Factors

Question 1

Finding how many different factors a number has

Answer 1

1. Find the prime factorization of the number
2. Add 1 to the value of each exponent of the prime factors, then multiply these results. The product is the answer

e.g., 2,160 = 2⁴ x 3³ x 5¹ → (4+1) x (3+1) x (1+1) = 40 factors

Question 2

Total number of prime factors ≠ Total number of unique prime factors

Answer 2

e.g., 32 has a total of 5 prime factors but 1 unique prime factor of 2

Question 3

“All the same factors” vs. “All the same unique prime factors”

Answer 3

What is the GCF of 24 and x?

1. 24 and x share all the same factors
2. 24 and x share all the same unique prime factors

Answer = A
→ “share all the same factors” means that the two numbers are the same

Question 4

The number of unique prime factors in a number doesn’t change when that number is raised to a positive integer exponent

Answer 4

25 and 625 both have 1 unique prime factor

• 25 = 5²
• 625 = 25² = 5⁴

Question 5

LCM Calculation

Answer 5

A prime factor is considered repeated when that prime factor is shared by at least 2 of the numbers in the set. It doesn’t need to be shared by all of the numbers to be considered a repeated prime factor

Question 6

LCM Tips

Answer 6

• If two positive integers, x and y, share no prime factors, the LCM of x and y is xy. Otherwise, LCM < xy.
• For any set of positive integers, the LCM will always be ≥ the largest number in the set
• The LCM provides all of the unique prime factors of some set of positive integers. Thus, the LCM provides all the unique prime factors of the product of the numbers in the set.
• The LCM can be used to determine when two processes that occur at differing rates or times will coincide
• If z is divisible by both x and y, z must also be divisible by the LCM(x, y). z must be at least the LCM.

Question 7

GCF Tips

Answer 7

• For any set of positive integers, the GCF will always be ≤ the smallest number in the set
• Two consecutive integers will never share the same prime factors, i.e., GCF(n, n+1) = 1
• When finding the GCF of more than 2 numbers, “common” means shared by all numbers
• The GCF doesn’t provide all the prime factors of 2 numbers. e.g., GCF(p,q) = 6, is pq/240 an integer? Insufficient info.

Question 8

Consecutive Integers: Common Factors

Answer 8

Two consecutive integers are co-prime, which means that they don’t share ANY common factor but 1.

Example:

Question 9

The LCM and GCF when a positive integer y divides evenly into positive integer x

Answer 9

• LCM = x
• GCF = y

“y divides into x” = x / y

Question 10

LCM + GCF Trick

Answer 10

xy = LCM(x, y) x GCF(x, y)

Question 11

Converting a remainder from decimal form to fraction form

Answer 11

When a division of two integers results in a decimal, such as x/y = 9.48, we can’t be sure what the actual remainder is. However, the actual remainder must be a multiple of the most reduced fractional remainder, i.e., 12 in this case since 48/100 = 12/25

Question 12

Remainder Operations

Answer 12

Remainders can be added (must correct for excess remainders at the end of addition) and subtracted (must correct for negative remainders). Just remember that when we add or subtract remainders, we need to keep them in order.

• Remainder of (12+13+17)/5?
• Remainders of 12, 13, 17 are 2, 3, 2
• True remainder = (2+3+2)/5 = 1 R2 (answer = 2)

Same concept applies to multiplication.

• Remainder of (12 x 13 x 17)/5?
• Remainders of 12, 13, 17 are 2, 3, 2
• True remainders = (2 x 3 x 2)/5 = 2 R2 (answer = 2)

Question 13

Remainder Algebra

Answer 13

Perform the Dividend = Divisor(Quotient) + Remainder equation when it can be set up that way. If using this method doesn’t yield a clear path, just test numbers.

If n is a positive integer, and r is the remainder when n³ - n is divided by 9, what is the value of r?

1. n has a remainder of 1 when it’s divided by 9 → n = 9k + 1 → plug in n(n+1)(n-1) → divisible by 9
2. n² has a remainder of 1 when it’s divided by 9 → n² = 9s + 1 → plug in n(n²-1) → divisible by 9

Question 14

Range of Possible Remainders

Answer 14

If a divisor is n, then possible remainders could be any integers ranging from 0 to (n-1), inclusive.

