Q3) Number properties

Question 1

What are “whole numbers”?

Answer 1

Non-negative integers aka 0 and positive integers

Question 2

What is 0^2 equal to?

Answer 2

0

Concept: Zero raised to ANY POSITIVE power is zero

Question 3

Does -0 exist?

Answer 3

Yes

Concept: Zero is the only number that is equal to its opposite

0 = -0

Question 4

Unique properties of the number 0

Answer 4

1) Zero is the only number that is neither positive nor negative

**2) Zero is the only number that is equal to its opposite (0 = -1)

3) All numbers are factors of zero / zero is a multiple of all numbers (reverse of property of one)
> zero can be a factor to itself
> HOWEVER, usually GMAT will ask you to find the factors of a POSITIVE INTEGER and NOT 0 (infinite number of factors of 0)
> 0 is the first nonnegative multiple

**4) Zero is the only number that is equal to all its multiples

**5) Any number (except zero) raised to the zero power is equal 1 (NOT zero)

6) Zero is considered an EVEN number

Question 5

Unique properties of 1

Answer 5

1) One is a factor of all numbers and all numbers are multiples of one (reverse of property of zero)

2) One is the only number with exactly 1 factor (not even zero can have this property)

3) 1 is NOT a prime number (recall that the first prime number is 2)

Question 6

Can decimals be even or odd?

Answer 6

No

Concept: All INTEGERS are even or odd (incl. zero)
> even = integer is divisible by 2 without remainder, therefore all even integers have even units digits (0, 2, 4, 6, 8) and all odd integers have odd units digits (1, 3, 5, 7, 9)

Question 7

How do you express even and odd integers in mathematical expression?

Answer 7

Even: 2n
Odd: 2n+1 or 2n-1

Question 8

What are the even / odd addition, subtraction, multiplication, and division rules?

Answer 8

Addition and Subtraction follow the same rules:
MUST BE BOTH EVEN or BOTH ODD to be even
E +/- E = E
O +/- O = E
O +/- E = O

*in other words: if two integers are EQUAL to the absolute value of each other, then the sum or difference will be EVEN

e.g., if | x | = | y |, sum or difference is EVEN (incl. 0) because x and y are either both even or both odd (sign does NOT impact even or odd)

Multiplication: If one number in the product is even, the whole product is even
Remember acronyms EEE, EOE, OOO
E * E = E
E * O = E
O * O = O

Division: Many rules
Remember acronyms EOE, OOO
O/E –> Not integer
E / E –> E or O
E / O –> E
O / O –> O

Question 9

What is the remainder of odd / 2?

Answer 9

Always 1

Question 10

What is the meaning of an absolute value?

Answer 10

Basically asking how far away is n from 0 on the real number line

Question 11

When you see exponents + variables, what type of concept might be tested?

Answer 11

Even and odd exponents versus positive and negative answers

Aka how do exponents impact the SIGN of numbers

Question 12

Formulaic expression of factor (divisor), k

Answer 12

*** If k is a FACTOR of positive integer x, then 1 <= k <= x

> factors of positive integer –> smallest factor is 1 and the largest factor is ITSELF

**> Also x / k = integer

DEALING WITH POSITIVE INTEGERS

Question 13

What is the definition of a multiple of an integer? What is the formulaic expression of a multiple, x?

Answer 13

A multiple is the PRODUCT of an INTEGER and any other integer

x is a multiple of a if and only if: x = a*n

Also means that x / a = integer n

e.g., multiples of 5 = 5n, where n is a non-negative INTEGER
y = nx

DEALING WITH NON-NEGATIVE INTEGERS

Question 14

Memorize: What are the first 25 prime numbers

Answer 14

First 10: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Next 10: 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
Next 5: 73, 79, 83, 89, 97

> 2 is the only even prime (all other even numbers have 2 as a factor)
Integers ending in 5 will have 5 as a factor
Most likely candidates for primes have unit digits of 1, 3, 7, and 9 —> DOUBLE CHECK not divisible by 3 or 7

91 is NOT a prime because it is a multiple of 7

Question 15

What is the formulaic expression of prime factorization of a number, x?

Answer 15

x = (prime number)^# * (prime number)^#

DEALING WITH POSITIVE INTEGERS

Question 16

** How do you calculate the TOTAL NUMBER OF FACTORS of a particular number?

Answer 16

1) Find the prime factorization
2) # of factors = (1 + exponent) * (1 + exponent) …
> add 1 to the value of each exponent
> then, multiply these results

Watch out: Don’t forget about exponents of 1

Question 17

What is the difference between “number of prime factors” vs “number of unique prime factors” vs “sum of prime factors”

Answer 17

Number of prime factors = total number of individual prime factors (disregard whether it is unique prime or not)
> in prime factorization form, simply add up all the exponents

Number of unique prime factors = number of prime factors that are different from each other

Both Qs differ from “what is the SUM of all the prime factors of X” –> add up the individual prime factors (e.g, 2^4 = 2 + 2 + 2 + 2 = 2*4 = 8)

Question 18

Will raising a number to a positive exponent change the number of unique prime factors that number has?

Answer 18

No

If some number x has y unique prime factors, then x^n (where n is a POSITIVE integer) will have the SAME y unique prime factors

In other words, raising a number to a POSITIVE EXPONENT does NOT change its number of unique prime factors

e.g., 5^2 vs 5^4 –> both have only 1 unique prime factor

Question 19

What is the fastest way to find the LCM of any set of positive integers?

Answer 19

Concept: LCM includes ALL UNIQUE prime factors across set of integers and we multiply the repeated prime and non-repeated prime factors
> so if a DS question asks whether you know how many unique prime factors there are in product A*B and you know the LCM of A and B, then it is sufficient
> LCM => connected to unique prime factors across set of integers (and therefore the PRODUCT of these integers)

Strategy 1) Prime factorize each integer.
> for each REPEATED prime factor shared by AT LEAST TWO of the numbers in the set, take the one with the LARGEST EXPONENT
> Take all non-repeated prime factors of integers
> Multiply together to get to the least common multiple

Note: a prime factor does NOT need to be shared by all of the numbers in the set to be considered a repeated prime factor

Background - for LCM, you want to make sure you account for ALL the prime factors across the set of positive integers (but don’t need to double count prime factors, so taking the highest power of a repeated prime factor is good)

Strategy 2) Write out all the multiples of each integer until you find the smallest common multiple

Think of a NET (trying to capture everything)

Other ways of referring to LCM:
> “the least possible number of x, given that x is divisible by 20 and 30”

Question 20

When will the LCM be equal to the product of two positive integers?

Answer 20

Only when those two integers share NO common prime factors (no duplication)

Question 21

What is the fastest way to find the GCF of any set of positive integers?

Answer 21

> Prime factorize the set of positive integers
Look at only the COMMON (repeated factors for ALL the integers) ** different from in LCM, where you need to have at least two integers share factor
Choose the one with the LOWEST exponent (needs to be common to all numbers)
Multiply together the lowest common prime factors

** IF NO repeated prime factors are found, the GCF is 1 (Not 0)

Think of a venn diagram

Other ways of referring to GCF:
> “largest integer that will divide into these positive integers”
> “largest shared divisor”

DS questions:
> If you know the two integers are CONSECUTIVE (e.g., n and n+1), you ALSO KNOW that the GFC = 1

Question 22

GCF vs LCM

Answer 22

LCM will always be equal to or GREATER than the LARGEST number in the set –> lower bound is the largest number
> tells you all the UNIQUE prime factors in a set

While GCF will always be equal to or LESS than the SMALLEST number in the set –> upper bound is the smallest number

Question 23

Terminology: What does it mean for a number to “divide evenly into x”?

Answer 23

Refers to the DIVISOR or FACTOR

e.g., 4 divides evenly into 12

Divide INTO a LARGER number

Question 24

If it is known that y divides evenly into x, then can you determine what the LCM and GCF are of x and y?

Answer 24

Yes –> means that y is a factor of x (x is a multiple of y)

1 <= y <= x

GCF (bounded by smallest number) = y
LCM (bounded by largest number) = x

Question 25

If you know the LCM and GCF of TWO positive integers, what do you know?

Answer 25

You know the PRODUCT of the two positive integers

This is because GCF and LCM are like two sides of the same coin and together, you cover ALL the prime factors

Recall:
> LCM consists of the largest prime factors + non-repeated prime factors
> GCF consists of the smallest common prime factors

Question 26

Terminology: What does this mean?

x is a dividend of y?

Answer 26

x = numerator
y = divisor

Factor = divisor
Dividend = multiple

Also can express as: x is divisible by y

MATHEMATICALLY: x / y = INT (remainder = 0)

INT often assigned k

Question 27

What should you think of when you see a question regarding divisibility / multiples / factors?

Answer 27

ALWAYS prime factorize both the dividend and divisor
> you can see if there is a remainder

AND turn into PRODUCT of numbers

Question 28

What does factors of factors rule imply?

Answer 28

If x and y are positive integers and x / y is an integer (or y is a factor of x), then x / any factor of y is also an integer

> A positive integer is divisible by all factors of a factor

DEALING WITH POSITIVE NUMBERS

Question 29

How to express this mathematically?
Is x a multiple of both 2 and 3?

Answer 29

Is x = 23m, where m is a non-negative integer?\

Or 2 and 3 are both factors of x

Question 30

Divisibility and LCM rule – what must be true in order to conclude that an integer z is divisible by LCM of two numbers?

Answer 30

If z is divisible by BOTH of the two numbers, then z must also be divisible by the LCM of the two numbers (which is NOT always the product of the two numbers because of overlapping factors)

Alternatively: z is divisible by the LCM of its factors

Please note: It is NOT always true that z is divisible by the PRODUCT of its factors because of overlapping factors

e.g., z is divisible by 15 and 20
The smallest value of z = LCM of 15 and 20 = 60, which does NOT equal the product 300

Question 31

Memorize list of divisibility properties

Answer 31

1) Factor of factors: A positive integer is divisible by all factors of a FACTOR
2) A positive integer is divisible by the LCM of ITS FACTORS (but not necessarily the product of its factors)
—> single integer
3) The PRODUCT of two or more integers is divisible by the PRODUCT of their factors
—> product of multiple integers

Question 32

Word problems involving divisibility

Answer 32

1) Must be given information about POSITIVE INTEGERS or WHOLE NUMBERS / NON-NEGATIVE numbers

2) You can determine what “could” be the value using divisibility rules or whether an unknown is “divisible by x” (in order to make one of the unknowns a whole number)

Strategy: Set up the equation and rearrange so that one side is one unknown integer value, and the other side has the actual unknown integer value you are trying to solve for
> then simplify
> Then compare the actual unknown integer value with the denominator

e.g., At a certain department store, a rug was originally priced at W dollars, where W is a whole number. During a liquidation sale, the rug was sold for 6 percent of its original price. Which of the following could be the sale price of the rug?
> Let P = sale price
> What value of P would make W a whole number?
> W = 50P/3 –> P must be a multiple of 3 and W must be a multiple of 50

Question 33

Divisibility rule for 4

Answer 33

If the last two digits are divisible by 4, then the entire number is divisible by 4

Includes -00 (all multiples of 100 are divisible by 4, because will have 4*25 as factors)

Question 34

Divisibility rule for 8

Answer 34

If the number is EVEN and the last 3 digits are divisible by 8 (including -000, all multiples of 1000 are divisible by 8 because 8*125 = 1000)

Question 35

Divisibility rule for 9

Answer 35

Sum of all digits is divisible by 9

Question 36

Divisibility rule for 11

Answer 36

Sum of the odd-numbered place digits - sum of the even-numbered place digits is divisible by 11

Odd-numbered place digits –> ones, hundreds, ten-thousands etc.
Even-numbered place digits –> tens, thousands, hundred-thousands etc.

