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Kunal K.

Divisibility - memorize

Shortcuts on divisibility for first 10 integers

What are the rules of divisibility for integers from 1-10?

2 - integer is even

3 - sum of digits is divisible by 3

4 - (a) divisible by 2 TWICE, (b) last two digits are divisible by 4

5 - ends in 0 or 5

6- divisible by both 2 and 3

8 - (a) divisible by 2 THREE TIMES, (b) last three digits are divisible by 8

9 - sum of integer's digits is divisible by 9

10 - integer ends in 0

Divisibility - rule

If you add OR subtract multiples of N, the result is a multiple of N

For instance:

(1) 35+21=56

(2) 35-21=14

(3) n is divisible by 341, and p is divisible by 341, so n+p is divisible by 341. Also, n-p is divisible by 341

(4) If two numbers share a factor, their sum or difference also shares the same factor.

This also works for factorials, e.g. 6! + 21 is divisible by 3. e.g. 10! + 15 is divisible of 15 since 10! has a 5 and a 3, and 15 is divisible by 5 and 3.

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Divisibility rule

If you add OR subtract a multiple of N to a non-multiple of N, the result is always a non-multiple of N

Other related, but less-important rules to be aware of:

(1) if you add two non-multiples of N, the results can go either way: either it's a multiple of N, or a non-multiple of N

(2) The exception to this rule is when N=2 - two odds always sum to an even.

Primes - memorize

What are the first 10 primes?

I should be able to do this quickly...

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Primes - process to figure out if a number is prime

Take 79 as an example:

Step 1: Approximate square root of 79, which is between 8 and 9

Step 2: Determine numbers with which to test - if 79 is composite, it will have a factor less than or equal to 8

Step 3: Evaluate. 2, 3, 4, 5, 6, 7, and 8 are not factors of 79, so 79 is prime.

Primes - prime factorization

How can I determine ALL of the possible factors of some number (integer) that is given to me?

Answer:

(a) break the integer/number down into its prime factors

(b) build all possible products of those prime factors

e.g. 72 = 6 * 12, where 6 = 2*3 while 12 = 2*2*3. Thus, since there are three 2's that appear in 72's prime factors, then 2*2*2=8 and 8 is a factor of 72!

This is cool since 8 did not "appear" in our previous breakdown with 12 and 6.

Primes

Let's say you are told that N is an integer and its only factors are 8 and 9.

How can you know if N is divisible by 27?

Easy!

8 = 2*2*2

9 = 3*3

27 = 3*3*3

->But I only have 2 3's, but I need 3 3's in order for N to be divisible by 27. Therefore, N is not divisible by 27!

**Note that each prime factor can only be used once when determining subsequent factors!

Primes - pop quiz on prime factors.

Let's say you are given (a) an integer N, and you are also told that (b) N is divisible by 10, and is divisible by 8.

What are the prime factors for N?

10 = 5*2, 8 = 2*2*2

The prime factors are 5, 2, 2, and 2. Note that there are only 3 2's, not 4 2's.

--> The extra 2 can't be counted as a prime factor because I could use one of the 2's I already have to make 10.

Therefore, there are only 3 2's that are DEFINITELY in the prime factorization of N, so the 4th 2 may be redundant (i.e. it may be the same 2).

Therefore, I can't say for sure that 16 is a factor of N.

Divisibility - Factor Foundation rule

The GMAT expects me to know this rule

The rule states: if A is a factor of B, and B is a factor of C, then A is a factor of C

e.g. 72 is divisible by 12 and 6, and 12/6 are each divisible by 2 and 3. Therefore, 2 and 3 are factors of 72.

Multiple/factors terminology

The GMAT will phrase information about divisibility in many many ways, just in an effort to confuse you. All of the following statements mean the exact same thing, so be ready to keep your eyes open, and be ready to translate as needed:

12 is divisible by 3

12 is a multiple of 3

12 = 3n, where n is an integer

12 / 3 is an integer

12 / 3 yields a remainder of 0

3 goes into 12 evenly

3 is a factor of 12, or 3 is a divisor of 12

12/3 yields a remainder of 0

12 items can be shared amongst 3 people so that each person has the same # of items

Primes

Is 1 prime?

