**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)

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