Question 37

Divisibility rule for 12

Answer 37

Number is divisible by both 3 and 4

Question 38

Divisibility rule for 6

Answer 38

Number is an EVEN number whose digits SUM to a multiple of 3 (so divisible by both 2 and 3, the factors of 6)

Question 39

Divisibility rules of the PRODUCT of CONSECUTIVE integers?

e.g., product of three consecutive integers, like 123

Answer 39

Rule: The product of N consecutive integers will always be divisible by N! AND all the factors of N!

Alternatively, the product of N consecutive integers is a MULTIPLE of N!
(does not have to be equal to N!)
> Also the product is divisible by all the FACTORS of N! / factor combinations (factors of factors rule)
> so (N-1))! is a factor, (N-2)! is a factor etc.

Rationale:
> every N numbers, the pattern repeats for N and for N-1, N-2 etc.

e.g., the product of any three consecutive integers will always be divisible by 3!, but also 1 and 2

e.g., the product of any four consecutive integers will always be divisible by 4!, but also factors of 4! (24) –> 1, 2, 3, 4, 6, 8, 12, 24

Many algebraic forms: (n-1)n(n+1); (n+10)(n+11); (n-100)(n-99)

Question 40

Hidden product of consecutive integers in algebraic expressions

Answer 40

Remember:
> Order of the terms in a product don’t matter
n^3 - n = n(n^2 - 1) = n(n + 1)*(n - 1) still a product of 3 consecutive integers if rearranged as so

Another example: n^5 - 5n^3 + 4n
= n(n^4 - 5n^2 + 4)
= n(n^2 - 1)(n^2 - 4)
= n(n + 1)(n - 1)(n - 2)(n+2)
= (n-2)(n-1)n(n+1)*(n+2)

Question 41

Divisibility rules of the PRODUCT of CONSECUTIVE EVEN integers?

e.g., product of three consecutive even integers, like 246

Answer 41

Rule: Product of N consecutive EVEN integers is divisible by 2^N * N!

Rationale:
> each even term will have a “2” factor (2^N)
> then multiply by N! to account by consecutive nature

Many algebraic forms (gap must be two AND you know whether even/odd): (n-1)*(n+1)

NOTE: there is NO special divisibility rule for the product of consecutive ODD integers

Question 42

What are the properties of three consecutive integers?

Answer 42

1. at least one even integer, so the product is EVEN —> rule to memorize (N! –> also divisible by factors of N!)
2. product is divisible by 3! –> 2, 3, and 6 —> rule to memorize (N!)
3. If the middle term is ODD, the product is divisible by 8 (two consecutive even integers)

Question 43

What is the algebraic expression of DIVISION with a remainder?

Answer 43

x / y = Q + r / y

Rearranged:
x = Qy + r
r = x - Qy

Keep in mind:
> The remainder MUST be NONNEGATIVE and be LESS than DIVISOR y
——> the only possible values of r can be 0 to one less the divisor y
= [0, 1, … up to y - 1]
> r/y is always LESS THAN 1
> start off every remainder question by writing out “y > r”
> In other words, the DIVISOR (not x) determines the range of possible remainders
> DO NOT SIMPLIFY THE FRACTION r / y —> remainders only make sense in the context of a DIVISOR
> Q = quotient = must be an integer (but could be in other forms like n^2 + 1)

Any integer can be expressed as the product of some integer quotient Q and some divisor y, plus some remainder

Question 44

You have two algebraic expressions for the same unknown integer in the form x = Qy + r. How do you combine the two information?

Answer 44

If you know the remainders when X is divided by TWO DIFFERENT DIVISORS, you can come up with a COMBINED EQUATION FOR X

Approach #1) Combine into one equation for X
x = LCM of divisors * Integer + Smallest Value of X

e.g., n = 15Q + 2 and n = 17Z + 2

LCM of 15 and 17 is 15*17 = 255
Smallest possible value of n that satisfies both equations = 2

So combined equation for n = 255*int + 2

NOTE: If the form has a NEGATIVE (e.g., n = 15Q - 2), just convert negative number into positive form by ADDING the divisor

e.g., 15Q - 2 = 15Q + 13

Approach #2) Set the two equations equal to each other to understand relationship between QUOTIENTS

Question 45

Remainder x factor question

Answer 45

x = Qy + r
(x - r) = Qy —> Q and y are factors of x - r —> list them out!
> combined with y > r, you should be able to figure out properties of y (divisor)

Question 46

Can remainders be expressed as decimals?

Answer 46

YES

HOWEVER, there are WATCHOUTS

(1) Converting FRACTION REMAINDER to DECIMAL REMAINDER
> It is easier to convert fraction remainder to decimal remainder, than from decimal remainder to fraction remainder
> with a fraction remainder, you already know the divisor

(2) Converting DECIMAL REMAINDER to FRACTION REMAINDER
> When converting from decimal remainder to fraction remainder, we need MORE INFO to determine the remainder BECAUSE there is an infinite number of remainders possible (basically any fraction that simplifies into decimal is possible)

REMEMBER: The “remainder” is really a non negative INTEGER (numerator part of fraction, all DEPENDENT ON THE DIVISOR)

e.g., x / y = 9.48 —> cannot say remainder is 48 (only true if divisor y = 100)

x / y = 948/100 = 9 + 48/100 —> remainder is 48
x / y = 474/50 = 9 + 24/50 —> remainder is 24

Without further information on x or y, what we can do is determine what the MOST REDUCED FRACTION COULD BE —> remainder must be a MULTIPLE of this number

e.g., 48/100 = 24/50 = 12/25 —> actual remainder must be a multiple of 12

(3) Converting DECIMAL REMAINDER to INTEGER (ONLY possible if you know the DIVISOR)

Integer remainder = Decimal component of the division * divisor y

Question 47

How do you find the remainder of a product of integers?

e.g., if x = 500600700, what is the remainder when x is divided by 8?

e.g., A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

Answer 47

1) Divide each of the integers by the divisor
2) Focus on the remainders –> MULTIPLY the remainders
3) Correct for any “excess remainders” at the end of the multiplication (if remainder is greater than the original divisor)

OR see if R=0 first based on divisibility rules

Algebraically:
(ab)/c –> Remainder = (Ra*Rb)/c (product of the remainders)

Alternatively, you may be asked about the product of two unknown integers and given expressions –> find the combined equation and divide by divisor to figure out remainder

e.g., n = 9Q + 6; m = 12k + 4
So nm = (9Q + 6)*(12k + 4)
= 108QK + 36Q + 72k + 24

nm/18 = int + int + int + 1 + 6/18 —> remainder = 6

Question 48

How do you find the remainder of the addition or subtraction of integers?

e.g., if x = 12 + 13 + 17, what is the remainder when x is divided by 5?

