Ch. 2_Divisibility

Prime 1

*Note: 1 is not prime.

Why LCM is important?

If x is divisible by A and by B, then x is divisible by the LCM of A and B, no matter what. A more general way to think of the LCM is: the LCM of A and B contains only as many of a prime factor as you see it appear in either A or B separately.

Factors of X with no common primes: combine

1. Create two factor trees to represent the given information. Initially, always write two given facts about a variable separately. That way you can think carefully about how to combine those facts. 2. When the primes from two trees are all different, you can put all the primes on one tree.

Organized way to make a table of factor pairs of 60

1. Label 2 columns "small" and "large." 2. Start with 1 in the small column and 60 in the large column. 3. The next number after 1 is 2. Since 2 is a factor of 60, then write "2" underneath the "1" in your table. Divide 60 by 2 to find 2's siblings in the pair. 4. Repeat this process until the numbers in the small and the large columns run into each other.

Every integer can be placed into one of two categories

Prime or not prime.

When two numbers don't share prime factors

Their LCM is just their product.

If you... Want all the factors of 60.

Then you... Make a table of factor pairs, starting with 1 and 60.

9 if the sum of the integer's digits is a multiple of 9.

This rule is very similar to the divisibility rule for 3. Take the number 144. 1+4+4=9, so 144 is divisible by 9.

Factor pairs

To find all the factors of a small number.

Divisibility

Travels up and down the factor tree.

If you get an integer out of the division, then the first number is divisible by the second.

18 is divisible by 3 because 18 divided by 3 = an integer. On the other hand, 12 is NOT divisible by 8, because 12 divided by 8 is NOT an integer.

The only even prime number

2

Primes less than 20 MEMORIZE

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

5 if the integer ends in 0 or 5.

75 or 80 are divisible by 5, but 77 and 84 are not.

10 if the integer ends in 0.

8,730 is divisible by 10, but 8,753 is not.

Prime factors are building blocks

Example: 150 Prime factors are 2, 3, 5, 5 5 appears twice on 150's tree. So you are allowed to multiply those two 5's together to produce another factor of 150, namely 25.

For large numbers

Generally start with the smallest prime factors and work you way up. Use your divisibility rules. You could also split up a large number (i.e., 630 to 63 and 10). *As you practice, you'll spot shortcuts. Either way will get you to the same set of prime factors.

Every number is divisible by the factors of its factors

If a is divisible by b, and b is divisible by c, then a is divisible by c as well. Example: 12 is divisible by 6, and 6 is divisible by 3. So then 12 is divisible by 3 as well.

When you combine two factor trees of x that contain overlapping primes Example

Is a number divisible by 18 also divisible by 6? Sure, because 6 goes into 18. Is a number divisible by 18 also divisible by 9? Sure, because 9 goes into 18. *This has two 3's and a 2. The trees provide information about how many of each kind of prime factors appear in x. Those pieces can overlap. *The two given facts are like statements given by two witnesses. Don't double-count the evidence.

When two numbers share prime factors

Their LCM will be smaller than their product.

If the facts are about different variables (x and y)

Then the facts don't overlap.

If you... Want all the factors of 96

Then you... Break 96 down to primes, then construct all the factors out of the prime factors of 96.

If you... Know two factors of x that have primes in common

Then you... Combine the two trees into one, eliminating the overlap = Know only that x is divisible by the LCM of the factors.

If you... Know two factors of x that have no primes in common

Then you... Combine the two trees into one.

Divisibility Rules For Small Integers MEMORIZE THESE AND THE FOLLOWING

These rules come in very handy. An integer is divisible by: -2 if the integer is even. -3 if the *sum* of the integer's digits is a multiple of 3. -5 if the integer *ends* in 0 or 5. -9 if the *sum* of the integer's digits is a multiple of 9. -10 if the integer *ends* in 0.

Factor foundation rule

Works in reverse to a certain extent. Example: If d is divisible by two different primes e and f, then d is also divisible by e x f. In other words, if 20 divisible by 2 and by 5, then 20 is also divisible by 2 x 5 (10).

Drill 8 32. 5 and 6 are factors of x. Is x divisible by 15?

Yes. For x to be divisible by 15, we need to know that it contains all of the prime factors of 15. 15 = 3 x 5. Therefore, 15 contains a 3 and a 5. x also contains a 3 and a 5, therefore, x is divisible by 15. *Got wrong second time*

Drill 2 9. If 2,499 is divisible by 147, is 2,499 divisible by 49?

Yes. The factor foundation rule is helpful here. It states that if 2,499 is divisible by 147, then 2,499 will also be divisible by all factors of 147. 147 is divisible by 49 (147/49 = 3). Since 49 is a factor of 147, 2,499 is also divisible by 49.

Drill 7 29. If integer a is not a multiple of 30, but ab is, what is the smallest possible value of integer b?

b = 2. For integer a to be a multiple of 30, it would need to contain all of the prime factors of 30: 2, 3, and 5. Since a is not a multiple of 30, it must be missing at least one of the these prime factors. So if ab is a multiple of 30, b must supply that missing prime factor. The smallest possible missing prime is 2. If b = 2 and a = 15 (or any multiple of 15), both of the initial constraints are met.

Primes are...

extremely important in any question about divisibility. ...are the DNA of a number. *Every number has a unique prime factorization.

Drill 3 14. Find four even divisors of 84.

