What is the units digit of 7^2024

Question

What is the units digit of $7^{2024}$ ?

Solution — Step by Step

We don’t need to calculate $7^{2024}$ — it’s astronomically large. Instead, we look at only the units digit, which follows a repeating cycle.

Compute units digits of successive powers of 7:

$7^1 = 7$ → units digit 7
$7^2 = 49$ → units digit 9
$7^3 = 343$ → units digit 3
$7^4 = 2401$ → units digit 1
$7^5 = 16807$ → units digit 7 (cycle repeats!)

The units digits cycle is: 7, 9, 3, 1, 7, 9, 3, 1, …

The cycle has period 4: every 4 consecutive powers of 7 have units digits 7, 9, 3, 1.

In general: the units digit of $7^n$ depends only on the remainder when $n$ is divided by 4.

Remainder (n mod 4)	Units digit of $7^n$
1	7
2	9
3	3
0	1

2024 \div 4 = 506 \text{ remainder } 0

So $2024 \equiv 0 \pmod{4}$ .

From our table: when the remainder is 0 (i.e., the exponent is divisible by 4), the units digit is 1.

\boxed{\text{Units digit of } 7^{2024} = 1}

We can verify with small cases: $7^4$ ends in 1, $7^8$ ends in 1, $7^{12}$ ends in 1, … $7^{2024}$ ends in 1. ✓

Why This Works

The units digit of a product depends only on the units digits of the factors. So $7^n$ mod 10 depends only on $7^{n-1}$ mod 10, and the sequence $\{7^n \text{ mod } 10\}$ is periodic because there are only finitely many possible remainders mod 10 (0–9).

The period must eventually repeat, and once we see a repeat (we see 7 again at $n = 5$ after $n = 1$ ), the cycle has length 4. This is guaranteed by modular arithmetic — it’s not just an observation, it’s a theorem (Fermat’s little theorem gives the upper bound on cycle lengths).

The same method applies to units digit of $3^n$ (cycle: 3, 9, 7, 1, period 4), $2^n$ (cycle: 2, 4, 8, 6, period 4), $4^n$ (cycle: 4, 6, period 2), $5^n$ (always 5), $6^n$ (always 6).

Alternative Method

$7^{2024} = (7^4)^{506} = 2401^{506}$ . Since $2401$ ends in 1, any power of $2401$ also ends in 1 (since $1^k = 1$ ). So the units digit is 1.

This is slightly faster when the exponent is divisible by the cycle length.

Common Mistake

Students often make an off-by-one error in reading the table: they compute $2024 \mod 4 = 0$ but then look up the row for “remainder 0” and misread it as “4th position in cycle = 1” but write “7” (confusing position 1 with remainder 1). Write the table carefully with the remainder in the left column, and trust it: remainder 0 → units digit 1.

For any base $b$ , the units digit cycle length divides $\phi(10) = 4$ (Euler’s totient function for 10), where $\gcd(b, 10) = 1$ . That’s why 3, 7, 9 all have period dividing 4. Bases ending in 2, 4, 5, 6, 8, 0 have their own (often shorter) cycles. This fact saves you from rediscovering the cycle each time.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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