Answer 48

1) Divide each integer by the divisor
2) Focus on the remainders –> add or subtract (based on the sign)
3) Correct for any “excess remainders” at the end of addition (by subtracting multiples of the divisor) // any “negative remainders” at the end of subtraction (by adding multiples of the divisor)

e.g., 12/5 –> R = 2
13/5 –> R = 3
17/5 –> R = 2

x/5 has a remainder = 2 + 3 + 2 = 7 –> after adjusting for excess remainder, we get R = 2

Algebraically:
(a + b)/c –> Remainder = (Ra + Rb)/c (sum of the remainders)
(a - b)/c –> Remainder = (Ra - Rb)/c (subtraction of the remainders)

Question 49

What creates a trailing zero? How do you determine the number of trailing zeros?

e.g., 520 (1 trailing zero) vs 5200 (2 trailing zeros) vs 520000 (4 trailing zeros)

Answer 49

Each factor of 10 or (2x5) PAIR in the prime factorization of that number
> so the limiting factor is the number of 2s and 5s present, whichever is fewer

Application Qs:
> Trailing zero rule can be used to determine the TOTAL # OF DIGITS in a number

**Other things to keep in mind:
> Any factorial >= 5! will have a zero in its units digit
> if we have even ONE 2x5 pair, the UNITS DIGIT will always be zero

Question 50

How do you determine the number of digits in an integer?

Answer 50

You can leverage the understanding that (2x5) pairs create trailing zeros

1) prime factorize the number
2) determine the number of trailing zeros after the non-zero digit (how many 2x5 pairs are there)
3) find the PRODUCT of the remaining factors (unpaired 2s or 5s along with other nonzero prime factors)
4) Number of digits = # of trailing zeros + # of digits in the product of the remaining factors

Question 51

What are leading zeros? How do you determine the number of leading zeros in a FRACTION (form 1/X)

Answer 51

2) X (denom integer) IS a perfect power of 10 (e.g., equal number of 2s and 5s)

MUST BE 1 / #

When dealing with decimals, it refers to the zeros that occur to the RIGHT of the decimal point, but before the first non-zero number

e.g., 0.02 has 1 leading zero
0.2 has no leading zeros
0.002 has 2 leading zeros

Basically do not count the 0 before the decimal

Approach:
#1) X (denom integer) is NOT a perfect power of 10 (e.g., unequal number of 2s and 5s, or non 2 and 5 integers)
> If X is an integer with k digits, and if X is not a perfect power of 10, then 1/X will have (k-1) leading zeros

e.g., 1/8 –> 1 digit, so 0 leading zeros (0.125)
e.g., 1/40 –> 2 digits, so 1 leading zero (0.025)

> If X is an integer with k digits, and if X is a perfect power of 10, then 1/X will have (k-2) leading zeros

e.g., 1/100 –> 3 digits and X is a perfect power of 10, so 1/X will have 1 leading zero (0.01)

Question 52

What is the best way to determine the largest number of a prime number that divides into a factorial, N!?

e.g., What is the largest possible integer value of n such than 21!/3^n is an integer

Answer 52

1) Divide N by each power of the prime (ignoring remainders) until the quotient equals zero
e.g., 21/3^1, 21/3^2 …

2) Add the quotients from the previous divisions = sum represents the number of prime number x in the prime factorization of N!

Rationale:
> trying to COUNT the number of individual instances / double instances / triple instances etc. of a prime number

e.g., 21/3 –> 7 instances where 3 shows up as a factor
21/9 –> 2 instances where 3 shows up AGAIN (as an extra factor)

Question 53

What is the best way to determine the largest number of times a non-prime number divides into a factorial, N!?

e.g., What is the largest possible integer value of n such than 40!/6^n is an integer
e.g., What is the largest possible integer value of n such that (500!500!500!)/500^n is an integer

Answer 53

Similar to the steps involved to determine the largest number of times a PRIME divides into a factorial, except you want to figure out the max number of PRIME NUMBER PAIRS that divide into the factorial

e.g., max number of (2x3) pairs that divide into 40

> The number of pairs is CONSTRAINED by the larger of the pair (in this case, the number of 3s) –> short cut is to just solve for # of the larger of the pair

> ** Make sure to adjust for EVERY NUMBER IN THE NUMERATOR
e.g., there are three 500!, so multiply the total number of 5s by 3 (you must do this BEFORE setting up the inequality)

> If there is an EXPONENT on the prime, remember to just do an extra step of solving for the unknown in the exponent after

e.g., 50!/25 –> 50!/(5^2n)

We set 2n <= # of 5s and solve for n

Keep in mind:
> you cannot just divide 40 by 6 because you are missing lone numbers that can form a 2x3 pair

Question 54

What is the best way to determine the largest number of times a power of a prime number divides into a factorial, N!?

e.g., What is the largest possible integer value of n such that 30!/4^n is an integer

Answer 54

** 1) First express the divisor as a power of a PRIME number
e.g., 4^n = 2^2n
e.g., 500^n = 2^2n * 5^3n

2) Determine the number of the base prime that divide into the factorial (Divide N! by each power of the prime (IGNORING REMAINDERS **) until the quotient equals zero) —> constrained by the largest prime

e.g., 30/2 = 15
30/4 = 7
30/8 = 3
30/16 = 1
Total number of 2s in 30! = 26

***** Make sure to adjust the answer for EVERY NUMBER IN THE NUMERATOR
e.g., there are three 500!, so multiply the total number of 5s by 3 (you must do this BEFORE setting up the inequality)

4) Set up an inequality using just the exponents

2n <=26 (trying to determine the max value of n)
n <= 13

Keep in mind:
> you cannot just divide 30 by powers of 4 because you are missing single values of 2 that can combine into 4

Question 55

What type of number has a prime factorization with only EVEN exponents?

