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You will learn: Replace large numbers by small remainders and recognise repeating power cycles.
Before you start: Division with remainder and integer powers.
Keep the remainder, not the whole number
When an integer a is divided by a positive integer m, it can be written a = mq + r, where q is an integer and 0 ≤ r < m. The number r is the remainder. For instance, 23 = 5 × 4 + 3 gives remainder 3 on division by 5.
The notation a ≡ b (mod m) means that m divides a − b. In words, a and b have the same remainder on division by m. Thus 23 ≡ 3 (mod 5), but 23 is not equal to 3. The symbol ≡ records a remainder relationship.
Add and multiply small remainders
If a = mq + r and b = ms + t, then a + b = m(q + s) + r + t. So the remainder of the sum comes from r + t; divide again if r + t is at least m. Also ab = m(mqs + qt + rs) + rt, so the product has the same remainder as rt.
For negative numbers, use the same required range. Since −2 = 5(−1) + 3, the remainder of −2 on division by 5 is 3, not −2.
Worked example 1: multiply without expanding
Find the remainder of 48 × 73 on division by 7. We have 48 ≡ 6 (mod 7) and 73 ≡ 3 (mod 7). Multiply these small remainders: 6 × 3 = 18. Since 18 = 7 × 2 + 4, the required remainder is 4. Writing 18 as the answer would miss the condition that a remainder is smaller than 7.
Worked example 2: the last digit of a power
Find the last digit of 3¹⁴. A last digit is a remainder on division by 10. The first four powers end in 3, 9, 7, 1. Multiplying by 3 starts the pattern again.
Since 14 = 4 × 3 + 2, three complete groups of four contribute a last digit of 1. The two remaining factors contribute the second digit of the cycle, 9. Equivalently, 3¹⁴ = (3⁴)³ × 3² ≡ 1³ × 9 ≡ 9 (mod 10).
Do not cancel a factor without checking
From 2 × 1 ≡ 2 × 4 (mod 6), it does not follow that 1 ≡ 4 (mod 6). Their difference is 3, not a multiple of 6. Addition and multiplication preserve congruences; division needs an extra condition. In this introductory lesson, work directly with the possible remainders rather than cancelling.
Practise at your next step
A six-question session chooses from nine questions. Two correct answers in a row without hints move you up a level; an incorrect answer brings a simpler next question where one is available. Hints keep you at the same level. This is a practice suggestion, not an exam score or proof of mastery.
Interactive practice loads here. You can also use the complete question set below.
Write a proof of your own
Prove that the square of every integer has remainder 0 or 1 when divided by 4.
Write your reasoning on paper before comparing. The practice checker does not grade a written proof.
Compare your proof with a full solution
Every integer n is even or odd. If n = 2k, then n² = 4k², with remainder 0. If n = 2k + 1, then n² = 4k² + 4k + 1 = 4(k² + k) + 1, with remainder 1. Both calculations use integers k, so together they cover all integers n.
Check: did you state the assumptions, explain the key step, and reach the requested conclusion?
All nine practice questions, hints and solutions
This complete set works without the interactive practice. Hide each solution until you have made an attempt.
1. What is the remainder when 38 is divided by 5?
Foundation
- 2
- 3
- 5
- 7
Hint
Find the largest multiple of 5 no larger than 38.
Answer and explanation
3. 38 = 5 × 7 + 3. Since 0 ≤ 3 < 5, the remainder is 3.
2. What is the remainder when −1 is divided by 4, using a remainder from 0 to 3?
Foundation
- −1
- 1
- 3
- 4
Hint
Choose a multiple of 4 below −1.
Answer and explanation
3. −1 = 4 × (−1) + 3. The remainder is 3 because it lies in the required range 0 ≤ r < 4.
3. Which pair has the same remainder on division by 6?
Foundation
- 14 and 20
- 14 and 21
- 14 and 22
- 14 and 23
Hint
Subtract the smaller number. Is the difference a multiple of 6?
Answer and explanation
14 and 20. 20 − 14 = 6, so these numbers have the same remainder. In fact, both leave remainder 2 when divided by 6.
4. What is the remainder of 29 + 47 on division by 7?
Core
- 1
- 3
- 5
- 6
Hint
Reduce both terms, then add their remainders.
Answer and explanation
6. 29 leaves remainder 1 and 47 leaves remainder 5 on division by 7. Their sum therefore leaves remainder 1 + 5 = 6, which is already smaller than 7.
5. What is the remainder of 17 × 19 on division by 5?
Core
- 1
- 2
- 3
- 4
Hint
Use 17 ≡ 2 and 19 ≡ 4 modulo 5.
Answer and explanation
3. The product has the same remainder as 2 × 4 = 8. Since 8 = 5 + 3, the required remainder is 3.
6. What is the last digit of 2⁸?
Core
- 2
- 4
- 6
- 8
Hint
The last digits of powers of 2 repeat as 2, 4, 8, 6.
Answer and explanation
6. The cycle has length 4. Since 8 is a multiple of 4, the eighth power uses the fourth position, with last digit 6. A remainder of 0 in the exponent count means the end of a complete cycle, not its first position.
7. Which remainder is impossible for an integer square on division by 4?
Stretch
- 0
- 1
- 2
- Both 0 and 1
Hint
Split the original integer into the even and odd cases.
Answer and explanation
2. An even integer 2k has square 4k², with remainder 0. An odd integer 2k + 1 has square 4(k² + k) + 1, with remainder 1. Therefore remainder 2 is impossible.
8. What is the remainder of 3¹⁰ on division by 7?
Stretch
- 1
- 2
- 4
- 6
Hint
Compute 3² and 3³ modulo 7, then use 3⁶.
Answer and explanation
4. 3² ≡ 2 and 3³ ≡ 6 ≡ −1 (mod 7). Squaring gives 3⁶ ≡ 1. Thus 3¹⁰ = 3⁶ × 3⁴ ≡ 1 × (3²)² ≡ 4 (mod 7).
9. If n is an integer and n ≡ 4 (mod 5), what is the remainder of n² + n on division by 5?
Stretch
- 0
- 1
- 3
- 4
Hint
Replace n by 4 before simplifying.
Answer and explanation
0. n² + n has the same remainder as 4² + 4 = 20. Since 20 is a multiple of 5, the remainder is 0.
Where to go next
Move on to the mixed proof-practice plan · Explore the wider Olympiad topic map
Teaching examples prepared for this guide, not attributed to a contest paper. Familiar elementary problems may appear in other learning materials. Report an unclear step or an error.