No! 1 is not a prime number!

Prime numbers start at 2, not 1!

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Greatest common factor / least common multiple (1/3)

What is the definition of Greatest Common Factor, and for Least common Multiple?

GCF = the largest divisor of two or more integers

LCM = the smallest multiple of two or more integers.

**GCF will always be smaller (or equal) to the starting integers.

**LCM will always be larger (or equal) to the starting integers.

Greatest common factor / least common multiple (2/3) - prime factor columns/table

What's a fast way to compute the GCF and LCM?

I write a table, and put the prime factors on top, and the integers on the side. Then, the middle is filled with # times each prime appears. Try drawing this out!

GCF --> take the LOWEST count (the lowest power) of each prime factor found across ALL the integers (i.e. look in each prime factor column, take smallest #).

LCM --> take the HIGHEST count of each prime factor found across ALL the integers. (i.e. look in each prime factor column, take largest #).

-->What is the GCF and LCM for 12 and 40?

GCF = 2*2=4; LCM = 2^3 * 3 * 5 = 120

-->What is the GCF and LCM for 100, 140, and 250?

GCF = 2*5=10; LCM = 2^2 * 5^3 * 7 = 3500

Greatest common factor / least common multiple (3/3) - Venn diagram method

How do you construct a venn diagram for GCF/LCM problems that can tell you the answer?

(ignore this if the other method makes more sense)

First, write out all the prime factors for one of the numbers in one circles. Then, write the other prime factors in the other circle.

BUT, before you write the second circle, you must remember to follow this rule - if a prime in the first # is found in the second #, then you would count the prime for the first number as the same prime in the second number, and then go on to the next.

GCF = product of factors in overlapping area

LCM = product of all factors in the entire diagram

Greatest common factor / least common multiple - ADVANCED

(1) For integers M and N, GCF × LCM = M × N.

(2) for M and N --> GCF ≤ (M-N)

(3) Consecutive multiples of N have a GCF of N.

Explanations:

(1) The product of GCM and LCM is the result of multiplying all factors together (no factors are omitted)

(2) A better way to think about it is that the difference between M and N has to be bigger than the GCF. Otherwise, you start at M, add the GCF, and you overshoot for N (which renders the GCF, not a GCF). Also, if M-N > GCF, then M-N would be the new GCF!!

(3) Take 8 and 12. They have a common factor of 4. 4 has to be the GCF (can't be 5 or else you overshoot)

GCF/LCF - be careful!

30 = 2*3*5

24 = 2*2*2*3

When calculating GCF, you only use the shared factors, so GCF = 2*3

But when calculating LCM, you only use the non-shared factors, so LCM = 2^3 * 3 * 5.

Be careful to not calculate the LCM as LCM = 2^4 * 3^2 * 5 ---> This calculation is wrong because you're including the shared factors! The extra 2 and the extra 3 are shared factors and you don't need those to calculate the LCM.

That's why the product of 30 and 24 is the product of the GCF and LCM!!

Keep in mind that the Venn diagram doesn't make this so clear...

Divisibility - advanced - pop quiz

Is p divisible by 168, given the fact that p is divisible by 12 and divisible by 14?

Answer is that you can't tell!

168 = 2*2*2*3*7

Prime factorization of 12 is 2*2*3.

Prime factorization of 14 is 2*7.

The product of all these primes is 168...

BUT!!! All I mentioned is that p is divisible by 12 and 14. It is possible that one of the 2's in the factorization of 12 and 14 is a REPEAT. Therefore, you are short one of the 2's needed to make 168. Thus u can't conclude!

Divisibility question

Is the integer z divisible by 6?

(1) The GCF of z and 12 is 3

(2) the GCF of z and 15 is 15

How would you solve this? Whats the answer?

