Divisibility Questions with Solutions

Divisibility questions are presented below with clear and detailed solutions. When an integer \( n \) is divided by a nonzero integer \( m \), the result can be written in the form

\[ n = qm + r \]

where \( n \) is the dividend, \( m \) is the divisor, \( q \) is the quotient, and \( r \) is the remainder.


Questions with Solutions

Question 1

If a positive integer \( n \) is divided by 5, the remainder is 3. Which of the following expressions yields a remainder of 0 when divided by 5?

A) \( n + 3 \)    B) \( n + 2 \)    C) \( n - 1 \)    D) \( n - 2 \)    E) \( n + 1 \)

Solution

If \( n \) leaves a remainder of 3 when divided by 5, then

\[ n = 5k + 3 \]

for some integer \( k \). Adding 2 to both sides gives

\[ n + 2 = 5k + 5 = 5(k + 1) \]

Thus, \( n + 2 \) is divisible by 5 and leaves a remainder of 0.

Answer: B


Question 2

If an integer \( n \) is divisible by 3, 5, and 12, what is the next larger integer divisible by all three numbers?

A) \( n + 3 \)    B) \( n + 5 \)    C) \( n + 12 \)    D) \( n + 60 \)    E) \( n + 15 \)

Solution

If \( n \) is divisible by 3, 5, and 12, then it must be a multiple of their least common multiple:

\[ \mathrm{lcm}(3,5,12) = 60 \]

Thus,

\[ n = 60k \]

The next larger integer divisible by all three is

\[ n + 60 = 60(k + 1) \]

Answer: D


Question 3

What is the smallest positive integer that is a multiple of 5, 7, and 20?

A) 70    B) 35    C) 200    D) 280    E) 140

Solution

The smallest number divisible by 5, 7, and 20 is their least common multiple:

\[ \mathrm{lcm}(5,7,20) = 140 \]

Answer: E


Question 4

When the integer \( n \) is divided by 8, the remainder is 3. What is the remainder when \( 6n \) is divided by 8?

A) 0    B) 1    C) 2    D) 3    E) 4

Solution

Since \( n \) leaves a remainder of 3 when divided by 8,

\[ n = 8k + 3 \]

Multiplying both sides by 6:

\[ 6n = 6(8k + 3) = 48k + 18 \]

Rewrite 18 as \( 16 + 2 \):

\[ 6n = 8(6k + 2) + 2 \]

Therefore, the remainder is 2.

Answer: C


Question 5

If \( n \) is an integer, when \( (2n + 2)^2 \) is divided by 4, the remainder is

A) 0    B) 1    C) 2    D) 3    E) 4

Solution

First expand the expression:

\[ (2n + 2)^2 = 4n^2 + 8n + 4 \]

Factor out 4:

\[ (2n + 2)^2 = 4(n^2 + 2n + 1) \]

Thus, the expression is divisible by 4 and the remainder is 0.

Answer: A


Question 6

What is the smallest positive two-digit integer divisible by 3 such that the sum of its digits is 9?

A) 27    B) 33    C) 72    D) 18    E) 90

Solution

Let the two-digit number be \( 10x + y \), where \( x \) and \( y \) are digits. Divisibility by 3 implies

\[ 10x + y = 3k \]

The sum of the digits is 9:

\[ x + y = 9 \]

Solving for \( y \) gives \( y = 9 - x \). Substitute:

\[ 10x + 9 - x = 3k \] \[ 9x + 9 = 3k \] \[ x = \frac{k - 3}{3} \]

Trying integer values of \( k \), the smallest valid digit is \( x = 1 \), giving \( y = 8 \). The number is 18.

Answer: D


Question 7

Which of the following numbers is not divisible by 3?

A) 339    B) 342    C) 552    D) 1111    E) 672

Solution

A number is divisible by 3 if the sum of its digits is divisible by 3.

A) \( 3 + 3 + 9 = 15 \) (divisible by 3)
B) \( 3 + 4 + 2 = 9 \) (divisible by 3)
C) \( 5 + 5 + 2 = 12 \) (divisible by 3)
D) \( 1 + 1 + 1 + 1 = 4 \) (not divisible by 3)

Thus, 1111 is not divisible by 3.

Answer: D


More References and Links

Divisibility Calculator
Divisibility Rules with Examples
Numbers and Fractions