Math
Mod 10. Olympiad 2026 . Nahuta Viktoriia
Step-by-step mathematics solution: Mod 10. Olympiad 2026 . Nahuta Viktoriia
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Problem Restatement
We are asked to determine the units (last) digit of the enormous number
[ 7^{2026}. ]
In other words, we need the remainder when (7^{2026}) is divided by (10).
Step‑by‑Step Solution
1. Understand the cyclicity of powers modulo 10
For any integer (a) that is coprime to (10) (i.e. ends in 1,3,7,9), the sequence of its last digits repeats with a period that divides (4).
This follows from Euler’s theorem because (\phi(10)=4).
So we only need to look at the first few powers of (7) modulo 10:
| Power | Value | Last digit (mod 10) |
|---|---|---|
| (7^1) | 7 | 7 |
| (7^2) | 49 | 9 |
| (7^3) | 343 | 3 |
| (7^4) | 2401 | 1 |
| (7^5) | 7·(7^4) | (7·1 = 7) |
Thus the pattern of last digits is (7, 9, 3, 1) and then it repeats every 4 exponents.
2. Reduce the exponent (2026) modulo the period
Since the pattern repeats every (4) powers, we compute the remainder of (2026) upon division by (4):
[ 2026 = 4 \times 506 + 2 \quad\Longrightarrow\quad 2026 \equiv 2 \pmod{4}. ]
So the exponent (2026) is equivalent to the exponent (2) in the repeating cycle.
3. Pick the corresponding digit from the cycle
The cycle (starting with exponent (1)) is:
[
\begin{aligned}
\text{exponent }1 &\to 7,
\text{exponent }2 &\to 9,
\text{exponent }3 &\to 3,
\text{exponent }4 &\to 1.
\end{aligned}
]
Because (2026 \equiv 2 \pmod{4}), we take the second element of the cycle, which is 9.
Formally:
[ 7^{2026}\equiv 7^{2}\equiv 49\equiv 9\pmod{10}. ]
4. State the final answer
The last (units) digit of (7^{2026}) is
[ \boxed{9}. ]
Common Mistakes
| Mistake | Why it’s wrong | How to avoid it |
|---|---|---|
| Forgetting that the cycle length is 4 and trying to compute a huge power directly. | Computing (7^{2026}) explicitly is impossible by hand; the cyclic property is essential. | Remember Euler’s theorem: (\phi(10)=4). Always reduce the exponent modulo 4 for bases ending in 1,3,7,9. |
| Using the remainder 0 instead of 4 when the exponent is a multiple of 4. | If the exponent were exactly divisible by 4, the last digit is the fourth element of the cycle (which is 1), not the “0‑th” element. | Treat “remainder 0” as meaning “use the fourth term of the cycle.” |
| Mixing up the order of the cycle (e.g., thinking the second term is 3). | The cycle for (7) is (7,9,3,1); swapping terms leads to an incorrect digit. | Write out the first few powers (at least up to the fourth) and keep the list handy. |
| Assuming the pattern changes for larger exponents. | The pattern is truly periodic; it never changes. | Verify the pattern for the first 4 powers; then trust the repetition. |
By keeping these points in mind, you can quickly find the last digit of any large power of a number ending in 1, 3, 7, or 9.
Original question: Mod 10. Olympiad 2026 . Nahuta Viktoriia on Mathematics Stack Exchange, licensed CC BY-SA.