To solve this, I must set up a prime table:

Number -- 2 -- 3

12 -- 2^2 -- 3^1

z -- ? -- 3^1

B/c GCF looks at the smallest count of prime factors, I know from the table I set up that there's 0 2's and a 3. Therefore, there's no 2, so statement 1 is SUFFICIENT..

For 15, you see that there's a 3, but 2 is never factored in. Thus I have no idea if 2 appears or doesn't appear in z. Therefore, statement 2 is INSUFFICIENT.

Divisibility question

If LCM of a and 12 is 36, what are possible values of a?

I will need a prime factor table for this one.

If I draw one out, I'll see that there are no more than 2 3's in a (since LCM is 36). But, I'll also know that there are no more than 2 2's in a (since LCM is 36).

Therefore, a is either 9, 18, or 36.

Counting factors and primes - advanced

The Gmat can ask 3 things:

(1) how many different prime factors / unique prime factors?

(2) How many total prime factors / what is the length?

(3) How many total factors?

---> What would the answers look like for 1400 (=2^{3} × 5^{2} × 7)

(1) 3 because there are 3 distinct/unique factors - 2, 5, 7)

(2) The length is 6 because there are 6 numbers that need to be multiplied to give your original #. Or, you can simply sum the exponents on the prime factors.

(3) Combinatorics method - add 1 to the exponent of each prime factor. Then, multiply exponents together.

With this approach, you're really just answering "how many ways can I choose a 2, and how many ways can I choose a 5?" Remember: Total # decisions possible = product of # individual decisions.

Number of factors

Most #'s have even number of factors because they are part of a pair that multiply together to get the original number (e.g. 7x6 = 42)...

But you get an odd # of factors when you are factoring a number that is a square of an integer. This is because the pair is the square root to itself. (e.g. 16 has 5 factors).

Another special case is the square of a prime number. These numbers always have 3 factors - 1, itself, and the prime number. (e.g. 25 has factors of 1, 5, and 25).

Basic math definitions - dividing

Dividend, divisor, quotient, and remainder (in the context of 8/5)

Dividend = the number being divided (8)

Divisor = the number that is dividing (5)

Quotient = the number of times that a divisor goes into the dividend COMPLETELY. The quotient is ALWAYS an integer. (1)

Remainder = what's left over. (3)

Basic rule on dividing

For the number 5.70:

0.70 = remainder / divisor.

Remainders

There are 3 ways to describe remainders:

(a) as integers

(b) as fractions (remainder / divisor)

(c) as decimals (remainder / divisor)

You MUST know how these different forms of remainders are connected to one another.

Problem: When positive integer A is divided by positive integer B, the result is 4.35. What could be the remainder when A is divided by B?

Step 1: Since A/B = 4.35, then the remainder R divided by the divisor B = 0.35, or R/B = 0.35.

Step 2: R/B = 35/100 = 7/20

Step 3: 7B = 20R

Step 4: The prime factors on both sides of the equation must be equal. B must be a multiple of 20, and R must be a multiple of 7. So R can be 7, 14, 21, 28, 35, etc.!!

Remainders

When you need to create numbers that give you a certain remainder, use the following formula.

E.g. When positive integer n is divided by 7, the remainder is 2.

Dividend = Quotient × Divisor + Remainder

n = (integer) × 7 + 2

Thus, the dividend is 2, 9, 16, 23, etc.

On the GMAT, I will need to be able to generate these dividends quickly!!!

Arithmetic with remainders - advanced / excessive

When you divide an integer by a positive integer N, the possible remainders range from 0 to (N-1).

This means that there are N possible remainders.

If a ÷ b yields a remainder of 5, and c ÷ d yields a remainder of 8, then what is the smallest value possible for b+d?