2, 4, 6, 12, 14, 28, 42, and 84. Got right but was unsure. Step 1: Identify the divisors (or factors) by breaking 84 down into prime numbers. The prime factors are 2, 2, 3, and 7. Because the question asks for even factors, only factors that are built with at least one 2 will be correct. Even divisors can be built using one 2. Alternatively, you can make a factor pair table.

Drill 9 39. If n is the product of 2, 3, and a two-digit prime number, how many of its factors are greater than 6?

4. Because we have been asked for a concrete answer, we can infer that the answer will be the same regardless of which 2-digit prime we pick. So for simplicity's sake, let's pick the smallest and most familiar one: 11. If n is the product of 2, 3, and 11, its factors are 1, 66 2, 33 3, 22 6, 11 In this case, we can simply use the right hand portion of our chart. We have four factors larger than 6: 11, 22, 33, and 66. *Got wrong second time*. I said 3. I did not count 66.

Divisibility

Has to do with integers. Integers have no decimals or fractions attached.

2 if the integer is even.

Any even number is, by definition, divisible by 2. Even numbers end in 0, 2, 4, 6, or 8.

Factor tree

Break down until you can't further. The numbers should all be prime. When you find a prime factor, that branch on the factor tree has reached the end. Circle prime numbers as you go. Is the best way to find a prime factorization. Purpose is to break integers down into primes.

When you combine two factor trees of x that contain overlapping primes

Drop the overlap. You're already covered.

Factors are divisors

Ex: 6 is divisible by 1, 2, 3, and 6. That means that 1, 2, 3, and 6 are FACTORS of 6. Learn all the ways you might see this relationship expressed on the GMAT.

Factors: built out of primes

Example: 60 All the factors of 60 (except 1) are different combinations of the prime factors of 60. In other words, every factor of a number (again, except 1) can be expressed as the product of a set of its prime factors. This relationship between factors and prime factors is true of every number.

Drill 9 37. 14 and 3 divide n. Is 12 a factor of n?

Maybe. For 12 to be a factor of n, n must contain all of the prime factors of 12. 12 = 2 x 2 x 3, so 12 contains two 2's and a 3. n also contains a 3 but only contains one 2 that we know of, so we don't know whether 12 is a factor of n. *Got wrong second time* I said no.

Drill 8 33. If q is divisible by 2, 6, 9, 12, 15, & 30, is q divisible by 8?

Maybe. To be divisible by 8, q needs three 2's in its prime factorization. Rather than combine all of the listed factors, we can just look through and see how many 2's we have. We can't simply count all of the numbers that contain 2, because we might have some overlapping factors. 6 is a multiple of 2 and 3, so the fact that q is divisible by both 2 and 6 tells us only that we have at least one 2 (and at least one 3); we don't necessarily have two factors of 2. Instead, we look for the largest number of 2's we see in one factor. 12 contains two 2's, so we know that q must be a multiple of 4, but we do not know whether q contains three 2's. It might or it might not. *Was unsure second time*

Drill 8 34. If p is a prime number, and q is a non-prime integer, what are the minimum and maximum numbers of factors they can have in common?

Minimum = 1; Maximum = 2. Start with the more contained variable p. Because it is prime, we know that it has exactly 2 factors - itself and 1. Therefore, our maximum number of "factors in common" cannot be more than 2. Can p and q have exactly 2 factors in common? Certainly; q can be a multiple of p. What about the minimum? Can p and q have absolutely no factors in common? Try some numbers. If we choose p = 3 and q = 10, then the two numbers don't have any prime factors in common, but notice that they are both divisible by 1. Any number is always divisible by 1. Therefore, our minimum possible number of factors is 1 and our maximum is 2.

Drill 7 28. If 7x is a multiple of 210, must x be a multiple of 12?

No. For x to be a multiple of 12, it would need to contain all of the prime factors of 12: 2, 2, and 3. If 7x is a multiple of 210, it contains the prime factors 2, 3, 5, and 7. However, we want to know about x, not 7x, so we need to divide out the 7. Therefore, x must contain the remaining primes: 2, 3, and 5. Comparing this to the prime factorization of 12, we see that x does have a 2 and 3, but we don't know whether it has two 2's. Therefore, we can't say that x must be a multiple of 12; it could be, but it doesn't have to be. *Got wrong second time*

Drill 6 24. Determine which of the following numbers are prime numbers. Remember, you only need to find one factor other than the number itself to prove that the number is not prime. 2 3 5 6 7 9 10 15 17 21 27 29 31 33 258 303 655 786 1,023 1,325

Prime numbers are: 2, 3, 5, 7, 17, 29, 31. Not prime: all of the even numbers other than 2. All the remaining multiples of 5 (15, 655, and 1,325). All of the remaining number whose digits add up to a multiple of 3 (9, 21, 27, 33, 303, 1,023). These are all divisible by 3.

3 if the sum of the integer's digits is a multiple of 3.

Take the number 147. Its digits are 1, 4, 7. 1+4+7=12, which is a multiple of 3. So 147 is divisible by 3.

Factors of x with primes in common: combine to LCM

The LCM of two numbers, say A and B, is the smallest number that is a multiple of both A and B.

Drill 9 36. The prime factorization of a number is 2 x 2 x 3 x 11. What is the number and what are all its factors?

The number is 132 and the factors are 1, 2, 3, 4, 6, 11, 12, 22, 23, 33, 44, 66, and 132. 2 x 2 x 3 x 11 = 132