terminate if

A

The denominator of the fraction in its most simplified form contains only the prime factors 2 And/Or 5 any other prime factor in the denominator will lead to a non-terminating decimal.

81
Q

Remainder range and patterns

A

If the Divisor = n then the remainder ranges from 0 to n-1 however if the divisor is kept constant and the numerator is increased exponentially or sequentially a recurring pattern emerges for example -

3/4 = 0 3/4, 9/4= 2 1/4, 27/4 = 6 3/4, 81/4 = 1/4
i.e Remainders repeat in a 3,1,3,1…pattern depending on whether 3 is raised to an even or an odd power.

82
Q

Units Digit and their patterns

A

Each digit from 0 to 9 exhibits a cyclic pattern when it is raised to consecutive integer powers know as cyclicity.

Only units digits can create units digits hence to determine the units digit of a Number X raised to any power check the units digit of the number X determine its Cyclicity and then divide the exponential power by the cyclicity (max value is 4) to determine the remainder based upon which we decide the units digit that is applicable.

eg- Since 3^1=3, 3^2=9, 3^3=27, 3^4=81,
The cyclicity of 3 is 4 i.e the pattern 3,9,7,1 repeats itself in cycles of 4s i.e 3^n+1=3, 3^n+2=9, 3^n+3=7, 3^n=1, where n is a multiple of 4 an +1,+2,+3 are remainders.

Therefore, the units digit of 3^47 = 3^44+3 =7 (44 is the multiple of 4 and 3 is the remainder)

83
Q

Recurring Patterns

A

Determine the number of entities it takes for the pattern to repeat itself and let that be its cyclicity.

To determine a specific point within the pattern divide the number of elements in the series by the pattern’s cyclicity i.e number of elements after which the pattern repeats and look for the remainder. Then match with the elements in the pattern depending on the remainder i.e cyclicity+1 remainder will be the first element in the pattern

Basically model the equation - Cyclicity x n + remainder
Where n is the number of complete cycles and r is the remainder which is used to count from the beginning of the pattern up to the element on which the partial cycle stopped.

84
Q

Remainders when dividing by powers of 10

A

When the divisor is then the remainder is the units digit of the numerator similarly when dividing by a power of ten i.e 10^x the remainder is the last x digits of the numerator.

i.e when 1234 is divided by 10^2 the remainder is 34

85
Q

Remainders when dividing 5 (2 properties)

A

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

i.e 9/5= 1 4/5, 19/5 = 3 4/5, 29/5 = 5 4/5 39/5= 7 4/5
hence since the units digit stays 9 the remainer is always 4

Similarly, when a number is raised to power we can use the principle of cyclicity to determine its units digit and dividing that units digit by 5 will give us the same remainder as the whole number divided by 5!

i.e The remainder when 3^123 is divide by 5 is 2 because of 3 has a cyclicity 4 i.e its units digits repeat in the pattern of 3,9,7,1 hence the units digit of 3^123 is 7 and 7/5 = 1 2/5

86
Q

3 types of consecutive Integer Sets -

A

1) Consecutive integers in which the common difference (CD) is one i.e (1,2,3,4,5,6,7) 4-3=5-4=6-5=1

Note -We can also have consecutive sets of even/odd integers i.e (0,2,4,6,8) or (1,3,5,7,9) i.e 6-4=5-3 = 2 (CD)

2) Multiples i.e Consecutive integers multiplied by a fixed intieger i.e (6,12,18,24,30) 12-6=24-18=30-24= 6 (CD)
Note - The common difference is the number by which each of the consecutive integers is multiplied.

3) Multiples + Constant i.e Consecutive integers multiplied by a integer + some constant i.e (3,7,11,15,19) 7-3=19-15= 4 (CD) and constant = 3
Note - The common difference is the number by which the consecutive numbers are multiplied and the constant can be found by dividing any element of the set by the common difference and checking the remainder.

87
Q

Consecutive integer properties (2)

A

1) Consecutive integers do not share any common prime factors.

2) The GCD of two consecutive integers is one
i. e GCD (n,n+1) = 1 since they do not have any other prime factors in common.

Note look for p - q = 1 or p = q + 1 i.e if p and q are positive they are consecutive integers.

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