Answer 55

Perfect square –> *** square root of a perfect square must be an integer
=> sqrt(perfect square) = integer

Note: 0 and 1 are the ONLY unique among perfect squares because they don’t have prime factors
> does not need to be represented as having an even exponent

May also be expressed as 230x = y^2

Question 56

What are the first 9 non-negative perfect cubes?

Answer 56

0
1
8
27
64
125
216
343
512

Question 57

What type of number has a prime factorization with only exponents that are divisible by 3?

Answer 57

Perfect cube
> has prime factorization form with exponents that are MULTIPLES OF 3 / divisible by 3

May also be expressed as:
450p = q^3

Note: 0 and 1 are unique among perfect cubes because they don’t have prime factors

Question 58

What is the remainder when 3^200 is divided by 5?

Answer 58

Strategy: Utilize the knowledge that REMAINDERS follow PATTERNS
> so we can determine the remainder when a very large number is divided by another number

Tactically, this means finding
(1) what the pattern of remainers is (pattern repeats every xth power)
(2) take exponent and RE-EXPRESS so it matches the pattern
(3) determine remainder
> similar approach is followed for UNITS DIGITS (but still different approaches)

e.g., What is the remainder when 3^200 / 5?

Step 1: Find the pattern in the remainders when powers of 3 are divided by 5
3^1 –> R = 3
3^2 –> R = 4
3^3 –> R = 2
3^4 –> R = 1
3^5 –> R = 3 (pattern repeats itself every 4th power). In other words, all powers that are a multiple of 4 has a remainder = 1

Step 2: Re-express exponent to match pattern
3^200 (already a multiple of 4)

Step 3: Determine remainder
> we know every exponent that is a multiple of 4 will have a remainder = 1
> BUT a multiple of 4 is also a multiple of 2 –> that’s okay, it is still a multiple of 4

Question 59

Patterns in units digits when a positive integer is raised to consecutive powers of positive integers

e.g., what is the units digit of 6,789^37
What is the units digit of 88^5 * 99^6 * 77^3

Answer 59

Units digit will follow a pattern:
> stay the same (regardless of power)
> two-number pattern (tied to whether you have an even vs odd exponent)
> four-number pattern (every power that is a multiple of 4 will have units digit = x)

Stay the same regardless of power: 0, 1, 5, 6

Two-number pattern: 4 and 9
> For 4 —> 4 for odd exponents and 6 for even exponents
> For 9 –> 9 for odd exponents and 1 for even exponents

Four-number pattern: 2, 3, 7, 8

Note: For base integers greater than 9 (not single digits), the same units-digit pattern holds true –> just ignore all other digits and focus on the units digit and the exponent

e.g., 123384^33 –> read as 4^33

Applications of this pattern:
> Short cut to PS (scan answer choices and pick answer based on matching units digit)
> Product of powers (units digit of the product = product of each term’s units digit)
e.g., 7^(8a + 9b + 10c) has what units digit?
Can re-express as 7^8a * 7^9b * 7^10 c
(if one of the units digit = 0, all 0)
> Sum of powers (e.g., 3^15 + 2^15 –> choose PS option set that matches unit digit)
> for DIVISION of powers –> usually you can re-express as a common base and simplify into a single term (e.g., 27^324 / 3^134 = 3^838)

Question 60

Other common patterns involving multiples

Ex #1) finding the 79th digit to the right of the decimal place in 2/11

Ex #2) Pattern is 2 morsels of wheat, 3 morsels of white, and 1 morsel of rye. If Robin drops 2000 morsels of bread in total, what are the last 3 morsels of bread that she drops

Answer 60

Ex #1) Finding the xth digit to the right of the decimal place in a fraction
> We can use our understanding of digit patterns because there is a consistent pattern in DIGITS to the right of the decimal point
> To solve, first do division to get a sense of the pattern

e.g., 2/11 = 0.181818… each digit in the ODD numbered place to the right is 1 and each even numbered place digit to the right of the decimal point is 8. So because 79 is an odd number, the 79th digit to the right of the decimal place must be 1

Ex #2) Other common pattern applications
> To solve, first right out the CORE PATTERN that repeats (it is best to write out each component separately, even for repeated elements after each other)
> e.g., Wheat, Wheat, White, White, White, Rye —> Pattern repeats every 6 items
> Use your knowledge of MULTIPLES to figure out how many COMPLETE ITERATIONS can occur within the total number of items
> e.g., 2000/6 –> 333 complete iterations, Remainder 2
> Figure out the pattern at the end of the total number of items

Question 61

Remainders after division by 10^n?

Answer 61

Just move the decimal point to the LEFT. The digits to the right of the decimal point represent the remainder

e.g., divide by 10 –> remainder will be the UNITS digit of the dividend
e.g., divide by 100 –> remainder will be the last two digits of the dividend

So if you KNOW the units digit of a number, you KNOW the remainder after division by power of 10

Application of this concept:
> typically combined with finding patterns in large powers (units digit)

Falls into category of special remainder rule (similar to division by power of 10)

Question 62

Remainders after division by 5?

Answer 62

For a given UNITS digit, the remainder after division by 5 is ALWAYS THE SAME

So if you KNOW the units digit of a number, you KNOW the remainder after division by 5 (sufficient)

Falls into category of special remainder rule (similar to division by power of 10)

Applications:
> Often combined with patterns in units digit of large powers

Question 63

What are all the ways that evenly spaced sets commonly appear on the GMAT?

Answer 63

1) consecutive integers (constant difference = 1)

2) consecutive MULTIPLES of a number (constant difference >1)

3) consecutive numbers with the same remainder when divided by some integer (constant difference > 1)
e.g., [1, 7, 13, 19, 25] –> d = 6, remainder when divided by 6 all equal 1

Question 64

If you know two integers are CONSECUTIVE, what can you deduce?

Answer 64

(1) Product is even // divisible by 2!
(2) Two integers do NOT have common prime factors // GCF = 1

Question 65

Q: If K = 148, which of the following numbers could be added to k to create a sum that is divisible by 12?

A) 0
B) 2
C) 4
D) 6
E) 8

Answer 65

Concept: Divisibility (multiples/factors)

OPTION 1 Algebraic way (factors/multiples)
1) Translate the prompt
Let x be the number we are trying to solve for?

(148 + x)/12 = int

2) To make the testing easier, first SIMPLIFY the expression so you take out the WHOLE number

148 = 144 + 4
So re-express as (144 + 4 + x)/12
= 12 + (4+x)/12

3) Then sub in values of x that produce an integer result

Ans E (8)

OPTION 2: Divisibility rules
> to be divisible by 12, a number must be divisible by both 3 and 4
> so try each number to see if the sum with 148 is divisible by 12

Question 66

If x and y are integers and xy = 660, x or y MUST be divisible by which of the following?

a. 3
b. 4
c. 6
d. 20
e. 30

Answer 66

Concept: Factors / multiples / divisibility
> We are trying to see what number must be a FACTOR of FACTORS
> it is true that x or y must be divisible by EVERY PRIME FACTOR of 660
> BUT x or y may not always be divisible by the PRODUCT OF A COMBINATION of prime factors (especially powers of primes can be split)

1) prime factorize 660 = 2^2 * 3 * 5 * 11

2) Process of elimination
a. 3 –> x or y must always be divisible by 3
e.g., 2^2 * 5 * 11 x 3 —> ONLY PRIME FACTOR in the option set

b. 4 –> x or y do NOT need to be divisible by 4
e.g., 2^2 * 5 * 11 x 3 (one number is divisible by 4)
e.g., 2 * 3 x 2 * 5 * 11 (neither number is divisible by 4)

Question 67

If a and b are integers and ab=/0, what is the units digit of ab?

1) a = 4!
2) b = 7!

Answer 67

Concept: Factorials and units digit

Question 68

If a and b are integers and ab=/0, what is the units digit of ab?

1) a = 4!
2) b = 7!

Answer 68

Concept: Factorials and units digit
> Any factorial >= 5! will have a zero in its units digit

(1) a = 4! –> NS because 4! = 24 but b could be 10 or any other number

(2) b = 7! –> S because any factorial >= 5! has a zero in its unit digit

B

Question 69

What can you tell about the remainder of 5y/7, if you know that when y is divided by 7, the remainder is 1?

Answer 69

RULE for remainders (addition, subtraction and multiplication):
(ab)/c –> Remainder = (Ra*Rb)/c (product of the remainders)

5/7 —> remainder is 5/7
y/7 –> remainder is 1/7

SO remainder of 5y/7 = (5*1)/7 = 5/7 (5 is the remainder)

Question 70

Memorize the following common factorials

4!
5!
6!

Answer 70

4! = 24
5! = 120
6! = 720

Question 71

Which of the following must be subtracted from 2^256 so that the resulting integer will be a multiple of 3?

Answer 71

Concept: Remainder patterns in powers (slightly different approach to pattern in unit digit of powers)

(1) Determine if 2^256 is a multiple of 3 // is 2^256 divisible by 3
> check if Remainder = 0

Remainders when powers of 2 are divided by 3:
2^1 / 3 –> R = 2
2^2 / 3 –> R = 1
2^3 / 3 –> R = 2
2^4 / 13 –> R = 1

Pattern: every odd exponent of 2 has R = 2, while every even exponent of 2 has R = 1

Since 256 is even, 2^256/3 has R = 1 (so 2^256 is NOT divisible by 3)

(2) If 2^256 is NOT divisible by 3, figure out what number you need to subtract from it so get remainder = 0

1

Question 72

What are the first five powers of 3

Answer 72

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

Question 73

If r, s, t are unique positive integers, what is the product of r and s?

(1) LCM of r and t is 48
(2) LCM of s and t is 80

Answer 73

Concept: LCM of two different pairs of integers, trying to figure out product of two integers

(1) No value of s, so NS

(2) No value of r, so NS

(3) Testing reveals two different answers, so NS

** COME UP WITH TEST VALUES FASTER

Statement 1: 48 = 2^4 * 3

Statement 2: 80 = 2^4 * 5

t is the SHARED integer between the two statements —> t must be 2^1 <= 2^4

LCM means HIGHEST POWER of shared factors and all non-repeated factors

Test #1: t = 2^4, r = 3, s = 5
rs = 15

Test #2: t = 2^4, r = 2^43, s = 2^45

rs = 2^8 * 15

Question 74

Terminating decimals - what are they and what number should you always test?

e.g., are there any terminating decimals, n, that fit 1/5 < n < 1/4?

Answer 74

A decimal is terminating if the REDUCED FRACTIONAL EQUIVALNET contains only powers of 2’s and/or powers of 5s in the denominator

Always test denom = 100 (yes terminating decimal)

e.g., n could be /100 such as 22/100

Question 75

Do powers influence even / odd properties of a number?

Answer 75

NO, exponents do NOT influence even / odd properties of a number

Powers are simply multiplying the number by itself, either an odd number or an even number

Question 76

If x, y and z are positive integers, is (x!)*(y!) > z!?

(1) x^y > z
(2) x + y = z

Answer 76

Concept: Factorials
> best strategy is to TEST
> Tip when testing: test small values AND large values, and weird value (0, 1, fractions if relevant)

(1) x^y > z

Test 1: x = 2, y = 3, z = 1
2!*3! > 1! is TRUE

Test 2: x = 2, y = 3 (KEEP ONE SIDE CONSTANT TO MAKE IT EASIER TO TEST), z = 6
2!*3! > 6! is FALSE

NS

(2) x + y = z

From this statement, we can see that z will ALWAYS be greater than x, and z will ALWAYS be greater than y

Logically, (x!)*(y!) > z! will always be FALSE

Test 1 (SMALL values): x = 1, y = 1, z = 2
1!*1! > 2! is FALSE

Test 2 (LARGE values): x = 10, y = 5, z = 15
10!*5! > 15! is FALSE

S

Question 77

How do you determine the number of trailing zeros in a large factorial?

e.g., Does 21! have more than 6 trailing zeros?

