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Last digits and cycles How can you find the last digit of 7²⁰²⁶ without working it out?

Maths olympiad · 45 minutes

Able students attack the last digit of 7²⁰²⁶ by trying small cases, see the cycles of last digits on a graph, prove that every cycle must repeat, and use remainders to finish. Slides to present, a vote that shows who's stuck, practice that climbs to olympiad problems, and homework.

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Each step is the last digit of aⁿ, for n = 1, 2, 3, … Start at a = 7, then move a. What do you see each time, and why must it happen?

36912

At x = 1, 2, 3 …: 7, 9, 3, 1, 7, 9, 3, 1, 7, 9, 3, 1

The idea in one minute

The last digit of a product depends only on the last digits of the numbers multiplied, so the last digits of 7, 7², 7³, … go 7, 9, 3, 1, then again 7, 9, 3, 1. They must repeat: there are only ten possible last digits, and each one fixes the next. To find the last digit of 7²⁰²⁶, divide 2026 by the length of the cycle, 4: the remainder, 2, says where in the cycle it lands. This is arithmetic with remainders, made concrete: olympiads call it working modulo 10.

The lesson is problem first. Students meet a number with 1,713 digits and are asked for its last one, before anyone explains. They try small cases, spot the cycle, see the cycles of every digit on a graph, and only then say why a cycle must come back and how a remainder picks the answer. Practice climbs from single powers to sums of powers and short written proofs, which the teacher marks.

By the end, students can

  • find the cycle of last digits of the powers of any whole number
  • explain why the last digits of powers must repeat
  • use the remainder of the power to find a last digit, including remainder 0
  • find the last digit of sums and products of large powers, and remainders on dividing by 2, 5 or 10
  • write a short proof by checking every last digit

Key words

  • last digit
  • cycle
  • remainder
  • divisible
  • modulo
  • proof by cases

Before this lesson

Students should know powers (7³ = 7 × 7 × 7), divide with a remainder, and tell from the last digit whether a number divides by 2, 5 or 10.

The lesson, step by step

45 minutes in total. The stretch is optional.

  1. 1 · Think5 min

    A number too big to write

    7²⁰²⁶ has 1,713 digits.

    No calculator can show them all.

    What is its last digit?

    No calculator. Start with smaller powers.

    Write here

    5 min 1 · Hook · students think · the slides say “Think”

    A number too big to write

    Students attack the last digit of 7²⁰²⁶ before anyone explains.

    Say

    “7²⁰²⁶ has 1,713 digits. No calculator shows them all. What is its last digit? You have three minutes, and no calculator.”

    Ask
    • What could you work out that is smaller?
    • Do you need all of 343 to find the last digit of 343 × 7?
    Watch for

    Students who multiply out whole numbers (16 807, 117 649 …) run out of time: ask which digits they really need.

    For the teacher

    Three minutes of silent work. The good move is to try small cases: 7¹, 7², 7³, … and look for a pattern.

    Ask for their lists and a guess. Then the real questions: does the pattern go on for ever, and how do we use it for 2026?

    Most begin 7, 49, 343, 2401. Collect their lists. Take a vote with Ask the class on the last digit: many will guess 7 or 1. Don't settle it yet.

    Shorter: Give 7, 49, 343, 2401 at once; vote straight away (−2 min).

  2. 2 · Watch12 min

    Only the last digit matters

    7³ = 343, and 7⁴ = 343 × 7 = 2401

    The last digit comes from 3 × 7 = 21 alone.

    Keep only last digits: 7, 9, 3, 1, 7, …

    Each last digit fixes the next one.

    The last digit is 1. What comes next, and next?

    Write here

    2 · Watch

    Last digits of 7ⁿ

    n = 1 to 4: 7, 9, 3, 1

    n = 5 to 8: 7, 9, 3, 1

    n = 9 to 12: 7, 9, 3, 1

    A cycle of length 4

    Will it ever stop repeating?

    2 · Watch

    Why every cycle repeats

    There are only ten last digits: 0 to 9.

    Each last digit fixes the next one.

    So within ten steps a digit comes round again,

    and from then on everything repeats.

    (Ten boxes, eleven steps: the pigeonhole idea.)

    Say it in your own words to your neighbour.