Question 15

Remainder Patterns in Powers

Answer 15

What number must be subtracted from 2⁵²⁶ so that the resulting integer will be a multiple of 3?

• Observe patterns. The remainders of 2¹÷3, 2²÷3, 2³÷3, 2⁴÷3 are 2, 1, 2, 1, respectively.
• 2⁵²⁶ ÷ 3 has a remainder of 1
• Hence subtract 1

Question 16

Shortcut for Determining the Number of Primes in a Factorial

Answer 16

To determine the (largest) number of a prime number x that divides into y!:

1. Divide y by x¹, x², x³, etc. Keep track of the quotients while ignoring any remainders. Stop dividing when y/x^k produces a quotient of 0.
2. Add the quotients from the previous divisions. The sum represents the number of prime number x in the prime factorization of y!.

THINK: How many x’s are in y!?

Question 17

What is the largest integer value of k, such that 400! / 5^k is an integer?

Answer 17

400/5<sup>1</sup> = 80
400/5<sup>2</sup> = 16
400/5<sup>3</sup> = 3… remainder
400/5<sup>4</sup> = 0… remainder

Answer = 80 + 16 + 3 = 99

Question 18

Shortcut for Determining the Number of Non-Primes in a Factorial

Answer 18

To determine the (largest) number of a non-prime number x that divides into y!:

1. Break x into prime factors
2. Use the largest prime factor of x (because there will be fewer of this), apply the factorial divisibility shortcut.
3. The quantity determined represents the largest number of x that divides into y!.

Question 19

What is the largest integer value of n, such that 40! / 6ⁿ is an integer?

Answer 19

6<sup>n</sup> = 2<sup>n</sup> x 3<sup>n</sup>
40/3<sup>1</sup> = 13…remainder
400/3<sup>2</sup> = 4…remainder
400/3<sup>3</sup> = 1… remainder
400/3<sup>4</sup> = 0… remainder

Answer = 13 + 4 + 1 = 18

Question 20

0! = ___

Answer 20

1

Question 21

If N²/M is an integer,

how many unique prime factors do N and M have relative to each other?

Answer 21

M must have fewer than or the same number of unique prime factors as N

Question 22

If n is a prime number between 0 and 100, how many positive divisors does n³ have?

Answer 22

4: n⁰, n¹, n², n³

Question 23

GCF and LCM: Hard Question

Answer 23

M and N are integers such that 6 < M < N. What is the value of N ?

1. The greatest common divisor of M and N is 6.
2. The least common multiple of M and N is 36.

Given that the greatest common divisor (GCD) of M and N is 6 and 6 < M < N, then it is possible that M = (6)(5) = 30 and N = (6)(7) = 42. However, it is also possible that M = (6)(7) = 42 and N = (6)(11) = 66; NOT sufficient.

Given that the least common multiple (LCM) of M and N is 36 and 6 < M < N, then it is possible that M = (4)(3) = 12 and N = (9)(2) = 18. However, it is also possible that M = (4)(3) = 12 and N = (9)(4) = 36; NOT sufficient.

Taking (1) and (2) together, it follows that 6 is a divisor of M and M is a divisor of 36. Therefore, M is among the numbers 6, 12, 18, and 36. For the same reason, N is among the numbers 6, 12, 18, and 36. Since 6 < M < N, it follows that M cannot be 6 or 36 and N cannot be 6. Thus, there are three choices for M and N such that M < N. These three choices are displayed in the table below, which indicates why only one of the choices, namely M = 12 and N = 18, satisfies both (1) and (2).

Question 24

Factor Hard Question

Answer 24

If n is a positive integer and n² is divisible by 72, then the largest positive integer that must divide n is: 12

• Since n² is divisible by 72, n² = 72k for some positive integer k.
• Since n² = 72k, then 72k must be a perfect square.
• Since 72k = (2³)(3²)k, then k = 2m² for some positive integer m in order for 72k to be a perfect square.
• Then, n² = 72k = (2³)(3²)(2m²) = (2⁴)(3²)m2 = [(2²)(3)(m)]², and n = (2²)(3)(m). The positive integers that MUST divide n are 1, 2, 3, 4, 6, and 12. Therefore, the largest positive integer that must divide n is 12.