Answer 77

Concept: Application of trailing zero rule

You need to figure out the number of (2x5) pairs in the factorial
> since there are fewer 5s than 2s factors, focus on determining the NUMBER OF 5s in the factorial

Question becomes - how many 5s are there in 21!?

Answer: Use the factorial shortcut

21/5^1 = 4 5’s
21/5^2 = 0 5’s

So there are 4 5’s in 21!

Question 78

Divisibility involving decimals:

If the price of a computer were reduced by 16%, which of the following could not be the final price of the computer?

A) $844.10
B) $896.70
C) 1,056.30
D) 1,136.10
E) 1,264.20

Answer 78

Concept: Divisibility involving decimals
> convert to INTEGERS (in this case, price in CENTS is always an integer)

Let F = final price in CENTS
P = original price in CENTS

F = P*(0.84)
F = P * (84/100)
P = 25F/21 —> F must be divisible by both 3 and 7

A gives a sum of 17, which is not divisible by 3, so A could not be the final price of the computer

Question 79

What is the greatest integer value n such than 9^n is a factor of 45?

Answer 79

WATCH OUT - this type of question is NOT the same as finding the number of primes in a factorial

1) express potential factor as a PRIME
e.g., 9^n = 3^2n
> you need to do this to account for lone 3’s that can combine into 9

2) Prime factorize the dividend
45 = 3^2 * 5

3) Count the number of 3’s –> 2 3’s

4) Set up inequality of the powers to solve for n

2n <= 2
n <= 1

Question 80

When you have two division equations representing two integers and are asked to find the possible values of m + n, what are the best approaches to solve?

e.g., The variables m and n are positive integers. When m is divided by 18, there is a remainder of 12, and when n is divided by 24, there is a remainder of 14. Which of the following are possible values of m + n?

I. 50
II. 70
III. 92

Answer 80

A) Algebraic method (leverages understanding of REMAINDERS)
> create the equation for n + m using information about the remainders

m = 18Q + 12
n = 24k + 14
So, n + m = 18Q + 24k + 26

> then, divide n + m by a number that divides into 18 and 24, but leaves a remainder when divided into 26 (e.g., 3)
The correct answer that matches the algebraic expression of n + m will have the same remainder when divided by 3

6Q + 8k + 8 + 2/3

I. 50/3 = 16 R2 (True)
II. 70/3 = 23 R1 (False)
III. 92/3 = 30 R2 (True)

I and III

ALSO you can set the combined equation EQUAL to each of the potential answer choices and see if you can find INTEGER VALUES for the variables that make the equation work, if yes, then it is a possible answer

Question 81

What are the unique divisibility rules associated with perfect squares?

Answer 81

RULE 1: Perfect squares are either divisible by 4 (R=0) or leave a R=1 when divided by 4

> Even perfect squares are always divisible by 4
(Proof: (2n)^2 / 4 = 4n^2 / 4 –> integer with 0 remainder)

> Odd perfect squares always have remainder = 1 if divided by 4
(Proof: (2n+1)^2 / 4 = (4n^2 + 4n + 1)/4 —> R=1

RULE 2: Perfect squares are also either divisible by 3 (R=3) or leave a R=X when divided by 3

> Perfect squares that are divisible of 3 (e.g., 9, 36, 81) leave R=0
Perfect squares that are NOT divisible by 3 (e.g., 1, 4,16, 25) leave R=1

Question 82

What are the unique divisibility rules associated with perfect squares?

Answer 82

RULE 1: Perfect squares are either divisible by 4 (R=0) or leave a R=1 when divided by 4

> Even perfect squares are always divisible by 4
(Proof: (2n)^2 / 4 = 4n^2 / 4 –> integer with 0 remainder)

> Odd perfect squares always have remainder = 1 if divided by 4
(Proof: (2n+1)^2 / 4 = (4n^2 + 4n + 1)/4 —> R=1

RULE 2: Perfect squares are also either divisible by 3 (R=3) or leave a R=X when divided by 3

> Perfect squares that are divisible of 3 (e.g., 9, 36, 81) leave R=0
Perfect squares that are NOT divisible by 3 (e.g., 1, 4,16, 25) leave R=1

Question 83

If n is a positive two-digit integer, how many different values of n allow n^3 - n to be a multiple of 12?

Answer 83

Concept: Factors and patterns involving consecutive integers

What we know:
> n is between 10 and 99, inclusive
> n^3 - n is equivalent to the product of 3 consecutive integers
> So the question is asking how many different values of n produce a remainder of 0 when n^3 - n is divided by 12
> the product of 3 consecutive integers is divisible by n! or 6 (so we already know that the product of 3 consecutive integers is divisible by 3, now we just need to determine when n^3 - n is divisible by 4!!

For these GROUPS of consecutive integer questions, it is easiest to ANCHOR TO THE MIDDLE NUMBER (every instance of the middle term will produce a grouping, just double check the ends)

ALSO note that only n needs to be between 10 and 99 inclusive; n-1 and n + 1 don’t follow this rule!! so no special adjustments are needed

Case 1) n is odd –> n^3 - n is divisible by 8 and therefore also by 4
> how many numbers between 10 and 99 are odd?

Total # of integers [10,99] = (99-10)+1 = 90 integers, half of which are odd
45 odd integers from 10 to 99 inclusive = 45 instances where n^3 - n is divisible by 12

Case 2) n is even, meaning n-1 and n+1 are odd and don’t have any factors of 2
> need to determine the number of two digit MULTIPLES OF 4 that can be n

Formula when the increment does not equal 1:
(Highest number divisible by the given number - lowest number divisible by the given number)/increment + 1

= (96 - 12)/4 + 1
= 21 + 1
= 22 instances where n^3 - n is divisible by 12

Putting it all together, we have 45 odd n + 22 even n = 67 outcomes where n^3 - n will be divisible by 12

Question 84

Perfect square of a PRIME number –> what should you remember about the FACTORS

Answer 84

Perfect square of a prime will always have only 3 factors
> 1, itself, and the prime it is raised to the power of 2

This is because in prime factorization form, the power is 2 and to find the total # of factors, 2+1 = 3

If an integer has only 3 positive factors (including 1) or 2 positive factors other than 1, what does this tell you?

