All Chapter Review

Question 1

If a and b are integers and
2(a^2)(b) + a^2 is even, which of the following must be odd? (could be multiple)

1. ab + 1
2. 2a(b^2) + b^2
3. (a^2)(b^2) - 1

Answer 1

1 and 3

When we factor the expression in the question stem we get: (a^2)(2b + 1)

That means (2b + 1) will always be odd, so a^2 and thus a must be even. (we don’t know what b is).

1. even*anything= even
   even + 1 = odd
2. factored is:
   (b^2)(2a +1)

we don’t know if b is even or odd, so we don’t know if b squared is even or odd. We can’t determine #2.

3. we know a squared is even. So a^2 * b^2 must be even. Even - 1 will always be odd.

Question 2

If 120m = n^2, all of the following must divide into n^2 except?

A. 20
B. 40
C. 720
D. 1,440
E. 1,800

Answer 2

D.

Prime factorize 120 and we get (5)(3)(2^3), this means the smallest possible value of n^2 is 3,600 (all prime factor exponents must be even).

All of the answer choices divide evenly into 3,600 except for D.

Question 3

When x is divided by 23 the remainder is 12. When x is divided by 29 the constant is Q and remainder is 12. Q must be divisible by which of the following numbers?

a. 6
b. 12
c. 23
d. 29
e. 52

Answer 3

C. 23

Set the 2 equations equal to each other and solve for the constant of the 1st equation (we can call it P). That allows us to see what integer we can divide Q by.

The equations simplify to p = 29Q/23