“FACTORS”, First get hold of it ! Part 2 (Practice questions )

Practice Questions:

1) The number of boxes in a warehouse can be divided evenly into 6 equal shipments by boat or 27 equal shipments by truck. What is the smallest number of boxes that could be in the warehouse?
(A) 27
(B) 33
(C) 54
(D) 81
(E) 162

2) How many odd factors does 210 have?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

3) If n is the smallest integer such that 432 times n is the square of an integer, what is the value of n?
(A) 2
(B) 3
(C) 6
(D) 12
(E) 24

4) How many distinct prime numbers are factors of 33150?
(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight

5) If n is a positive integer, then n(n + 1)(n – 1) is
(A) even only when n is even
(B) odd only when n is even
(C) odd only when n is odd
(D) always divisible by 3
(E) always one less than a prime number

Answers

1) C
2) E
3) B
4) B
5) D

Explanations

1) This tells us that the number of boxes is evenly divisible by both 6 and 27; in other words, it’s a common multiple of 6 and 27. The question says: what’s the smallest value it could have? In other words, what’s the LCM of 6 and 27? (This question is one example of a real-world set-up where the question is actually asking for the LCM.)

Step (a): 6 = 2*3 27 = 3*3*3

Step (b): 6 = 2*3 27 = 3*3*3 GCF = 3

Step (c): 6 = 3*2 27 = 3*9

Step (d) LCM = 3*2*9 = 54

Thus, 54 is the LCM of 6 and 27.

Answer: C.

2) Start with the prime factorization: 210 = 2*3*5*7

For odd factors, we put aside the factor of two, and look at the other prime factors.
set of exponents = {1, 1, 1}
plus 1 to each = {2, 2, 2}
product = 2*2*2 = 8
Therefore, there are 8 odd factors of 210. In case you are curious, they are {1, 3, 5, 7, 15, 21, 35, and 105}

Answer: E.

3) The prime factorization of a square has to have even powers of all its prime factors. If the original number has a factor, say of 7, then when it’s squared, the square will have a factor of 7^2. Another way to say that is: any positive integer all of whose prime factors have even powers must be a perfect square of some other integer. Look at the prime factorization of 432

432 = (2^4)*(3^3)

The factor of 2 already has an even power —- that’s all set. The factor of 3 currently has an odd power. If n = 3, then 432*n would have an even power of 2 and an even power of 3; therefore, it would be a perfect square. Thus, n = 3 is a choice that makes 432*n a perfect square.

Answer: B.

4) Start with the prime factorization:

33150 = 50*663 = (2*5*5)*3*221 = (2)*(3)*(5^2)*(13)*(17)

There are five distinct prime factors, {2, 3, 5, 13, and 17}

Answer: B.

5) Notice that (n – 1) and n and (n + 1) are three consecutive integers. This question is about the product of three consecutive integers.

If n is even, then this product will be (odd)*(even)*(odd) = even

If n is odd, this this product will be (even)*(odd)*(even) = even

No matter what, the product is even. Therefore, answers (A) & (B) & (C) are all out.

Let’s look at a couple examples, to get a feel for this

3*4*5 = 60
4*5*6 = 120
5*6*7 = 210
6*7*8 = 336
7*8*9 = 504

Notice that one of the three numbers always has to be a multiple of 3: when you take any three consecutive integers, one of them is always a multiple of 3. Therefore the product will always be divisible by 3.

Therefore, Answer: D.

BTW, for answer choice E of that question, you will notice that for some trios of positives integers, adding one to the product does result in a prime, but for others, it doesn’t.

3*4*5 + 1 = 61 = prime
4*5*6 + 1 = 121 = 11^2 (not prime)
5*6*7 + 1 = 211 = prime
6*7*8 + 1 = 337 = prime
7*8*9 + 1 = 505 = 5*101 (not prime)

cat questions on prime numbers
factors questions and answers
how many factors of 16x125x2401 are odd numbers
how many factors are even numbers
2iim number system questions
the number of factors of n 3 5 5 3 7 2
if a two digit number pq has 3 divisors then how many divisors does the number pqpq have
formula for number of even factors