Question 85

Can you determine the tens digit of a large power?

e.g., what is the tens digit of 7^1415?

Answer 85

FIND THE PATTERN in the TENS DIGIT (ignore what’s going on in the units digit) using smaller exponents (don’t give up just because the number is getting very large)

e.g., Tens digit
7^1 = 0
7^2 = 4
7^3 = 4
7^4 = 0
7^5 = 0
7^6 = 4

Pattern in the tens digit of powers of 7 = 0, 4, 4, 0

1415 = R 3 when divided by 4, so tens digit = 4

Question 86

How do you figure out whether x and y are even or odd from this format (binomial):

(x+y)^2

Answer 86

Express as a product of two x+y’s:
(x+y)*(x+y) = even or odd

Question 87

LCM question involving figuring out an unknown divisor given LCM of set of integers

e.g., If n is a positive integer, what is the smallest positive integer that is divisible by 6, 15, and n?
(1) 40 is the smallest integer that is divisible by 8 and n
(2) 60 is the smallest integer that is divisible by 12 and n

Answer 87

How to solve: For each LCM clue provided …
> Prime factorize the integers
> Remember that LCM = non-repeated prime factors * highest power repeated prime factors shared by at least two of the integers
> for the unknown integer, determine what the EXPONENTS for each prime factor COULD BE (e.g., 2^0to3)

Example:
6 = 23
15 = 35
N = ??

(1) 40 = 2^3*5
8 = 2^3
N = 2^0to3 * 5 (must have five)

NS because N could be 5, 10, 20, 40, which changes the LCM of 6, 15 and N

(2) 60 = 2^235
12 = 2^2 * 3
N = 2^0to2 * 3^0to1 * 5

NS because N could be 5, 10 etc.

(3)
NS because N = 2^0to2 * 5 = 5, 10, 20, whcih changes the LCM of 6, 15 and N

Question 88

What does this statement mean? “The value represented by which of the following expressions is equal to its own opposite?”

Answer 88

The ONLY number that is EQUAL TO ITS OPPOSITE is ZERO

So this statement is asking for the expression that EQUALS zero

Question 89

PS: What is a fast way to figure out a possible divisor for a large factorial?

e.g., A certain number N is divisible by the product of the first fourteen positive integers. Which of the following must be a factor of N?

(A) 1690
(B) 1210
(C) 726
(D) 625
(E) 616

Answer 89

Factorials

LONG WAY is to prime factorize 14! then compare to the prime factorized forms of each answer

SHORTER WAY is to look at the OPTIONS FIRST and compare starting from LARGEST factor to smallest factor

(A) 1690 is 2513^2 —> we know that 14! only has one 13, so this is WRONG

(B) 1210 is 2511^2 –> we know that 14! has only one 11, so this is also WRONG

(C) 726 is 2311^2 –> we know that 14! has only one 11, so this is also WRONG

(d) 625 is 5^4 –> instead of prime factorizing 14!, figure out how many 5s are in 14! using the factorial division SHORTCUT

14/5 = 2
14/25 = 0

Max 2 5’s in 14!

So this is also wrong

(keep going until you realize that ans E is right)

Question 90

Tips for choosing test cases for DS (involving GCF/LCM/divisibility)

Answer 90

> Create test cases that produce Y and N option or produce multiple different values
When creating Y and N cases, try to keep most of the numbers the SAME and move only ONE number to compare differences

e.g., W is a multiple of 5 and W and Z share only ONE common prime. What is their GCF?

Test case 1) W = 5 * 7, Z = 7 —> GCF is 7
Test case 2) W = 5 * 3, Z = 3 —> GCF is 3

NS since different values

e.g., W is a multiple of 5, Z is a multiple of 9, and W and Z share only ONE common prime. What is their GCF?

Test case 1) W = 5 * 7, Z = 3^2 * 7 —> GCF = 7
Test case 2) W = 5 * 13, Z = 3^2 * 13 —> GCF = 13

NS since different values

Question 91

Parts of a division equation:

Dividend
Quotient
Remainder
Divisor

Answer 91

Dividend / Divisor = Quotient + remainder / divisor

Divisor = what you divide by = denominator
Dividend = numerator
Quotient = answer after division
Remainder = what does not get evenly divided

Question 92

For positive integers a and b, the remainder when a is divided by b is equal to the remainder when b is divided by a. Which of the following could be a value of ab?

(I) 24
(II) 30
(III) 36

Answer 92

Remainders

THINK ABOUT TWO CASES of the relationship between a and b:

Case 1: a = b
> a/b = a/a = 1 R = 0
> b/a also has R = 0
> Case 1 fits the statement that a / b has the same remainder as b / a
> therefore looking for a PERFECT SQUARE (a^2) = ab = b^2
> Only III = 36 is a perfect square

Case 2: a > b
> b / a has a R = b (because b / a < 1)
> a / b has a R < b (remainder rule is 0 <= R < divisor b)
> does not match the statement

Case 3: a < b
> a / b has a R = a
> b / a has a R < a
> does not match the statement

Therefore only when a = b is the statement true

Question 93

Whether the sum or difference of numbers is a multiple of an integer

Answer 93

If a and b are both multiples of an integer > 1, then a + b and a - b will also be a multiple of the same integer
Example: a=6 and b=9, both are divisible by 3 —> a+b=15 and a−b=−3

If only one of the integers is a multiple of an integer > 1, then a + b and a - b will not be a multiple of that integer
Example: a=6 and b=5 → a+b = 11 and a-b = 1, neither is divisible by 3

If a and b are both not multiples of an integer > 1, then a + b and a - b may or may not be a multiple