Gmat Number Properties

Front 1

Divisibilty Rules:

An integer is divisible by:

Back 1

2 if the integer is even

3 if the sum of the integers digits is divisible by 3

4 if the integer is divisible by 2 twice

5 if the integer ends in 0 or 5

6 if the integer is divisible by both 2 and 3 (i.e. it is even and the sum of the digits is divisble by three)

8 if the number is divisble by 2 three times

9 if the sum of the digits are divisble by nine

10 if the integer ends in 0

Front 2

zero is

Back 2

a multiple of every integer

Front 3

a factor is a

Back 3

`POSITIVE integer that divides evenly into an integer. so 1, 2, 4, 8 are all factors of 8

Front 4

a multiple of an integer is

Back 4

formed by multiplying the integer by any whole number... including zero.

Front 5

Sum and Difference Rule

Back 5

If 2 numbers have a common divisor/factor there SUM and DIFFERENCE retain the common divisor. I.E. 8 is a factor of 40 and 64. 8 is also a factor of 64-40=24 and 64+40=104

Front 6

y=xq+r

Back 6

104=8*13
24=8*3

y=dividend
x=divisor
q=quotient
r=remainder

Front 7

primes

Back 7

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, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199

Front 8

factor formula:

how many factors does 18 have

Back 8

step one break into primes:

2 * 3^2

add one to each power and multiply through

2*3=6 factors

long way:
1 * 18
2 * 9
3 * 6

Front 9

factor foundation rule:

Back 9

if an integer is divisible by some number n, it is also divisable by all factors of n.

i.e. if 72 is divisble by 12, it is also divisble by 1,2,3,4,6)

Front 10

Is 27 a factor of 72?

Back 10

Procedure:

break 27 into primes:
3^3

break 72 into primes

2^3 3^2

we can not make 27 from the prime factors of 32 therefore 27 is not a factor of 72

Front 11

Given that the integer n is divisble by 3, 7, and 11 what other numbers must be divisors of n

Back 11

since we know that n has AT LEAST 3,7, and 11 as primes we can say that it is also divisble by 3*7=21 3*11=33 11*7=77
and 3*7*11=231

Front 12

Lowest Common Multiple

Back 12

the smallest number that is a multiple of two integers

procedure:

us a prime box and find the product of all the prime numbers in each box, using the higher power of any primes that might appear in both boxes

Front 13

Greatest Common Factor:

Back 13

the largest number by which two integers can be divided

procedure:

us a prime box, and find the product of the primes that are common to both numbers to the lower power

Front 14

What is the GCF/LCM of 24 and 30

Back 14

24=
2^3, 3

30=
2,3,5

GCF = 2*3=6
LCM = 2^3*3*5=120

Front 15

terminating decimals

Back 15

if the denominator has only factors of 2 or 5 you are garunteed a terminating decimal

note 2^0 or 5^0 still counts

ie a denominator of 5, 2, 25

Front 16

if 80 is a factor of r is 15 a factor of r

Back 16

PROCEDURE:

USE PRIME
BOX

80 = 2^4*5

15=3*5

SINCE 3 IS NOT IN THE PRIME BOX OF 80 WE CAN NOT MAKE 15 OUT OF THE FACTORS OF 80... SO WE DO NOT KNOW IF 15 IS A FACTOR OF R (IT COULD BE) REMEMBER THAT THE PRIME BOX IN
THIS SITUATION COULD REPRESENT ONLY A PARTIAL LIST OF R'S FACTORS.

Front 17

given that 7 is a factor of n and 7 is a factor of p is 7 a factor of n+p

Back 17

yes. use the sum and difference rule. if x is a factor of integers n and m, x will be a factor of n+m and n-m

Front 18

ODDS AND EVENS

Back 18

Odd +/- Odd = Even
Odd +/- Even = Odd
Even +/- Odd = odd.
Even +/- Even = Even

Odd X Odd = Odd
Odd X Even = Even
Even X Even = Even

Front 19

-|x| = x for all x <=0


for x = -5

-|-5| = -5

Back 19

also |x| = -x for all negative x

x=-5

abosolute value of x = 5, 5 = -(-5)

Front 20

what is 66 divided by -33 times |-9|

Back 20

be careful of wording or text as sometimes it is easy to think this problem means


66/(-33*|-9|) but it really means

(66/-33)*|-9|

the answer is -2*9 = -18

Front 21

consecutive integers follow one another without skipping from an even starting point

Back 21

consecutive Integers:

-1,0,1,2

consecutive even

0,2,4

consecutive primes

2,3,5,7...

Front 22

how many integers are between 14 and 765 inclusive

Back 22

procedure:

subtract and add one 765-14+1=752

Front 23

how many factors of seven are there from 8 49 inclusive

Back 23

arithmetic progression formula

an = a1 + (n-1)d

an= nth term >> end number
a1= first term >> first number divisable by factor
n= number of terms
d= common difference >> factor

49 = 14 +(n-1)7
49 = 14 +7n -7
n =6 six multiples

Front 24

How many integers between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5

Back 24

there are (150-100)+1 = 51 numbers between 100 and 150 inclusive.

using Arithmetic Progression formula:

150 = 105 + (n-1)*15 (numbers that are devisable by 3*5) n= 4
150 = 102 + (n-1)*3 (numbers that are devisable by 3) n = 17
150 = 100 + (n-1)*5 (numbers that are devisable by 5) n = 11

since we are looking for numbers that cannot be evenly divided by 3 nor 5, then:

51-17-11+4 = 27

Front 25

sum of consecutive integers (ie 2,3,4... only integers not consecutive even, odd, etc...)

what is the sum of all the integers from 20 too 100 inclusive

Back 25

1: find the middle:

(100+20)=60

2:
find the number of terms:

100-20+1=81

multiply the average by the number of terms:

81*60=4860

Front 26

what is the average term in a consecutive set?

what is te average of all the integers from 20 to 100 inclusive?

Back 26

if you know only first and last term:
(100+20)/2 = 60

if you know the sum of the set and number of terms in the set:

sum of all integers/# of terms = average

4180/81 =60

Front 27

The product of n set of consecutive integers will always have n as a factor.

Back 27

1*2*3 = 6
2*3*4 = 24

both have 3 has factor

5*6*7*8= 1680 is divisible by four

Front 28

Special Sums:

For any set of CONSECUTIVE INTEGERS with an odd number of terms, the sum of the terms is always a multiple of the # of terms

Back 28

1+2+3=6
8+9+10=27

both multiple of 3

4+5+6+7+8=30
13+14+15+16+17=75

both multiples of 5.

THIS APPLIES ONLY TO A SET OF CONSECUTIVE INTEGERS WITH AN ODD NUMBER OF TERMS.

consider three consecutive integers:

n, n+1, n+2
=3n+3 is divisble by 3

4n+6 is not

Front 29

if x^3-x=p and x is even is p divisble by 4?

Back 29

x-1=odd
x=even
x+1=odd

factor the problem

x(x^2-1)
x(x+1)(x-1)
e * o * o
answer:

we know it is even but we can not say for certain it is divisable by 4.

Front 30

list six factors of the product of five consecutive even integers

Back 30

all we can say is that it has a prime box of 5 2's

so

2
2*2=4
2^3=8
2^4=16
2^5=32

Front 31

beware of negative exponents as they can hide the the base

Back 31

x^2=4

x= -2 or 2

Front 32

x^2 = x^6 = x^8

x=?

Back 32

x = 0,1,or -1

Front 33

(3^6)^2=
(x^a)^b

Back 33

3^12
x^(ab)

Front 34

when can you combine exponents?

Back 34

if two expressions have a base in common or an exponent in common you can combine them and ONLY for multiplication and division:

SAme Base:

x^a * x^b = x^a+b
3^2 * 3^3 = 3^5

x^a/x^b = x^a-b
3^3/3^4 =3^-1 = 1/3

Same exponent:
x^a*b^a=xb^a
3^3 * 5^3 = 3*3*3*5*5*5
15^3

x^a/b^a = (x/b)^a
9^3/3^3=
(9*9*9)/(3*3*3)=3*3*3 =3^3

Front 35

When can you combine squareroots

Back 35

during multiplication you can split larger roots into separate factors:

sqrt400 =sqrt(25*16) =5*4=20

During division you can split a larger quotient into dividend and remainder

sqrt(144/16) = sqrt144/sqrt16
=12/4=3

YOU CAN NOT SPLIT OR COMBINE ADDITION/SUBTRACTION

Front 36

Imperfect squares do not yield an integer

Back 36

you can estimate:

sqrt(52) is between sqrt49 and sqrt64 i.e. between 7 and 8

or:

sqrt52 = (sqrt4*13) = 2sqrt13