Let p and q be positive integers such that 18 is a factor of p and 45 is a factor of q. Wh… — GRE General Test

Let $p$ and $q$ be positive integers such that $18$ is a factor of $p$ and $45$ is a factor of $q$ . Which of the following statements MUST be true? Select all such statements.

The sum $p + q$ is divisible by $9$ .Answer
The product $p \cdot q$ is divisible by $270$ .Answer
C
The sum $p + q$ is divisible by $18$ .
D
The product $p \cdot q$ is divisible by $1620$ .
The sum $p^2 + q^2$ is divisible by $81$ .Answer

Answer

The statements asserting that the sum

p + q

is divisible by

9

, the product

p \cdot q

is divisible by

270

, and the sum

p^2 + q^2

is divisible by

81

MUST be true.

The correct options are those stating that the sum

p + q

is divisible by

9

, the product

p \cdot q

is divisible by

270

, and the sum

p^2 + q^2

is divisible by

81

. Each of these statements can be proven algebraically by substituting

p = 18a

and

q = 45b

into the expressions and factoring out the required divisor.

Step-by-Step Solution

Express

p

and

q

in terms of their prime factorizations and integer multipliers.

p = 18a = (2 \cdot 3^2)a

and

q = 45b = (3^2 \cdot 5)b

for positive integers

a

and

b

Establishing explicit algebraic representations allows verification of divisibility for any combination of

p

and

q

Evaluate the divisibility of the sum

p + q

p + q = 18a + 45b = 9(2a + 5b)

Factoring out

9

proves that

p + q

is always divisible by

9

. However,

2a + 5b

is not necessarily even (e.g., if

a=1, b=1

2a+5b=7

), so

p+q

is not guaranteed to be divisible by

18

Evaluate the divisibility of the product

p \cdot q

p \cdot q = (18a)(45b) = 810ab = 270(3ab)

Since

810ab

is a multiple of

270

, the product is always divisible by

270

. However, it is not necessarily a multiple of

1620

unless

ab

contains an additional factor of

2

Evaluate the divisibility of the sum of squares

p^2 + q^2

p^2 + q^2 = (18a)^2 + (45b)^2 = 324a^2 + 2025b^2 = 81(4a^2 + 25b^2)

Factoring out

81

proves that

p^2 + q^2

is always divisible by

81

Key Concept

Divisibility rules and algebraic factoring of linear and quadratic integer expressions

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic