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2026-04-17
Problem — 2026-04-17
Easy
Easy
number_theory
What is the last digit of
7
2025
7^{2025}
7
2025
?
Enter a single digit
Show solution
The last digits of powers of
7
7
7
cycle with period
4
4
4
:
7
1
=
7
7^1 = 7
7
1
=
7
,
7
2
=
49
7^2 = 49
7
2
=
49
,
7
3
=
343
7^3 = 343
7
3
=
343
,
7
4
=
2401
7^4 = 2401
7
4
=
2401
,
7
5
=
16807
7^5 = 16807
7
5
=
16807
, etc. The cycle is
7
,
9
,
3
,
1
,
7
,
9
,
3
,
1
,
…
7, 9, 3, 1, 7, 9, 3, 1, \ldots
7
,
9
,
3
,
1
,
7
,
9
,
3
,
1
,
…
. Since
2025
=
4
⋅
506
+
1
2025 = 4 \cdot 506 + 1
2025
=
4
⋅
506
+
1
, we have
2025
≡
1
(
m
o
d
4
)
2025 \equiv 1 \pmod{4}
2025
≡
1
(
mod
4
)
, so the last digit of
7
2025
7^{2025}
7
2025
is
7
7
7
.