15

Arithmetic with remainders - advanced / excessive

There are two tricks when dealing with arithmetic operations with remainders, just so long as you have the SAME DIVISOR throughout

(1) You can add/subtract remainders directly, just as long as you correct excess remainders at the end

e.g. so if you get a remainder of 4 and 6 when dividing x/7 and y/7, then (x+y) / 7 gives a remainder of 3

(2) You can multiply remainders as long as you correct excess remainders at the end

In the above case, x*y / 7 gives a remainder of 3

Odds/even rules

Basic rules for odd and even numbers when performing basic math calculations (1/2)

ADDITION/SUBTRACTION

EVEN Numbers result when you add/subtract TWO ODDS, or TWO EVENS

ODD Numbers result when you add/subtract ONE ODD WITH ONE EVEN

Odds/even rules

Basic rules for odd and even numbers when performing basic math calculations (2/2)

MULTIPLICATION

EVEN Numbers result when you multiply any set of numbers together AS LONG AS <<AT LEAST ONE>> of those numbers is EVEN

ODD Numbers result when you multiply any set of numbers together AS LONG AS <<NONE>> of those numbers is EVEN (i.e. all numbers are ODD)

Summary of odds/evens rules

Odd ± Even = ODD

Odd ± Odd = EVEN

Even ± Even = EVEN

Odd × Odd = ODD

Even × Odd = EVEN

Even × Even = EVEN (and divisible by 4 -- can you figure out why?)

eDivisibility rules

There are very few guaranteed outcomes in division. Many outcomes are possible.

For the most part, there are no rules/guaranteed outcomes in division.

That being said, there's just 1-2 rules (or better yet, "guidelines") to be aware of...

1. Odd ÷ any integer ≠ any even integer

2. Odd ÷ even number ≠ an integer -- the reason why this is true has to do with the fact that a factor of 2 is "concealed" in the even number

Even ÷ Even = Even or Odd -- can't be determined!!**

Odd ÷ Odd = Odd**

Even ÷ Odd = Even**

** Doesn't have to be an integer

Odds/evens

When dealing with an odd number, it would greatly help me out if I knew how to algebraically write an odd number.

How can I algebraically express any odd number?

Hint: all even numbers are divisible by 2.

So, ANY arbitrary even number can be written as 2N, where N is any given integer.

Then, any odd number is either:

2n+1

2N-1

If you're testing odd and even cases, then be structured and make a table to keep track of different scenarios where one variable is odd and the other is even!

This will prevent me from missing problems!

Thus, be systematic!

Positives/negatives - quick tip

When I need to plug in variables to solve a problem, there are 7 numbers that I should try:

N>1

N=1

0 < N < 1

N=0

-1 < N < 0

N= -1

N < -1

Sum of two primes

What are some rules to know about the sum of two primes?

(1) If the sum of two primes is odd, then ONE of the numbers HAS to be 2

(2) If the sum of two primes is even, then NONE of the numbers is 2

Factorials to memorize

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 720

Combinatorics 101

All combinatorics questions I will encounter fall into 3 categories

1. Making decisions (how many possibilities?)

2. Arranging groups (how many ways can this be done?)

3. A combination of these two categories

I must break all problems down into these two components:

(a) am I making (i.e. arranging) groups? If so, then use combinatorics

(b) am I making decisions? if so, then add/multiply them together?

Breaking problems down like this will help me nail these questions!

Combinatorics - Basic rules of thumb

This is only applicable when I"m merely counting the # of possibilities

1. The word "or" means "add" (decision 1 OR decision 2)

e.g. a side dish of soup or salad; an entree of chicken, bacon, or steak

2. The word "and" means "multiply" (decision 1 AND decision 2); e.g. choose an entree and a side dish

3. It doesn't matter if decisions are made sequentially or simultaneously. You still treat the decision the same way!

e.g. (steak OR chicken OR salmon) AND (soup OR salad) = (1 + 1 + 1) x (1 + 1) = 6 possible set of options.

Combinatorics - Basic rules of thumb

This is only applicable when you are arranging groups, AND every member of the group is UNIQUE

The # ways to arrange N distinct objects, if there are no restrictions, is N!

This might apply when you are dealing with, e.g. figuring out how many ways there are to arrange 4 people in 4 chairs in a row?

Combinatorics - Basic rules of thumb

This is only applicable when you are arranging groups, AND some/all members of the group is treated as if they are IDENTICAL

Case Question: How many arrangements are there of the letters in the word "Eel?"

