Quant (eGmat)

Mode

The value that occurs most frequently in a given data set. -A given data set may have more than one mode -A given data set may have no mode

If each of the numbers of a set are increased by the same number N then the mean

then the mean increasing by N

Remainder Rules

Always >= 0 Always < Divisor

Solve: Find the least number of apples that can be equally distributed either among 8 girls or among 6 boys.

"Least" - Should prompt you to think LCM LCM of 8 & 6 Steps 1. Prime Tree 8: 2 * 2 * 2 6: 2 * 3 2. Prime Factorization Equation 8: 2^3 6: 2^1 * 3^1 3. Pick the GREATEST power of each prime factor for both 8 and 6 2^3, 3^1 4. Multiply and solve 8 * 3 = 24

(A^2 + B^2) (A^2 + B^2)

(A^4 + B^4)

Divisibility Rules 3:

- If the sum of the values in a number is divisible by three - the sum of three consecutive numbers is divisible by 3 - the product of three consecutive numbers is divisible by 3

Greatest Common Divisor (GCD)

- Will never be greater than either of the given numbers - Could equal one of the given numbers

How to determine whether a number is prime

1. All prime numbers except 2 and 5 end in 1, 3, 7 or 9. So if it ends in some other digit it's not prime. If #1 does not solve the problem, then move to the following steps: 2. Find the square root of the given number 3. List the prime numbers lesser than or equal to the square root of the given number Ex: 43 = sort of 43 = 6.43 Primes: 2, 3, 5, 7 4. Check whether the prime numbers (2,3 ,5, 7) divide the number (43) -If all primes do not divide the number, then the given number is prime (43)

Total Number Of Factors

1. Create Prime Factorization Tree 2. Prime Factorization Equation P1^a * P2^b * P3^c 3. Total number of factors = (a+1)(b+1)(c+1)

Total Number of Prime Factors

1. Create Prime Factorization Tree 2. Prime Factorization Equation P1^a * P2^b * P3^c P1, P2, P3 = 3 total prime factors

What is the total number of Prime factors of 90?

1. Create a prime factorization tree to determine primes 3, 3, 5, 2 2. Write the prime factors in exponential form 3^2 * 5^1 * 2^1 3. Count the number of distinct prime factors = 3 (2, 3, 5)

If n is the product of integers from 1 to 10, exclusive, how many prime factors does n have?

1. Create prime factorization tree to determine primes. For this problem we are given: N = 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 The only numbers that are not broken down to primes are 4, 6, 8, 9 4: 2 * 2 6: 2 * 3 8: 2 * 2 * 2 9: 3 * 3 N = 2 * 3 * (2*2) * 5 * (2*3) * 7 * (2*2*2) * (3*3) 2^7 * 3^4 * 5^1 * 7^1 N has 4 prime factors

If A and B are positive integers, is A/B an integer? 1. Every factor of B is also a factor of A 2. Every prime factor of B is also a prime factor of A

1. Every factor of B is also a factor of A This means that every number that divides B also divides A - Sufficient 2. Every prime factor of B is also a prime factor of A -Use the prime factorization equation to visualize (P^M * P^N) / (P^m * P^n). The only way for A/B to be an integer is if M>m & N>n. This statement only tells us the prime factors, but not the power of each prime factor. This statement is insufficient. Answer: A

What is the greatest prime factor of 9^13 - 3^35

1. Find a common base (3^2)^19 - 3^35 2. Simplify exponents 3^38 - 3^35 3. Factor out like terms 3^35(3^3 - 3^0) 3^35(27 - 1) 3^35(26) 3^35(13*2) Prime Factors: 2, 3, 13 Greatest Prime = 13

If K is the positive integer less than 205 and 26K/84 IS also an integer, then what is the maximum possible value of k?

1. Find prime factors using factorization tree (2, 3, 7) - confirm no other prime factors are valid (if we add 5 as a prime factor then p = 210 which is too big) 2. The minimum possible value of k = 2 * 3 * 7 = 42 So the other values of K are multiples of 42 that are less than 205: 42, 84, 126, 168 Answer: 168 is the maximum value of K

How to find the # of trailing zeros If N is a product of 24 * 36 * 125 * 65 * 18. How many trailing zeroes does N have?

1. Find the highest power of 10 in N 24: 2 * 2 * 2 * 3 36: 2 * 3 * 2 * 3 125: 5 * 5 * 5 65: 5 * 13 18: 2 * 3 * 3 N: 2^6 5^4 The highest power of 10 in N is 4.

What is the units digit of 23^99 * 14^352 + 9002^1003 * 918^437986

1. Find the units digit of each of the terms in the expression 3^99 + 4^352 + 2^1003 + 8^437986 2. Find the cyclicity of each unit digit 3 - C = 4 4 - C =2 2 - C = 4 8 - C = 4 3. Express the units digit power in terms of cyclicity 3^99 99 in terms of 4 is (4*24) + 3 4^352 352 in terms of 2 is 2m (because 352 is even) 2^1003 in terms of 4 is 4m+3 (because 1003 is even) 8^437986 in terms of 8 is 4m + 2 (when finding remainder when a number is divided by 4, you only need to consider the last two digits 4. Find the units digit by comparing the cycle written in step 2 3^4m+3 = 3^3 = 27 Units digit 7 4^2m = 4^2 = 16 Units digit is 6 2^4m+3 = 2^3 = 8 Units digit is 8 8^4m+2 = 8^2 = 64 Units digit is 4 5. Find the sum of the unit digits 7 * 6 + 8 * 4 42 + 32 (PEMDAS) 2 + 2 = 4

To find the LCM

1. Prime Factor Tree 2. Prime Factorization Equation 3. Pick the GREATEST power of each prime factor for each # 4. Then multiply the powers selected Ex: Find the LCM of 135 and 147 1. 135: 3 * 3 * 3 * 5 147: 3 * 7 * 7 2. 135: 3^3 * 5^1 147: 3^1 * 7^2 3. 3^3, 5^1, 7^2 4. 81 * 5 * 49

To find the GCD

1. Prime Factor Tree 2. Prime Factorization Equation 3. Pick the LOWEST power of each prime factor for each # 4. Then multiply the powers selected Ex: Find the LCM of 135 and 90 1. 135: 3 * 3 * 3 * 5 90: 2 * 3 * 3 * 5 2. 135: 3^3 * 5^1 * 2^0 90: 2 * 3^2 * 5^1 3. 3^2 5^1 2^0 4. 9 * 5 * 1 = 45

If k is a positive integer less than 205 and 26k/84 is also an integer, then how many different positive prime factors does k have?

1. Prime factorization tree of 26 and 84 26: 13 * 2 84: 2 * 2 * 7 * 3 2. Rewrite given fraction (13 * 2 * K) / (7^1 * 2^2 * 3^1) This simplifies to: 13k / (7*2*3). So K must be divisible by 2, 3, 7. But are these the only prime factors? No! Because the question says k < 205 2 * 3 * 7 = 42. Therefore, p could have more prime factors to make it a number closer 205. Now, add the lowest possible prime into the equation to determine if it is valid. 2 * 3 * 7 * 5 = 210. Therefore, the next lowest prime cannot be a factor of p because 210 is not less than 205 Answer: 2, 3, 7 are the only prime factors of K

If d = 1 / (2^3 * 5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

1. Try to get the denominator to have powers of 10 by multiplying what would make the 2 & 5 have the same exponent 2^4 * 1 / 2^4 * 2^3 * 5^7 -multiple by 2^4 so 2 and 5 have same power of exponents 2^4 / (2*5)^7 16 / 10^7 =0.000016 Answer = 2 (because there are 2 nonzero digits 1 & 6)

If the number 13 completely divides x, and x = a^2 * b, where a and b are distinct prime numbers, which of these numbers must be divisible by 169?

169 = 13^2 Given that X is a multiple of 13, one of the numbers a or b must be 13. to be Only one can be 13 because a and b are distinct primes We do not know if a or b is 13 So we must make an equation that will work if either a or b is 13 Answer: A^2 * B^2

Consecutive Odd Integers

2n + 1, 2n + 3, 2n + 5

Average Speed (round trip)

= 2xy / x + y x: average speed one direction y: average speed of the other direction *This formula ONLY works if the distance of directions is the SAME

(X+A)(X+B)

=X^2 + (A+B)X + AB

Percent Problems A is P% greater than B

A = B + P% of B -What ever is after the "than" is the base

Compound Interest Problems

A = P(1+r/100)^t A = total amount P = starting investment amount r = interest rate or rate of return t = time *You must express r & t in the same term of period compounding EX: If compounded semi-annually for 2 years, r = r/2 t = 4

Perfect Square

A perfect square will have an odd number of factors (and vice versa) A number with an odd number of factors will always be a perfect square

Consecutive Even Integers

Even integers that come on after the other 2n, 2n+2, 2n +4

Cyclicity of 1

Cyclicity: The number of powers after which a number repeats itself 1

Cyclicity of 2

Cyclicity: The number of powers after which a number repeats itself 2^1: 2 2^2: 4 2^3: 8 2^4: 16 2^5: 32 (units digit repeats after 4) Cyclicity of 2: 4 2^4m+X

Cyclicity of 3

Cyclicity: The number of powers after which a number repeats itself 3^1: 3 3^2: 9 3^3: 27 3^4: 81 3^5: 3 * 81 (1*3 =) (units digit repeats after 4) Cyclicity of 3: 4 3^4m+X

Cyclicity of 4

Cyclicity: The number of powers after which a number repeats itself 4^1: 4 4^2: 16 4^3: 64 Cyclicity of 4: 2 4^2m+X

Cyclicity of 5

Cyclicity: The number of powers after which a number repeats itself 5^1: 5 5^2: 25 5^3: 125 Cyclicity of 5: 1 5^m

Cyclicity of 6

Cyclicity: The number of powers after which a number repeats itself 6^1: 6 6^2: 36 - Repeat 6^3: 216 Cyclicity of 6: 1 6^m

Cyclicity of 7

Cyclicity: The number of powers after which a number repeats itself 7^1: 7 7^2: 49 7^3: 343 7^4: 2401 7^5: 2401 * 7 (7*1 = 7) - Repeat Cyclicity of 7: 4 7^4m

Cyclicity of 8

Cyclicity: The number of powers after which a number repeats itself 8^1: 8 8^2: 64 8^3: 512 8^4: 4096 8^5: 4096 * 8 (6*8 = 48) - Repeat Cyclicity of 8: 4 8^4m+1

Cyclicity of 9

Cyclicity: The number of powers after which a number repeats itself 9^1: 9 9^2: 81 9^3: 729 Cyclicity of 9: 2 9^2m

When positive integer n is divided by 12, the remainder is 5. What is the remainder when n is divided by 4?

Dividend = Divisor * Quotient + Remainder N = 12k + 5 - confirm that the remainder is > 0 Yes - 4 is a factor of 12 n = (3 * 4)k + 5 -confirm that M < D, failed, 5 is not less than 4. format remainder in aD + b 4a + b = 5 4+ 1 = 5 b = 1 Remainder = 1

When positive integer n is divided by 12, the remainder is 5. What is the remainder when n is divided by 6?

Dividend = Divisor * Quotient + Remainder N = 12k + 5 - confirm that the remainder is > 0 Yes -6 is a factor of 12 n = (6 * 2)k + 5 - confirm that M < D Remainder = 5

Even/Odd Multiplication

E * E = E O * O = O E * O = E If given N which is the product of 10 factors, then if one of the factors is even then N is even

Even/Odd Addition & Subtraction Operations

E + E = E O + O = E E + O = O (same for subtractions)

Is A/B an integer?

For A/B to be an integer: 1. B must completely divide A OR 2. Every number that divides B also divides A (All of B's factors are also A's factors)

The product of the GCD and LCD of 135 and 90

GCD: 3^2 * 5 LCM: 2 * 3^3 * 5 (3^2 * 5) * (2 * 3^3 * 5) (45) * (270) = 12150 135 * 90 = 12150 So, the product of the GCD and LCM of two numbers is equal to the product of the two numbers

If A and B are positive integers, is A * B^4 a multiple of 2? 1. (2A + B)^3 is an even number 2. A + 3B is an odd number

Given: A, B > 0, Integers Question: Is AB^4 even? For AB^4 to be even, at least A or B must be even (or both) 1. (2A+B)^3 is an even number Since, (X)^3 = Even (2 * 2 * 2) Then 2A + B = E E + B = E B must be even. Sufficient 2. A + 3B is an odd number A + 3B = O If A = 1 and B = 2 1 + 6 = 7 (pass) If A = 2 and B = 1 2 + 3 = 5 (pass) If A = 1 and B = 3 1 + 9 = 10 (fail) Therefore, either A or B must be even for statement 2 to be true Sufficient Answer D

Prime Numbers

Have only 2 positive divisors (1 and the number itself) 2 is the smallest prime and the only prime that is even 2, 3, 5, 7, 11, 13, 17, 19 (these are consecutive PRIME numbers) 2 and 3 are the only consecutive numbers that are prime All positive integers are comprised of primes, or prime factors prime^2 = perfect square X^n has the same prime factors as X NX may have the same or more prime factors as X

If n is a positive integer, is n divisible by 2? 1. 7n - 8 is divisible by 20 2. 3n^2 + 2n + 5 is a prime number

If n is a positive integer, is n divisible by 2? Given: N > 0, Integer Question: Is N even? 1. 7n - 8 is divisible by 20 7n - 8 = [20,40, 60, 80] This means that 7n-8 = Even E - E = E E - O = O O- O = E Therefore, for 7n - 8 = Even then N must be Even. Sufficient 2. 3n^2 + 2N + 5 is a prime number Since N > 0, N must be >= 1. 3(1)^2 + 2(1) + 5 = 10 Therefore, 3n^2 + 2n + 5 >= 10 "is a prime number" = All prime numbers > 10 are odd. 3n^2 + 2n +5 = Odd 3n^2 + 2n = Odd - 5 3n^2 + 2n = Even 3n^2 + E = E 3n^2 must be Even. Therefore, N must be even. Sufficient. Answer D

Divisibility Rule 8:

If the last three digits of the number together are divisible by 8

Divisibility Rule 4:

If the last two digits together are divisible by 4

Divisibility Rule 6:

If the number is divisibility by both 2 and 3 - If the sum of the digits is divisible by 3 - If the units digit is even then it is divisible by 2

How to determine if a number has a terminating zero (or decimal that has finite number of nonzero digits)

If the only denominator only contains 2 and/or 5 primes then it is terminating -If the denominator contains 2, 5, and another prime number, check to see if it can be reduced - you may need both statements to determine if it can be reduced -Always solve for 1 & 2 in DS problems - if the denominator and the numerator can be reduced into an integer then the number has terminating zero -in DS if one statement is simply an integer (like 3) do not rule out C - if the numerator and demoninator reduce into an integer then it has terminating zeros

Divisibility Rule 9:

If the sum of its digits is divisible by 9

Even Numbers

Integers that are multiples of 2 Represented as 2N Terms with even coefficients (6x, 10x, 24x) are always even even^(any positive integer) = even The product of three consecutive integers is always even

Odd Numbers

Integers that are not multiples of 2 Represented as 2n + 1 or 2n - 1 odd^(any positive integer) = odd The sum of a odd number of odd numbers is odd The sum of a even number of odd numbers is even

The smallest common divisor of two integers

Is always 1

6n-1, 6n+1

Represents all prime numbers > 3 2, 3 are not represented in this equation

How many liters of a 60% alcohol solution must be added to 50 liters of a 40% alcohol solution to result in 50% alcohol solution?

Solution A = .6 alcohol Solution B = 20L alcohol (40% of 50) X= Amount to be added .6X + 20/50 + X = 1/2 .6X + 20 is the alcohol liters from solution A + the alcohol liters from solution B 1/2 represents that 50% of the final solution is alcohol -The amount of X from input solutions A & B must equal the amount of X in the final solution

What is the remainder when N is divided by 12? N = 12k + 14

Steps 1. Confirm format is in n = Dk + m (if yes, use process A) 2. Is M < D M = 14, D = 12 M is not less than D... so move to step 3 3. Reformat the equation to make it so M = D 12k + 12 * 1 + 2 n = 12(k + 1) + 2 n = 12k + 12 + 2 Remainder = 2

What is the remainder when N is divided by 12? N = 12k - 14

Steps 1. Confirm format is in n = Dk + m (it is not, use process B) N = Dk - m 2. Write m in the form ad - b 14 = 12 * 2 - 10, 10 < 12 n = 12 - (12 * 2 - 10) n = 12(k-2) + 10 n = Db + m 10 < 12 Remainder = 10

What is the remainder when N is divided by 12? N = 12k + 8

Steps: 1. Confirm format is in n = Dk + m (if yes, use process A) 2. Is M < D? M = 8 D=12 8 < 12 Remainder = 8

The amount of X from input solutions A and B must equal

The amount of X in the final solution

Standard Deviation

The difference between elements in a given data set from the data sets mean. -The more values are spread away from the mean, the greater the standard deviation -The sum of the difference between the mean and element of a data set is zero -(Standard Deviation)^2 = Variance

Range

The difference between the highest and lowest value in a given data set. -The difference of any two numbers in a data set cannot be greater than the Range

Finding the units digit of the multiplication/addition of different numbers

The last digit of the sum or multiplication of two or more numbers only depends on the last digit of each. Ex: What is the units digit of 1999 * 31452 * 60003 * 718437886 9 * 2 * 3 * 6 18 * 18 8 * 8 64 Answer: 4

Does y have a factor p such that 1 < p < y ?

This is asking whether Y is prime of not

Average Speed =

Total Distance / Total Time

What is the value of integer n if the integers 2, 5, 7 are the only prime factors of N?

WIth the given information you cannot determine the value of N. If you were given a contstraint, such as N < 100 then you could because 2*5*7 = 70.

(X+A)^2

X^2 + 2XA + A^2

(X+A)(X-A)=

X^2 - A^2