    2 · Watch

    Move a: every cycle

    Each step: the last digit of aⁿ

    a = 2: 2, 4, 8, 6

    a = 9: 9, 1

    a = 5: 5, 5, 5, …

    a = 12 looks like a = 2. Why?

    Which a have cycles of 4? 2? 1?

    2 · Watch

    Use the cycle: the remainder

    7ⁿ: the cycle 7, 9, 3, 1 has length 4

    2026 = 4 × 506 + 2: remainder 2

    Remainder 2: the 2nd in the cycle, so 9

    Remainder 0: the last in the cycle, not the first

    Dividing by 4? The last two digits say: 26 leaves 2.

    Name it: 7²⁰²⁶ ≡ 9 (mod 10), 'the same last digit as 9'.

    12 min 2 · Explain · you explain · the slides say “Watch”

    Discover it, then name it

    Only the last digit matters; every cycle must repeat; a remainder picks the answer.

    Say

    “Only the last digit can change the last digit. So keep only last digits, and see what happens.”

    Ask
    • The last digit is 1. What will the next one be?
    • Why must a digit come round again within ten steps?
    • 2026 divided by 4 leaves what? And 2024?
    Watch for

    Remainder 0 read as the first in the cycle: it is the last, as 7⁴, 7⁸, 7¹² all end in 1.

    Only the last digit

    7⁴ = 343 × 7 = 2401

    Only 3 × 7 = 21 matters: last digit 1

    The tens of 343 can't change the units

    Use the cycle

    7ⁿ ends in 7, 9, 3, 1, then again

    2026 = 4 × 506 + 2: remainder 2

    The 2nd in the cycle: 7²⁰²⁶ ends in 9

    For the teacher

    First slide: 7⁴ = 343 × 7. Its last digit comes from 3 × 7 = 21 alone: the tens and hundreds only add tens and hundreds. So we keep only last digits: 7, then 7 × 7 = 49 → 9, then 9 × 7 = 63 → 3, then 3 × 7 = 21 → 1, then 1 × 7 → 7 again.

    Second slide: the dots show the last digits of 7ⁿ for n = 1 to 12: 7, 9, 3, 1 three times.

    Third slide: why it must repeat. There are only ten last digits, and each one fixes the next. So within ten steps a digit comes again, and from then on everything repeats. (This is the pigeonhole idea.)

    Fourth slide: move a. Bases ending in 2, 3, 7, 8 repeat every 4; in 4 or 9 every 2; in 0, 1, 5, 6 never change. a = 12 looks just like a = 2: only the last digit of the base matters.

    Fifth slide: use it. 2026 = 4 × 506 + 2, so 7²⁰²⁶ lands on the 2nd in the cycle: 9. The remainder by 4 can be read from the last two digits, because 100 is a multiple of 4: 26 leaves 2. Remainder 0 lands on the last in the cycle. Olympiads write this as 7²⁰²⁶ ≡ 9 (mod 10).

    Vote before moving a (Ask the class): which a will have a cycle of length 1? Give the name 'modulo 10' only after students have said 'only the last digit matters' themselves.

    Shorter: Skip moving a; show the four kinds of cycle on one slide (−4 min).

  3. 3 · Your turn8 min

    Worked, then yours

    I do: the last digit of 3²⁰²⁶

    Cycle 3, 9, 7, 1; 2026 leaves 2; the 2nd: 9

    You do: the last digit of 2²⁰²⁶

    The cycle, the remainder, the place.

    Write here

    3 · Your turn

    Two more pairs

    I do: 7²⁰²⁴: remainder 0, the last in the cycle: 1

    You do: the last digit of 8¹⁰⁰

    I do: 3⁵⁰ + 4⁵⁰ ends like 9 + 6 = 15: in 5

    You do: the last digit of 7²⁰²⁶ + 9²⁰²⁶

    How long is each cycle?

    Write here

    Its answers' slide comes with your copy.

    8 min 3 · Your turn · students write

    Worked, then yours

    Each worked example beside a twin that students solve themselves.

    Say

    “Watch me: the cycle, the remainder, the place in the cycle. Then you do its twin.”

    Ask
    • How long is the cycle?
    • What is the remainder, and where does it land?
    Watch for

    Adding last digits and writing 15: only the last digit of the sum counts.

    One power

    3²⁰²⁶: the cycle is 3, 9, 7, 1

    2026 leaves 2 on dividing by 4

    The 2nd in the cycle: 9

    Then theirs

    The last digit of 2²⁰²⁶

    Remainder 0

    7²⁰²⁴: 2024 leaves 0 on dividing by 4

    Remainder 0: the last in the cycle

    7, 9, 3, 1: so 1

    Then theirs

    The last digit of 8¹⁰⁰

    A sum of powers

    3⁵⁰ + 4⁵⁰: 50 leaves 2 on dividing by 4

    3⁵⁰ ends in 9; 4ⁿ has a cycle of 2, and 50 is even: 6

    9 + 6 = 15. Keep the last digit: 5

    Then theirs

    The last digit of 7²⁰²⁶ + 9²⁰²⁶

    For the teacher

    Work each example aloud in three moves: the cycle, the remainder, the place. Students solve its twin while you watch.

    The second pair is the one that catches most: remainder 0 is the end of the cycle. The third pair adds two powers whose cycles have different lengths (4ⁿ and 9ⁿ repeat every 2), so check the cycle's length for each base.

    Open the answers slide once most have finished their twins. Ask for the cycle, the remainder and the place for each.

    Shorter: Do the first two pairs only (−3 min).

  4. 4 · Vote and practise15 min

    Vote first

    What is the last digit of 3¹⁰⁰?

    A: 3

    B: 1

    C: 9

    D: 0

    Choose on your own, then say why.

    4 · Vote and practise

    Climb: from fluent to olympiad

    1. The last digit of 2¹⁰?

    2. The cycle of last digits of 4ⁿ?

    3. The last digit of 9²⁰²⁷?

    4. The last digit of 13²⁰²⁷?

    5. Why must the last digits of 7ⁿ repeat?

    Stuck? Write the cycle and number its places.

    Write here

    4 · Vote and practise

    Reason, then prove

    6. The remainder when 7²⁰²⁶ is divided by 5?

    7. Prove that no square ends in 2, 3, 7 or 8.

    8. The last digit of 2²⁰²⁶ + 3²⁰²⁶ + 4²⁰²⁶?

    9. The last digit of 1 + 3 + 3² + … + 3²⁰²⁵?

    10. Prove n⁵ − n ends in 0 for every whole n.

    A proof by cases: check every last digit, 0 to 9.

    Write here

    15 min 4 · Who's stuck? · students vote · the slides say “Vote and practise”

    Vote, then climb

    One question shows who's stuck; then practice from single powers to olympiad sums and proofs.

    Say

    “Choose an answer on your own. Nobody talks until everyone has chosen.”

    Ask
    • What is 3⁴'s last digit? So what about 3⁸, 3¹², …?
    Watch for

    Remainder 0 read as the first in the cycle; an even power read as the square; a last digit copied from the exponent.

    Show your working: the cycle, the remainder, the place in the cycle. The last questions are proofs: check every possible last digit.

    The question that shows who's stuck

    1. What is the last digit of 3¹⁰⁰?
      1. 3
      2. 1
      3. 9
      4. 0
    In your Classroom, as they answer (an example)
    • A9
    • B11
    • C5
    • D2

    Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.

    1. What is the last digit of 2¹⁰?
    2. The last digits of 4¹, 4², 4³, 4⁴ are ___, ___, ___, ___: a cycle of length ___
    3. What is the last digit of 9²⁰²⁷?
    4. What is the last digit of 13²⁰²⁷?
    5. Why must the last digits of 7, 7², 7³, … repeat?
      1. Because 7 is a prime number
      2. There are only ten last digits, and each one fixes the next
      3. Because 7⁴ = 2401 ends in 1
      4. They needn't: we only checked the first twelve
    6. What is the remainder when 7²⁰²⁶ is divided by 5?
    7. Prove that no square number ends in 2, 3, 7 or 8.
    8. What is the last digit of 2²⁰²⁶ + 3²⁰²⁶ + 4²⁰²⁶?
    9. Stretch: what is the last digit of 1 + 3 + 3² + 3³ + … + 3²⁰²⁵?
    10. Stretch: prove that n⁵ − n ends in 0 for every whole number n.

