 ## Thursday, 2 July 2015

1. If a is an even integer, b is an integer and 2 is not a factor of a+b, which of the following could be
the value of b?
(1) 0  (2) 2  (3) 4  (4) 10  (5) 15
Solution: (5)
If 2 is not a factor of a+b, then a +b is odd. Because a is even , the only way to make the sum odd is to
make b odd. The numbers in choices (1), (2), (3) and (4) are all even. The answer is choice (5).

2. If a is an integer and b = 5a + 3 , which of the following can be a divisor of b ?
(1) 5  (2) 10  (3) 13  (4) 17  (5) 25
Solution: (3) Note that 5a is divisible by 5 but that 3 is not. Therefore, 5a + 3 cannot be divisible by 5 or
any multiple of 5.That eliminates choices (1),(2) and (5). Try Plugging in a few values for a. If a = 1, then b
= 8.However, none of the answers is a divisor of 8. If a = 2, then b = 13, which is divisible by 13.The

3. If x and y are integers and x2 – y2 is odd, which of the following must be odd ?
(1) x  (2) y   (3) x2  (4) x2 + 1  (5) x + y
Solution: (5)

Try to find counterexamples for each answer choice. For choices (1) and (3), if x = 4 then y could
equal3.Since x does not have to be odd, choices(1) and (3)can be eliminated. For choice (2), if y =4, then
x could equal 5. Since y does not have to be odd, choice (2)can be eliminated .For choice (4), if x = 5,
then x2 + 1 = 26 and y could equal 3. Eliminate choice (4). So, choice (5) must be correct. Moreover, note
that x2 – y2 = (x+y) (x-y) which means that both (x+y) and (x-y) must be odd. The answer is choice (5).

4. When m is divided by 5, the quotient is 2 and the remainder is b. When n is divided by 7, the
quotient is 2 and the remainder is also b. If mn = 221,what is the value of b ?
(1) 2  (2) 3  (3) 4  (4) 13  (5) 27
Solution: (2)
Since the remainder cannot be larger than the divisor, choices (4) and (5) can be eliminated. Since
choice (2) is the middle value of the remaining answers, start there. If b = 3, then m = ( 5x2) + 3 = 13 and
n=(7x2)+3 =17.Because mn = 13 x 17 = 221, b= 3. The answer is choice (2)

5. Which of the following is a multiple of 4! + 6?
(1) 2! + 3  (2) 4! + 12  (3) 5! + 6  (4) 5! + 30  (5) 8! + 12
Solution: (4)
A multiple of 4! + 6 can be factored into 4! + 6 and an integer. So, one way to determine a multiple is to
divide the answer choices by 4! + 6. For choice (4), 5!+30 / 4!+6 = 5(4!+6) / 4! +6) = 5. Be careful of
choice (5) : 4! X 2 ≠ 8 ! The answer is choice (4).