GMAT CH 5: Divisibility and Remainders

Binomial Theorem

A theorem that gives the expansion formula for (a+b)^n

IF given an LCM GCD question - always try and convert original nos into prodcuts of their prime factors

AKA PRIME FACTORIZATION

List the roots (R) in decimal form (2-10)

R2 = 1.41; R3 = 1.73; R5 = 2.24; R6 = 2.45; R7 = 2.65; R8 = 2.83; R10 = 3.16

(a+b)^3

a^3+3a^2b+3ab^2+b^3

Question: If a and b are positive integers such that a/b = 97.16, which of the following cannot be the remainder when a is divided by b?

convert 97.16 into mixed fraction => 97*(16/100) = 97*(4/25) => this means remainder is 4 when divided by 25=> look at the answer options and see what is a multiple of 4 to check if it can be a remainder.

x div by y gives remainder z type questions?

no algebra, just solve multiples and get trial and error

N = integer when div by 3 gives remainder 2 => 3Q + 2 = n

this can also be written as 1 is less than next multiple oof 3 or => 3Q - 1 = n

In an evenly spaced set (constant difference in sequence)

mean = median

(2) Find the trailing zeros in a factorial

-> similar to (3) with prime as 5; first find n when 5^n =/< the term of factorial (f) -> compute divisor (before the decimal) of f/5^(n-1) + ...f/5^n -> add up the divisors and you get no of powers of a prime factor in factorial (or if you have use p=5 then no of trailing zeroes)

Pattern for repeating decimal conversion

notice the pattern!

(4) Find the last digit of (xyz)^n

this is the same as asking what is z^n; if n is large use cyclicity: pattern of repetition of numbers in multiples/squares/etc

use binomial remainder trick on 7^(11)/4

= [(8 - 1)^(11)] = remainder = -1 = 3

(3) Find powers of a prime factor in factorial

use method (2) but instead of 5 insert any prime factor the questions asks

whenever the average of the numbers will be an integer:

we will be able to write them as PRODUCTS OF a+b and a - b because one number will be some number more than the average and the other will be the same number less than average.

Binomial Theorem remainder trick

Every term of (26+1)^ 28 will be divisible by 26 except for the last one. The last term will be 1^28 = 1. Hence, when you divide 27^28 by 26, the remainder will be 1. (start by looking for a power one less or one more). If you cant manipulate the exponent directly then consider breaking up the base power - ex: 3^ab = 3^a * 3^b

Even roots have only a positive value on the GMAT.

So root of x is positive. But if x^2 = n is given then two roots (-ve and +ve exist)

Question: When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

TRICK: find first common n and then add LCM of divisors to it.

Successive Division

Trick for these nos is FIRST - UNDERSTAND THE CONCEPT Solution: Let's find one number, say n, which when divided successively by 4 and 5 leaves remainder 1 and 4 respectively. When you divide n by 4, you get a quotient and remainder 1. When you divide the quotient by 5, you get the remainder 4. What can the quotient be (when you divide by 5) for the remainder to be 4? The first one that comes to mind is that the quotient could be 4 itself. If you divide 4 by 5, the remainder is 4. Now, if in the previous step when you divided n by 4, if the quotient was 4 and the remainder was 1, then the number n must have been 4*4 + 1 = 17 (n = Quotient*Divisor + Remainder). Now, what will be the remainder when you divide 17 by 20? It will be 17.

co prime attribute

Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

When can we write a number as difference of squares?

When the number is odd OR When the number has 4 as a factor

coommon remainder two eqns

When we have a common remainder,the smallest value of n would be the common remainder. Say, if n were of the forms: (3a + 1) and (7b + 1), the smallest number of both these forms is 1

in E - O questions that have complex division

convert the division equation into multiplication equation.

(5) Cyclicity: z^n given n is some large no

cyclicity is when does the last digit of z^1 repeat itself in further exponents; for (4) after finding cyclicity ©-> n/; use remainder (r) to compute z^r -> last diigit of this is last digit of large number raised to large numbner question

negative remainders conversion v

(1 more than multiple of 3 is two less than next multiple of 3)

Binomial Theorem (a+b)^n

Notice that every term above is divisible by 'a' except for the last term b^n. Every term but the last has 'a' as a factor. T

Pascals triangle

Replaces nCk in binomial expansion = go to the n term row and the (k+1)th entry on the row

Successive Division trick

SECOND - Trick: List divisors in ascending List remainders in ascending The last remainder multiplied with the previous divisor added to the previous remainder = ANS => Algebra approach is also easy for two divisions

Odd no doubling -> result on factors

When odd number n is doubled, 2n has twice as many factors as n.That's because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2.

What is meant by - 'a number when divided successively by 4 and 5 leaves remainder 1 and 4 respectively'?

divide and then divide second by quotiient

tests of divisibility > 5

6 = if no is div by 3 and 2; 7 = if (double of last digit)-(rest of the two no) is div b 7 11: if (last digit) - (rest of the two no) is div b 11 8 = if last 3 are div by 8; 9 = if sum is div by 9; 12: If div by 4 and 3; 25: if last two digits are 00, 25, 75; NOTE: 4,8 and 3,9 and 6,12 = similar

Question 1: What is the remainder when 1555 * 1557 * 1559 is divided by 13?

(13*119 + 8)*(13*119 + 10)*(13*119 + 12) => 8*10*12 = 960 => 960/13 = remainder = 11 (can solve in easier way using negative remainders method)

(1) Find the no of factors in a large integer

-> prime factorize, add 1 to exponents of primes, multiply exp

RANGE OF LCM AND GCD VALUES

1 < = GCD of a set of integers < = Magnitude of the smallest integer of the set Magnitude of the largest integer of the set < = LCM of a set of integers < = Product of magnitude of all integers in the set

Question: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30? IS THERE A TRICK?

Answer options are the remainders. Divide each by 5 and 6 and see if the remainder is 4 and 3. Whichever satisfies both conditions is the correct answer.