1. The # ways to arrange N distinct objects, if there ARE NO restrictions, is N!

2. Since there ARE restrictions on N, then divide the total # arrangements by M! (where M is the # of identical objects, i.e. objects that can't be distinguished)

3. Rule #2 applies for EACH SET OF IDENTICAL NUMBERS. Therefore, it's easily possible that there are multiple M's to divide N! by (see following flash card)

Combinatorics - how to approach a specific type of problem

Tonight and tonight only, Darin has 20 white girls who are willing to stick their finger up his butt and penetrate his "he-spot." However, Darin can only hook up with a maximum of 7 girls tonight. How many different groups of women could turn Darin into what he is so close to being - a woman?

These problems are common when selecting people who are either in or out of some group.

You can rationalize the answer using the anagram method (i.e. Y's and N's). There are 20 spots, with 7 Yes' and 13 No's. Thus, the answer is 20! ÷ (7! 13!)

---> You need to divide by 13! to account for repeats!!

If confused as to why, then think of it like this: only 7 girls will get to do action A, and 13 girls will get to do action B. Thus, there are 2 sets of identical members!!

Combinatorics - general tip on questions (1/3)

When I am dealing with a question where I am arranging members of a set, and where I am counting the # of possibilities, then the best practice is to break the problem down, then proceed about it per my normal strategies for combinatorics questions.

Warning: Be careful about when I need to MULTIPLY (decision 1 AND decision 2) and when I need to ADD (either decision 1 OR decision 2)

Combinatorics - general tip on questions (2/3)

Mo Pete the Slamma' is picking bitches to fuck. There are 7 bitches to fuck - Indian, Asian, American, European, African, Australian, and Latino. How many different combos of bitches are possible for Mo Pete's fuckfest if he can only select at most 2 bitches?

This question involves multiple decisions. Since there are "at most" 2 bitches that Mo Pete can select, he can select 1 or 2 bitches.

Thus, split the problem into these components:

1 bitch = 7! / (1! × 6!) = 7

2 bitches = 7! / (2! × 5!) = 21

Since you can select 1 OR 2, you add these up (OR means ADD).

So, there are 21 + 7 = 28 possible combos of bitches!

Combinatorics - general tip on questions (3/3)

Mo Pete the Slamma' must cap 3 of 5 Bloodz members, and 2 of 7 Cripz members. How many different cappings is possible?

Note that Mo Pete has 2 separate decisions to make. He is capping 3 Bloodz's and 2 Cripz's. Therefore, this is AND so you will multiply the possibilities.

Bloodz: 5! / (3! × 2!) = 10

Cripz: 7! / (5! × 2!) = 21

--> 21 × 10 = 210 possibile combinations of cappings

Combinatorics - shortcut / tricks

When I'm working a combinatorics problem, I can make it easy to think about + understand by doing one simple trick.

Turn it into a question of "how many anagrams are possible"

Demonstration - I have a bin in which to blend 2 ingredients, and there are 4 possible ingredients. How many possibilities are there?

---> I'm essentially asking "how many ways can I rearrange the letters in the word YYNN?"

This greatly simplifies my thinking of this problem.

4 letters + 2 repeating sets --> 4! ÷ (2! × 2!)

Probability

Probability and combinatorics are similar in the way they treat the word "And" and "Or"

In probability, as well as in combinatorics, the word "AND" means "MULTIPLY," and the word "OR" means "ADD."

+ For example, what is the probability that 2 tosses of a coin yields heads both times?

+ For example, if there's a 40% chance of sun, 25% chance for rain, and 35% chance for hail, what is the probability that it hails or rains today?

If I am having trouble breaking down a probability problem, then try breaking the problem down into a series of "AND" and "OR" events.

Probability

The only trick that you need to know for probability is as follows:

This might seem obvious, but sometimes it's faster + easier for me to calculate the probability of the event NOT happening, and then subtract that from 1 to get your answer.

Keep this strategy in mind when working probability problems. I can approach it 1 of 2 ways:

+ The probability of something happening

+ The probability of something NOT happening

General tip. - advanced

Recognizing combinatorics problems can be tricky. They may seem like they have little to do with combinatorics, and instead have to do with probability. The combinatorics comes into play when counting #possibilities.

