Divisibility Questions with Step-by-Step Solutions

Practice remainders, LCM, and algebraic divisibility — 7 questions with full reasoning

Understanding the Division Algorithm

Below is a set of seven divisibility practice questions — with clear, detailed solutions — covering remainders, the least common multiple (LCM), and algebraic reasoning. As a quick review, remember that when an integer \( n \) is divided by a nonzero integer \( m \), the result can always be written as:

\[ n = qm + r \]

where:

Try each question first, then expand View Solution to compare your reasoning with the full step-by-step method.

Practice Questions

Question 1 — Remainder after adding to n

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) \( n + 1 \)

View Solution

Since \( n \) leaves a remainder of \( 3 \) when divided by \( 5 \), we can write:

\[ n = 5k + 3 \]

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

\[ n + 2 = 5k + 3 + 2 = 5k + 5 = 5(k + 1) \]

Because \( 5(k + 1) \) is a multiple of \( 5 \), the expression \( n + 2 \) is divisible by \( 5 \) with remainder \( 0 \).

Answer: B

Question 2 — Next common multiple of 3, 5, and 12

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) \( n + 15 \)

View Solution

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

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

Thus \( n = 60k \) for some integer \( k \). The next larger integer divisible by all three numbers is the next multiple of \( 60 \):

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

Answer: D

Question 3 — Least common multiple of 5, 7, and 20

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

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

View Solution

The smallest number divisible by \( 5 \), \( 7 \), and \( 20 \) is their least common multiple. Because \( 20 \) is already a multiple of \( 5 \), we only need \( \mathrm{lcm}(7, 20) \). Since \( 7 \) and \( 20 \) are coprime:

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

Answer: E

Question 4 — Remainder of 6n modulo 8

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) \( 4 \)

View Solution

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

\[ n = 8k + 3 \]

Multiply both sides by \( 6 \):

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

Split \( 18 \) into a multiple of \( 8 \) plus the remainder (\( 18 = 16 + 2 \)):

\[ 6n = 48k + 16 + 2 = 8(6k + 2) + 2 \]

The expression is a multiple of \( 8 \) plus \( 2 \), so the remainder is \( 2 \).

Answer: C

Question 5 — Remainder of (2n + 2)² modulo 4

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

A) \( 0 \)     B) \( 1 \)     C) \( 2 \)     D) \( 3 \)     E) \( 4 \)

View Solution

Expand the expression:

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

Factor out a \( 4 \):

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

Because the entire expression is a multiple of \( 4 \), it is divisible by \( 4 \) with remainder \( 0 \).

Answer: A

Question 6 — Smallest two-digit number with digit sum 9 that is divisible by 3

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) \( 90 \)

View Solution

By the divisibility rule for \( 3 \), a number is divisible by \( 3 \) whenever the sum of its digits is divisible by \( 3 \). If the digit sum is \( 9 \), the number is actually divisible by \( 9 \) — and therefore also by \( 3 \).

The two-digit numbers whose digits sum to \( 9 \) are:

\( 18, 27, 36, 45, 54, 63, 72, 81, 90 \).

The smallest of these is \( 18 \).

Answer: D

Question 7 — Which number is not divisible by 3?

Which of the following numbers is not divisible by \( 3 \)?

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

View Solution

A number is divisible by \( 3 \) if the sum of its digits is a multiple of \( 3 \). Test each option:

  • 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
  • E: \( 6 + 7 + 2 = 15 \) — divisible by 3

Thus \( 1111 \) is not divisible by \( 3 \).

Answer: D

Quick Reference: Divisibility Rules

DivisorRuleExample
2Last digit is even344 ✓
3Digit sum is a multiple of 3552 → 12 ✓
4Last two digits form a multiple of 41116 → 16 ✓
5Last digit is 0 or 5135 ✓
9Digit sum is a multiple of 9738 → 18 ✓
10Last digit is 0980 ✓

Frequently Asked Questions

What is the division algorithm?

For an integer \( n \) and nonzero integer \( m \), there exist unique integers \( q \) and \( r \) with \( 0 \le r < m \) such that \( n = qm + r \). Here \( q \) is the quotient and \( r \) is the remainder.

How do I know if a number is divisible by 3?

A number is divisible by \( 3 \) if the sum of its digits is a multiple of \( 3 \). For example, \( 552 \to 5 + 5 + 2 = 12 \), so \( 552 \) is divisible by \( 3 \).

How do I find the least common multiple (LCM) of several numbers?

Find the prime factorization of each number, take the highest power of every prime that appears, and multiply. For \( 3 \), \( 5 \), \( 12 \): \( \mathrm{lcm} = 2^2 \times 3 \times 5 = 60 \).

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Last updated: October 17, 2026