BNR_GMAT_QUANT_#PROPERTIES&FRACT.DECIMAL&%'S.

rules of divisiblity 4 certain integers

#3: if sum if the integers DIGITS is divis by 3. example: 681. #4: if integer is divisble by 2 twice OR if the last TWO digits are divisible by 4. eg: 781964 is divisible by 4 b/c 64 is divisible by 4.... #5- when the number ends in 0 or 5 ONLY. #6: if integer's divis by both 2 AND 3. #7-no easy way. #8- if integer is divis by 2 three times or if its last three digits are divisble 8 example: 2120. #9- if the sum of the integers digits is divisble by 9 eg: 792... 10 (obvious) number properties book

1. / 9.

.1111 or 11.1%

1. / 6.

.167 or 16.7%

prime #s to know

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 number properties book

counting factors and primes

GMAT may ask how many diff prime factors or how many prime total factors (aka length) or how many total factors. For DFP: count each repated prime factor only once... for TPF: length is the # of primes whose product is the # theyr asking you about. Add exponents of the prime factors. if no exponent, count it as 1. total factors: includes all factors, not just prime ones. number properties book

When integer X is divided by integer Y, the remainder is 9. if x/y=96.12 what is y?

Set the remainder over the divisor = to the decimal (or fraction) so 9/y=.12 therefore Y=75.

term

definition number properties book

multiples

a # formed by multiplying an integer by any integer so 24 and 32 are multiples of 8 for xample. number properties book

perfect squares & cubes etc

all perf squares have an odd # of factors and any integer that has an odd # of total factors must b a perfect square. All other non square integers have an even # of factors. This is b/c all squares and cubes have 1 # that is a factor more than once, so the # of factors is odd. pereects squares are formed from the prdouct of 2 copies of the same prime factors. aka 90^2 is just (2x3^2x5)^2 aka 2^2x3^4x5^2. thus, prime factorization of a perf square contains only even powers of primes. also any # whose prime factorization conatins only even powers of primes must be a perfect square. if a numbers prime factorization contains any odd poewrs of primes, then the # isnt a perf suare. same logic extends to pefect cubes and other perfect powers. perfect cubes are formed from 3 identical sets of primes, so all the poers of primes are multiples of 3 in a factorization of a perfect cube number properties book

prime factorization

all the prime factors of a #. ex: 96 - (48,2) breaks down to (24,2) which goes to (12,2) whcih goes to (6,2) which goes to (3,2) so 96=2x3x2x2x2x2. big tool on gmat since when you get the prime factors of a # you can get all its factors even for the biggest of #s. also helps to determine if one # is divisible by another. helps you determine the GCF of 2 #s. fraction reduction & finding the LCD, simplifying square roots and determing the exponent on 1 side of an equation if there are integer constraints.... ex of how it it helps: given that integer N is divisible by 5, 11 and 13, what other #s r divisors of N? answer: 55, 65, 143 & 721 number properties book

combinatorics (words and/or)

and means mulitiply, or means add....when it decision 1 or decision 2 you add, when its dec 1 and dec 2 you multiply. ex: you must create a 5 digit lock code the first and last digit must be odd how many combos? digit 1: 5 choices, digit 2: 8 choices. digit 3: 7 digit 4:6 and last 1 4 choices. so 8x7x6x5x4 is 6720. number properties book

sums of CI's and divisbility

any set of CI's w/ odd # of terms, the sum is always a multiple of the # of items (eg: 4+5+6+7+8=30 and 30 is a multiple of 5 and 6)... but w/ even # of terms, its NEVER a multiple of the # of items....use prime boxes to keep track of factors for consecutive integers. number properties book

1-x probability trick

calculating likelihood that something doesnt occur b/c its easier to caclulate that than calculating the opposite (The event that does occur) since that event is the majority occurrence. e.g. what is probablity that on four rolls of a dice, at least 1 roll will be a 4? calculate the odds that NO roles would be a 4 which is 5/6*5/6*5/6*5/6 aka 625/1296 so odds 1 role would be a 4 are: 671/1296. number properties book

multiplying remainders

can be done as long as you correct 4 excess ones at the end. So if x has a remainder of 4 when divided by 7 and z has remainder of 5 when divided by 7, 4x5=20, so then 7 goes into 20 2r6, so remainder = 6. number properties book

arrangements w/ constraints

complex issues where say 1 person won't sit next to another. Like if G,M,P,J,B,C go to a movie and sit next to each other but if M&J wont sit nxt to each other, how many possible arrangements are there? So w/out contraints its 6! Or 720, but since you have constraints you need to use the glue method. glue method is 4 problems in which items or ppl must be next to each other, pretend the items are stuck together and are actually 1 large item. since theyre "stuck together" you treat it as a 5!. however, since it can be m-j or j-m its 5!x2 so 240. number properties book

interest formulas

compound interest= P( 1 + r/n) ^ nt

testing odd/even cases

create grid with scenarios for each variable and what would have to occur for each to be odd or even. Once you create the grid you can see which variables need to be odd/even (or neg/pos) to make the question stick as true. number properties book

remainder terms

dividend: in 8/5 8 is dividend. divisor in that # is 5. the quotient is the # of times the divisor goes into the dividend. in this case its 1. the remainder is whats left over so 8/5 is 1R3. examp: when positive integer A is divided by positive integer B, result is 4.35 what could be the remainder?: A/B means B goes into A 4.35 times. so 4=quotient and R=.35 Remainder/divisor=.35... other egs: when N is divided by 7, remainder is 2. this means n/7= a quotient w/ r 2. relationship: dividend=quotientxdivisor + remainder. number properties book

odds/evens w/ division

even/even can be o/e/NI. even/odd can be even/NI but NOT odd. odd/even is NEVER an integer. odd/odd can be odd or NI> number properties book

disguised combinatorics

ex of it: how many 4 digit integers have digits w/ some specificed properties, or how many paths from point A to point B in a given diagram, or how many pairings etc. perfect example: if alicia lives on UES grid and her school is 4 blocks north and 4 blocks west of a her house, and she only walks north OR south on her way to school, how many possible routes can she take to school? easy so she has to walk 8 blocks but you have 2 groups so its 8!/(4!)(4!) so its 8x7x6x5x4x3x2x1 / 4x3x2x1 (4x3x2x1) and obviously the 4321's cancel and then for the left over one, simplify 4 into 8 3 into 6 and calculate. number properties book

when determining the # of things in an order

find the diff of last term - first term and add 1. (ticket at the deli, you took the #25 when getting online, there are 24 people ahead of you, how many people are on the line? last person online is you #25, first person in line is #1, 25-1 + 1 = 25.)

pos/neg disguised as inequalities:

if (a-b)/c <0 is a>b? 1. c<0 2:a+b<0... term 1: if c is less than zero and the whole term is less than zero that means a-b is a positive # so then A is greater than B. term 2. in sufficient b/c C could also be negative. number properties book

counting total factors

if asked say how many total factors does 2000 have? That takes quite a while, but you can do the prime factors which are 2^4 & 5^3. so the PF contains four 2's, this means there are 5 possibilities for the # of 2s in any factor of 2000 (0-4 inclusive) then for 5, there are 3 5's, so 4 possibilties for the # of 5s (0-3 inclusive).. thus, any # w/ more than four 2s in its prime box cant be a factor of 2000. so in general, if a PF appears to the Nth power, then there are (N+1) possibilities for the occurences of that prime factor. this is true for each of the individual prime factors of any #. additionally, you can borrow from combinatorics called FCP to simplify the calc of the # of prime factors in 2000. since FCP states that if you must make a # of sepaarete decisions, then multiple the # of ways to make each individual decision to find the # of ways to make all the decisions. the # of 2s & # of 5s to iclude in a factor of 2000 are 2 individual decisions you must make. these 2 choices are indy of each other so the total # of factors of 2k must be (4+1)(3+1)=5x4=20. number properties book

divisibliity & addition/subtraction

if you add a multiple of N to a non-multiple of N, the result's always a non multiple of N (and the same is true w/ subtraction). two-non multiples of N added together can produce either a multiple of N or a non-multiple of N. number properties book

probablity

is # of desired or successful outcomes over the # of possible ones. number properties book

GCF

largest divisor of two or more integers. for ex: the GCF of 32 and 40 is 8. number properties book

LCM

least common multiple. smallest mulitple of 2 or more #s. also its the prodcut of all primes b/t 2 #s. the LCM b/t 12 and 40 is 120 b/c it breaks down to 2x2x2x3x5 number properties book

heavy division shortcut

look at #s, see how you can round and simplify etc.

last digit shortcut

multiply out the #s theyre asking for. Then just take the last digit of those #s and multiply them by themselves for ex: 25^2 * 13^2 * 11^2 you get 625 169 and 121 so then just do 5x9x1 and you get 45 so answer is 5.

combinatorics w/ domino effect

not that diff than stuck together. Like if a candy machine has 7 red bears 5 blue and 4 green, what is prob that it dispenses 1 of each color while dispensing 3 at random? (its 7/16 * 5/15 * 4/14 and that gives you 1/24 but b/c you have six diff orders of how they come out, its that # times 6) number properties book

odds/evens

odd+odd & even+even always give you EVEN. o+e=odd. oddxodd=odd. oddxeven=even. evenxeven=even number properties book

percent inc/dec

orignal + change = new. Percent change: change in value / original value... remember don't add successive percents.

benchmark values

recognizing when a fraction is close to another like question what is 10/22 of 5/18 of 2000 just do what is 1/2 of 1/3 of 2000 and you get 333.

arithmetic w/ remainders

remainders could be added and subtracted directly, as long as you correct extra or negative remainders. Extra remainders are those larger or equal to the divisor. To correct them (and neg remainders) just add/subtract the divisor. So if X leaves a remainder of 4 after being divided by 7, and y leaes a remainder of 2 after being divided by 7, x+y leaves a remainder of 6 after being divided by 7. dont pick #s. simply R4+R2=R6.... however if x leaves R4 when divided by 7 and Z leaves R5 when divided by 7, obviosuly adding them together gives you 9, which is too high. so you take 7 out of the excess and in this case R4+R5=R9=R2.... similarly, X leaves R4 when divided by 7 and Z leaves R5, when you subtract you get -1 which you know doesnt work, so when you have a negative you add the divisor back to it so r4-r5=r-1=R6 number properties book

advanced remainders

remember dividend/divisor= quotient plus remainder/divisor... remainders can be vizualized by constructing a remainder ruler aka a number line marked off w/ large and small tick marks. Large tick marks r 4 the multiples that you care about, aka multiples of 7... this way you can see where the remainders lie etc, and you will be able to know that a number with say 3 extra small tick marks is added to one with 2 extra small tick marks, you get 5. these # lines are all positive btw. ex: if a/b ields a remainder of 5, c/d yields a remainder of 8, and a,b,c and d are all aintegers, whats the smallest value of b+d? b/c R is smaller than divisor, five must be less than B and 8 is smaller than D. also D must be an integer so D is at least 9 and B is at least 6 so answer's 15. number properties book

testing neg/pos cases

same as above, create the grid and test/figure out which need to be pos/neg to make question hold up number properties book

factors

same as divisors. those pos integers that divide evenly into an integer eg: 16's factors are 1 16 2 8 and 4. number properties book

to determine sum of consecutive integers

take the average of those integers and multiply that by # of terms in the set (e.g: the sum of all integers 1-29 inclusive: you know you have 29 #s in the set and to get avg its 1+29/2 which is 15, so 15x29=435.)

concrete vs relative values

tend to need more info to solve for concrete values. Questions for relative values can be solved w/ less info provided.if a DS ? Asks 4 the relative value of 2 pieces of a ratio, any statement that gives relative value of ANY 2 pieces of the radio will be sufficient. if a DS ? asks 4 concrete value of 1 element of a ratio you'll need BOTH concrete value of another element of ratio AND the relative value of 2 elements of the ratio.