One way to avoid getting confused by these problems during the GMAT is to recognize what a combinatorics problem looks like.

Keep in mind that combinatorics will probably be involved in a question that asks "How Many?..."

--How many 4-digit integers have digits with xyz property?

--How many paths exist from point A to point B in a given diagram?

--How many diagonals, triangles, lines, etc. exist in a given geometrical figure?

-How many pairings (eg handshakes, flights) exist?

Example on combinatorics - advanced

Bitch be walking on a grid system from her crib, which is on a corner, to her man's crib, which is 3 blocks south and 3 blocks east. How many possible routes can she take?

Bitch is simply making a decision to walk south for one block 3 times, and walk east for one block 3 times.

I'm arranging 6 decisions, but there are repeats. Therefore, the answer is: 6! ÷ (3! × 3!)

Alternatively, I could have used the anagram method!! (YYYNNN)

Combinatorics - advanced

Let's say I have 6 people sitting down. I have 6! ways that they can sit down in 6 chairs. BUT, I have a constraint - two of them don't want to sit next to each other. How do I figure out the total # possible arrangements?

Hint: Treat the item "stuck together" as one larger item.

This means that there are 5! ways for everyone to sit down so that the two people are sitting next to each other.

BUT, remember that the two people could simply swap places with one another (instead of Bob-Janet, it's Janet-Bob). Therefore, there's 2 × 5! possibilities.

Thus, answer is 6! - (2 × 5!)

Combinatorics and "Domino Effect" - this is about making questions easier when there's symmetry

Domino effect occurs when the probability of one event affects the probability of a subsequent event

Question: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses 3 gumballs at random, what's the probability that it dispenses one gumball of each color?

This problem is tricky at first because it involves the domino effect, and because there's many possibilities.

Let's say you calculate probability 1-by-1.

Step 1: calculate for BGR: (7/16) × (5/15) × (4/14) = 1/24

Step 2: calculate for GRB: (5/16) × (4/15) × (7/14) = 1/24

--> Notice that we get the same probability for each sequence...that's because the order doesn't matter!

Step 3: Use combinatorics to figure out how many sequences there are in total: 3! = 6 different cases

Step 4: 6 × (1 / 24) = 1 / 4 -- final answer!

Perfect squares, cubes, etc. -- ADVANCED

The GMAT occasionally tests perfect squares, cubes, etc.

One special property is that all perfect squares have an odd # of total, distinct factors.

Also, an integer with an odd # of total factors MUST be a perfect square.

Also, all other non-square integers have en even # of factors.

Also, any number which is a perfect square contains an even power of primes (i.e. the exponent in the prime factorization are all even).

e.g. 144 = 2^4 × 3^2. That's because perfect squares are formed from the product of two copies of the same prime factors (here, one "copy" would be 2^2 × 3)

If you see an odd prime factorization, then it isn't a perfect square.

Perfect cubes

Question: if k^3 is divisible by 240, then what is the smallest possible value of integer k?

12, 30, 60, 90, or 120??

Step 1: k^3 = k × k × k

Step 2: 240 = 2^4 × 3 × 5

Step 3: Recognize that the prime factors of k^3 should be the prime factors of k appearing in sets of 3

Step 4: Distribute the prime factors of k^3 into 3 columns/groups, where each group is the number k

Step 5: Trick - notice how you have 4 2's. This means that k is made of 2 2's.

Step 6: Thus, k is 2 × 2 × 3 × 5.

Step 7: Answer is 60!

Counting total factors

How many total factors are there in 2000?

2000 = 2^4 * 5^3

For a prime factor with Nth power, there are N+1 possibilities for the occurrence of that prime factor.

The Fundamental Counting Principle says that if you must make a number of separate decisions, then multiply the number of ways to make each individual decision to find the number of ways to make ALL the decisions.

--> Thus, 5 * 4 = 20.

--> Same process goes when there's >2 prime factors

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