Remainders repeat — find the cycle and the big power stops mattering.
The problem
What is the remainder when 210 is divided by 3?
you will not need a calculator
The remainders of 2ⁿ ÷ 3 cycle 2, 1, 2, 1, … Even powers land on 1, and 10 is even. (Check: 2¹⁰ = 1024 = 3 × 341 + 1. ✓) 2 is the trap: 10 ÷ 2 leaves 0, and a leftover of 0 means the last slot of the loop (1), not the first (2). 4 cannot be a remainder at all when dividing by 3. Find the cycle, then locate the exponent within it.
It is whatever is left after making as many full groups as you can.
two full groups of 3, and 2 left over
8 = 3(2) + 2
the leftover is the remainder
You could grind out 210 by hand. The same question with 2100 would be impossible, so learn the fast way now.
A remainder when dividing by 3 can only ever be 0, 1 or 2. That is already a big clue.
Start at the bottom and watch the remainder column.
| Power | Value | ÷ 3 | Remainder |
|---|---|---|---|
| 21 | 2 | 0 groups | 2 |
| 22 | 4 | 1 group | 1 |
| 23 | 8 | 2 groups | 2 |
| 24 | 16 | 5 groups | 1 |
| 25 | 32 | 10 groups | 2 |
| 26 | 64 | 21 groups | 1 |
You stop as soon as the remainders start repeating. Two rows would have been enough here — four more just prove it.
n = 1, 2, 3, 4, 5, 6 — this pair repeats forever
Each remainder is decided only by the one before it. Once a value comes back, the whole run comes back with it.
10 ÷ 2 = 5 remainder 0
exponent ÷ cycle length
a leftover of 0 means n = 10 sits on the last slot
2
1
shortcut: an even exponent gives 1, an odd one gives 2
2 is one short of 3 — it leaves the same remainder as −1 does.
2 = 3 − 1
so when dividing by 3, a 2 behaves like −1
210 behaves like (−1)10
(−1)10 = 1
ten minus signs cancel in pairs
Same answer as the table — but this route still works when the exponent is 100.
Shrink the base first: the remainder when 23¹⁸ is divided by 7.
23 = 7 × 3 + 2, so 23 behaves like 2
replace the base by its remainder
2ⁿ ÷ 7 leaves 2, 4, 1 (length 3)
8 ÷ 7 leaves 1, so the loop closes
18 ÷ 3 = 6 r 0 → last slot = 1
check: 2¹⁸ = 262,144 = 7 × 37,449 + 1 ✓
What is different: a base above the divisor would give huge rows in the table. Reducing it first keeps every number small.
A product, not a power: the remainder when 38 × 43 × 59 is divided by 6.
38 → 2, 43 → 1, 59 → 5
each remainder on division by 6
2 × 1 × 5 = 10
multiply the remainders
10 ÷ 6 leaves 4
reduce once more
What is different: no cycle at all — the remainders multiply just as they add. The full product, 96,406, gives the same remainder.
Counting backwards: today is a Friday. What day was it 150 days ago?
150 = 7 × 21 + 3
21 full weeks change nothing
3 days before Friday: Thu, Wed, Tuesday
step back, starting with yesterday
What is different: the remainder moves us back, not forward. Counting 3 days forward gives Monday — the wrong direction.
What is the remainder when 331 + 431 is divided by 5?
3n ÷ 5 leaves 3, 4, 2, 1 (length 4); 31 ÷ 4 r 3 → 2
the third slot
4 = 5 − 1 behaves like −1; (−1)31 = −1 → 4
odd power
2 + 4 = 6, and 6 ÷ 5 leaves 1
add the remainders, then reduce again
Today is a Wednesday. What day will it be 200 days from today?
200 = 7 × 28 + 4
the week is a cycle of length 7
28 full weeks change nothing
only the remainder moves you
4 days after Wednesday: Thu, Fri, Sat, Sunday
count from tomorrow
A sum of remainders can reach the divisor or pass it — 6 is not a remainder on division by 5, so reduce it to 1. And on the calendar, “4 days after” starts counting at the next day; counting Wednesday itself as day 1 lands on Saturday.
Remainder items use a big power, a sum of two powers, or a calendar or repeating-pattern story. Build the cycle of remainders (or spot a base that is one less than the divisor, which acts like −1), divide the exponent by the cycle length, and reduce any sum once more.
The table and the shortcut both said 1.
Find the remainder
What is the remainder when 325 is divided by 5?
3ⁿ ÷ 5 cycles 3, 4, 2, 1 — length 4. Dividing 25 by 4 leaves remainder 1, so 3²⁵ sits at the first position: 3. A assumes the cycle has restarted, which would need a remainder of 0.
Hint: this cycle is longer than 2. List powers of 3 until a remainder repeats, then divide 25 by that cycle length.
