Q3) Number properties Flashcards
Integers, zero and one, even and odd numbers, positive and negative numbers, evenly spaced numbers, divisibility, remainders, prime numbers, factors and multiples
What are “whole numbers”?
Non-negative integers aka 0 and positive integers
What is 0^2 equal to?
0
Concept: Zero raised to ANY POSITIVE power is zero
Does -0 exist?
Yes
Concept: Zero is the only number that is equal to its opposite
0 = -0
Unique properties of the number 0
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
Unique properties of 1
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)
Can decimals be even or odd?
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)
How do you express even and odd integers in mathematical expression?
Even: 2n
Odd: 2n+1 or 2n-1
What are the even / odd addition, subtraction, multiplication, and division rules?
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
What is the remainder of odd / 2?
Always 1
What is the meaning of an absolute value?
Basically asking how far away is n from 0 on the real number line
When you see exponents + variables, what type of concept might be tested?
Even and odd exponents versus positive and negative answers
Aka how do exponents impact the SIGN of numbers
Formulaic expression of factor (divisor), k
*** 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
What is the definition of a multiple of an integer? What is the formulaic expression of a multiple, x?
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
Memorize: What are the first 25 prime numbers
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
What is the formulaic expression of prime factorization of a number, x?
x = (prime number)^# * (prime number)^#
DEALING WITH POSITIVE INTEGERS
** How do you calculate the TOTAL NUMBER OF FACTORS of a particular number?
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
What is the difference between “number of prime factors” vs “number of unique prime factors” vs “sum of prime factors”
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)
Will raising a number to a positive exponent change the number of unique prime factors that number has?
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
What is the fastest way to find the LCM of any set of positive integers?
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”
When will the LCM be equal to the product of two positive integers?
Only when those two integers share NO common prime factors (no duplication)
What is the fastest way to find the GCF of any set of positive integers?
> 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
GCF vs LCM
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
Terminology: What does it mean for a number to “divide evenly into x”?
Refers to the DIVISOR or FACTOR
e.g., 4 divides evenly into 12
Divide INTO a LARGER number
If it is known that y divides evenly into x, then can you determine what the LCM and GCF are of x and y?
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
If you know the LCM and GCF of TWO positive integers, what do you know?
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
Terminology: What does this mean?
x is a dividend of y?
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
What should you think of when you see a question regarding divisibility / multiples / factors?
ALWAYS prime factorize both the dividend and divisor
> you can see if there is a remainder
AND turn into PRODUCT of numbers
What does factors of factors rule imply?
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
How to express this mathematically?
Is x a multiple of both 2 and 3?
Is x = 23m, where m is a non-negative integer?\
Or 2 and 3 are both factors of x
Divisibility and LCM rule – what must be true in order to conclude that an integer z is divisible by LCM of two numbers?
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
Memorize list of divisibility properties
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
Word problems involving divisibility
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
Divisibility rule for 4
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)
Divisibility rule for 8
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)
Divisibility rule for 9
Sum of all digits is divisible by 9
Divisibility rule for 11
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.
Divisibility rule for 12
Number is divisible by both 3 and 4
Divisibility rule for 6
Number is an EVEN number whose digits SUM to a multiple of 3 (so divisible by both 2 and 3, the factors of 6)
Divisibility rules of the PRODUCT of CONSECUTIVE integers?
e.g., product of three consecutive integers, like 123
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)
Hidden product of consecutive integers in algebraic expressions
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)
Divisibility rules of the PRODUCT of CONSECUTIVE EVEN integers?
e.g., product of three consecutive even integers, like 246
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
What are the properties of three consecutive integers?
- at least one even integer, so the product is EVEN —> rule to memorize (N! –> also divisible by factors of N!)
- product is divisible by 3! –> 2, 3, and 6 —> rule to memorize (N!)
- If the middle term is ODD, the product is divisible by 8 (two consecutive even integers)
What is the algebraic expression of DIVISION with a remainder?
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
You have two algebraic expressions for the same unknown integer in the form x = Qy + r. How do you combine the two information?
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
Remainder x factor question
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)
Can remainders be expressed as decimals?
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
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?
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
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?
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)
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)
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
How do you determine the number of digits in an integer?
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
What are leading zeros? How do you determine the number of leading zeros in a FRACTION (form 1/X)
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)
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
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)
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
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
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
** 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
What type of number has a prime factorization with only EVEN exponents?
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
What are the first 9 non-negative perfect cubes?
0
1
8
27
64
125
216
343
512
What type of number has a prime factorization with only exponents that are divisible by 3?
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
What is the remainder when 3^200 is divided by 5?
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
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
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)
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
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
Remainders after division by 10^n?
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)
Remainders after division by 5?
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
What are all the ways that evenly spaced sets commonly appear on the GMAT?
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
If you know two integers are CONSECUTIVE, what can you deduce?
(1) Product is even // divisible by 2!
(2) Two integers do NOT have common prime factors // GCF = 1
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
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
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
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)
If a and b are integers and ab=/0, what is the units digit of ab?
1) a = 4!
2) b = 7!
Concept: Factorials and units digit
If a and b are integers and ab=/0, what is the units digit of ab?
1) a = 4!
2) b = 7!
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
What can you tell about the remainder of 5y/7, if you know that when y is divided by 7, the remainder is 1?
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)
Memorize the following common factorials
4!
5!
6!
4! = 24
5! = 120
6! = 720
Which of the following must be subtracted from 2^256 so that the resulting integer will be a multiple of 3?
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
What are the first five powers of 3
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
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
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
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?
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
Do powers influence even / odd properties of a number?
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
If x, y and z are positive integers, is (x!)*(y!) > z!?
(1) x^y > z
(2) x + y = z
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
How do you determine the number of trailing zeros in a large factorial?
e.g., Does 21! have more than 6 trailing zeros?
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!
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
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
What is the greatest integer value n such than 9^n is a factor of 45?
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
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
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
What are the unique divisibility rules associated with perfect squares?
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
What are the unique divisibility rules associated with perfect squares?
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
If n is a positive two-digit integer, how many different values of n allow n^3 - n to be a multiple of 12?
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
Perfect square of a PRIME number –> what should you remember about the FACTORS
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?
Can you determine the tens digit of a large power?
e.g., what is the tens digit of 7^1415?
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
How do you figure out whether x and y are even or odd from this format (binomial):
(x+y)^2
Express as a product of two x+y’s:
(x+y)*(x+y) = even or odd
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
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
What does this statement mean? “The value represented by which of the following expressions is equal to its own opposite?”
The ONLY number that is EQUAL TO ITS OPPOSITE is ZERO
So this statement is asking for the expression that EQUALS zero
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
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)
Tips for choosing test cases for DS (involving GCF/LCM/divisibility)
> 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
Parts of a division equation:
Dividend
Quotient
Remainder
Divisor
Dividend / Divisor = Quotient + remainder / divisor
Divisor = what you divide by = denominator
Dividend = numerator
Quotient = answer after division
Remainder = what does not get evenly divided
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
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
Whether the sum or difference of numbers is a multiple of an integer
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
