Watch your signs and be sure to only count distinct products. Be methodical: start at the left and multiply that number by each of the other numbers, then do the same thing for the next number. Don't write down any numbers that aren't distinct.

-7: 35, 21, 0, -7, -21, -35, -49

-5: 15, 5, -15, -25

-3: 3, -3, 9

0: no distinct products

1: -1

3, 5, and 7 have no distinct products. The number of distinct products is 17 (B). You could write out the pairs: first match A with the other letters to get AB, AC, AD, AG, AF, and AG. A will repeat after 6 matches, so it's out. B forms 5 pairs (BC, BD, BE, BF, BG), C forms 4 pairs, D forms 3 pairs, E forms 2 pairs, and F forms 1 pair. Add them all up, 6+5+4+3+2+1=21. Plug in values for a and b. Because ab must be less than 10, you should start with the lowest possible values for each. The smallest possible value for b is 0, and the smallest possible value for b is 0, so ab=0. Plug in 1&2 for a, which produces new values for ab of 4 and 8, respectively, both of which are less than 10. If we try a=3, however, ab is 12, which is more than 10. There are no more distinct values. So the answer is (D), three. The easiest way to solve this problem is to PLUG IN numbers for y and z. Let's say y=-2 and z=3. The left side of the inequality in A is (-2)²(3)²=(4)(9)=36, in B is (-2)(3)²=(-2)(9)=18, in C is (-2)²(3)=12, in D is (-2)(3)=-6, and E is (-2)³(3)=-24. Since (B), (D), and (E) are all true, you can eliminate them. Now let's try y=-2 and z=-3. A is (-2)²(-3)²=36, and the left side of C is (-2)²(-3)=(4)(-3)=-12. Since (C) is now true, you can eliminate it as well, and therefore, through process of elimination, (A) is the answer. The positive odd integers less than 6 are 1, 3, and 5, so those are the possible values of a, b, and c. The question mentioned that a, b, and c, are distinct, which means they're each a different integer. Now try out all the possible arrangements:

1(3⁵)=243

1(5³)=125

3(1⁵)=3

3(5¹)=15

5(1³)=5

5(3¹)=15

The question asks for how many different values there are, so don't count a number twice. There are five different values (D). The elements of set f are all the integers less than 20 that are the products of exactly two different prime numbers. Which of the following is set f?

A. {2,3,5,7}

B. {2,3,4,5,7,9}

C. {6,10,14}

D. {6,10,14,15}

E. {4,6,9,10,14,15} PRIME FACTORS: STARS WITH 2, NOT 1

(A) {(1*2),(1*3),(1*5),(1*7)} ---> 1 is not a prime number = X

(B) {(1*2),(1*3),(1*4),(1*5),(1*7),(1*9)} ---> 1 is not a prime number = X

(C) {(2*3),(2*5),(2*7)} ---> good, but is it ALL? Let's see.

(D) {(2*3),(2*5)(2*7)(2*5)} ---> good, but again, is it ALL?

(E) {(2*2)..... wait = not two DISTINCT prime numbers

We're left with (D). In this problem, I can make up the value of n to solve it much more quickly. Since the value of m depends on the value of n, and since the side with n is more complicated than the side with m, I want to focus on the complicated side first.

say n=1

m = 1-2-2 = -5 = I works!

next, say n=-1

m = -1+2-2 = -3 = II works!

Does III work?

If I try moving to the next integer, -2, I realize that the last part of the equation (2/n²) would be a fraction.

If I try 3, then both of the (2/n) portions give non-integers. 4? 5? Worse and worse! I quickly realize that when making up possible values for n, anything other than -1 or 1 will not give me an integer value for m.

Thus, the answer is (B), I and II only.