    Support · Write the cycle out, number its places 1, 2, 3, 4, and say which place remainder 0 is. For the proofs: list every possible last digit, 0 to 9, and check each one.

    For the teacher

    Everyone chooses before anyone explains. Then ask for the cycle, the remainder and the place in the cycle behind each answer: remainder 0, even powers and the 0 of 100 each pull some students to a different digit.

    Questions 1–2 are single powers; 3–4 change the look; 5–6 ask why and link remainders; 7 is a proof by cases. 8–10 are olympiad level: sums of powers, a long sum, and a proof for every whole number.

    Students write proofs on paper or the board; the quiz keeps their answers for you to mark and comment on.

    If the class is split, write 3⁴, 3⁸, 3¹² and their last digits before practice. Long answers (7, 10) are marked by you: one mark for the idea (only the last digit matters), one for checking every case, one for the conclusion. Model answers are in each question's explanation.

    Shorter: After the vote, questions 1–5 only (−6 min).

    Longer: Swap and mark: pairs mark each other's proof of question 7, then 10 (+10 min).

  5. 5 · Show what you know5 min

    One more power

    What is the last digit of 8²⁰²⁶?

    Say how the cycle tells you.

    The cycle, the remainder, the place.

    Write here

    5 min 5 · Exit · students show · the slides say “Show what you know”

    One more power

    One last digit, with the reason, shows who can use a cycle.

    Say

    “Find the last digit of 8²⁰²⁶. Then say how the cycle tells you.”

    Watch for

    A right digit with no cycle or remainder in the reason.

    A digit, then a sentence or two.

    1. What is the last digit of 8²⁰²⁶?
    2. Say how the cycle tells you the last digit of 8²⁰²⁶.
    For the teacher

    What is the last digit of 8²⁰²⁶?

    Say in a sentence or two how you know.

    Wrong digits show who still miscounts the place in the cycle: start next time with remainder 0. Set the homework.

    Longer: Then: the last digit of 8²⁰²⁶ − 2²⁰²⁶ (+5 min).

One to one: let the student work out 7¹ to 7⁶ on paper, keeping only the last digit, and ask them to spot when it starts again. Ask them to say the remainder, and where in the cycle it lands, before each answer.

Homework: cycles and remainders

For each, write the cycle, the remainder and the place in the cycle. Question 5 is a proof; question 7 is a challenge.

  1. What is the last digit of 4²⁰²⁶?
  2. What is the last digit of 7¹⁰⁰?
  3. What is the last digit of 17²⁰²⁶ + 18²⁰²⁶?
  4. What is the remainder when 3²⁰²⁶ is divided by 5?
  5. Prove that 6ⁿ − 1 is divisible by 5 for every whole number n from 1 up.
  6. What is the last digit of 1¹ + 2² + 3³ + … + 10¹⁰?
  7. Challenge: what is the last digit of 7 to the power 7⁷? (7⁷ = 823 543 is the power, not a number to multiply by.)

Print the worksheet with working space and homework. Teacher answers come with your Library copy.

Teaching last digits and cycles: common questions

Why do the last digits of powers repeat?

The last digit of a product depends only on the last digits of the numbers multiplied, so each last digit of aⁿ fixes the next one. There are only ten possible last digits, so one must come round again within ten steps, and from then on the pattern repeats for ever.

Is this modular arithmetic?

Yes, made concrete. 'The same last digit' is what olympiad books write as ≡ (mod 10), and 'the remainder on dividing by 4' is working mod 4. The lesson names it once students have used it, so the notation means something when they meet it.

Which competitions is this lesson useful for?

Last digits and remainders come up often in maths challenges and olympiads for students aged about 11 to 16, such as the UKMT Junior and Intermediate Mathematical Challenges and their follow-on olympiads. The problems are our own, written in that style, not taken from past papers.

My student gets the right digit but can't explain it. What helps?

Ask for three things every time: the cycle, the remainder, the place in the cycle. For the proofs, ask for the reason only the last digit matters and then every case, 0 to 9. The long questions are marked by you in Quizzes, with a comment, so the write-up gets feedback, not only the answer.

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