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36 Cards in this Set
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Divisibilty Rules:
An integer is divisible by: 
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 

zero is

a multiple of every integer


a factor is a

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


a multiple of an integer is

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


Sum and Difference Rule

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 6440=24 and 64+40=104


y=xq+r

104=8*13
24=8*3 y=dividend x=divisor q=quotient r=remainder 

primes

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


factor formula:
how many factors does 18 have 
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 

factor foundation rule:

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) 

Is 27 a factor of 72?

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 

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

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 

Lowest Common Multiple

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 

Greatest Common Factor:

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 

What is the GCF/LCM of 24 and 30

24=
2^3, 3 30= 2,3,5 GCF = 2*3=6 LCM = 2^3*3*5=120 

terminating decimals

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 

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

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. 

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

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 nm


ODDS AND EVENS

Odd +/ Odd = Even
Odd +/ Even = Odd Even +/ Odd = odd. Even +/ Even = Even Odd X Odd = Odd Odd X Even = Even Even X Even = Even 

x = x for all x <=0
for x = 5 5 = 5 
also x = x for all negative x
x=5 abosolute value of x = 5, 5 = (5) 

what is 66 divided by 33 times 9

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 

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

consecutive Integers:
1,0,1,2 consecutive even 0,2,4 consecutive primes 2,3,5,7... 

how many integers are between 14 and 765 inclusive

procedure:
subtract and add one 76514+1=752 

how many factors of seven are there from 8 49 inclusive

arithmetic progression formula
an = a1 + (n1)d an= nth term >> end number a1= first term >> first number divisable by factor n= number of terms d= common difference >> factor 49 = 14 +(n1)7 49 = 14 +7n 7 n =6 six multiples 

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

there are (150100)+1 = 51 numbers between 100 and 150 inclusive.
using Arithmetic Progression formula: 150 = 105 + (n1)*15 (numbers that are devisable by 3*5) n= 4 150 = 102 + (n1)*3 (numbers that are devisable by 3) n = 17 150 = 100 + (n1)*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: 511711+4 = 27 

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 
1: find the middle:
(100+20)=60 2: find the number of terms: 10020+1=81 multiply the average by the number of terms: 81*60=4860 

what is the average term in a consecutive set?
what is te average of all the integers from 20 to 100 inclusive? 
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 

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

1*2*3 = 6
2*3*4 = 24 both have 3 has factor 5*6*7*8= 1680 is divisible by four 

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 
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 

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

x1=odd
x=even x+1=odd factor the problem x(x^21) x(x+1)(x1) e * o * o answer: we know it is even but we can not say for certain it is divisable by 4. 

list six factors of the product of five consecutive even integers

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 

beware of negative exponents as they can hide the the base

x^2=4
x= 2 or 2 

x^2 = x^6 = x^8
x=? 
x = 0,1,or 1


(3^6)^2=
(x^a)^b 
3^12
x^(ab) 

when can you combine exponents?

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^ab 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 

When can you combine squareroots

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 

Imperfect squares do not yield an integer

you can estimate:
sqrt(52) is between sqrt49 and sqrt64 i.e. between 7 and 8 or: sqrt52 = (sqrt4*13) = 2sqrt13 