additional prime # facts

there's an infinite # of prime #s, and there's no simple pattern. Remember # w/ only 2 factors (1 and itself) is always prime and number w/ more than 2 factors is never prime number properties book

multiple groups

treat each group separately then see if its and/or. remember its the total # in the top as factorial and in the bottom its whats picked factorial multiplied by whats not picked factorial. number properties book

advanced GCF/LCM

use prime columns by: calculating prime factors of each integer, creating column for each prime factor found w/in any of the integers, create a row for each integer, in each cell of the table, put prime factor raised to a power, this power counts how many copies o the columns prime factor appear in the prime box of the row's integer. then to calc the GCF, take lowest count of each prime factor ound across all the integers, which ounts the shared primes, to caclulate the least common multiple, take the highest count of each prime factor found across all integers this counts all the primes less the shared primes. smallest count means lowest power in any column. ex problem: is integer Z divis by 6? 1) the GCF of z and 12 is 3 2) GCF of z adn 15 is 15. #1 so you have Z and 12 (2x2x3)..make a prime box (12- 2^2, 3) and z (3 and we dont know what else). both 12 and z have a GCF of 3. 3 has no 2's as a factor, so therefore Z is not divisible by 2. if so, it cant be divisible by 6. therefore, statement 1 is sufficient. statement 2 tells you that z and 15 have a gcf of 15 meaning, both z and 15 have 1 3 and 1 5 in them. this doesnt tell you whetehr Z contains any 2's. Z must contain at least 1 2 and 1 3 in its prime factoirziation for it to be divisble by 6. if z had 2 as a prime factor, we'd be fine but 15 has no 2's. therefore you cant tell whether Z has a 2 in its prime factorization or not. another ex: if the LCM of a and 12 is 36, what are the possible values of A? so first A cant b bigger than 36, so max value is 36. then factor out 36 and you have 2^2 and 3^2. so the LCM of 12 and A both contain 2 2's. since rules of LCM's state that they contain each prime factor to the power that it appears most, and since 12 also has 2 2's, A can't have more than 2 2's, but doesn't need any so its 0 1 or 2 2s. then, the LCM of 12 and a both has two 3's, though 12 only has 1 three. since 12 has 1 three and 36 has 2 3's, you know that A contains exactly 2 3's. since a must have 2 3's and anywhere from 0-2 2's, A could be 9, 18 or 36. number properties book

arranging groups

w/out restrictions for N distinct objects simply n!. number properties book

smart numbers

when given fractions w/ diff denominators, use LCD or a common denominator of the 2. but if 1 concrete value is given you can skip smart #s and go off of that. For smart #s, pick 100 for unspecified value amount and work from there. Best way to success on GMAT is always pick 100

powers and roots

when raising a decimal to a larger power you can always rewirte decimal as product of the integer and a power of 10. such .5^4 ... rewrite .5 as 5 x 10^-1, then take that to the 4th. So you have 5^4 x 10^-4 which you then get 625 * 10^-4 aka 0.0625... or if theyre asking for the cube root of .000027 thats the same as (27 x 10^-6)^1/3... ends up as .03.

arranging groups w/ repetition

when you have members of a group that are identical its total # of arrangements! over m! so if you have 6 cars, but 4 are white and you want to make groups of 6 its 6!/4!....can also use an anagram grid where the # of columns in the grid is always = to the # of members of the group. this is very helpful. like if you had 9 ppl in a bike rice and the winner gets gold, the next 3 get silver the next 4 get bronze and last place gets brass its 9!/4!*3! which simplifies to 9*8*7*6*5/3*2*1 and you can break that down further too. so THE # OF WAYS OF ARRANGING N DISCINTCT OBJECTS, W/ NO RESTIRCITIONS IS N FACTORIAL. IF M MEMBERS OF GROUP ARE IDENTICAL, ITS N FACTORIAL OVER M FACTORIAL. number properties book