Remainder Shortcuts and Tricks: Find the Remainder Fast
Find the remainder quickly without long division. Learn the shortcuts for dividing by 2, 5, 9, 10 and 11, the power cycle trick for exponents, and the bitwise method for powers of 2.
Remainder shortcuts are rules that return the remainder of a division without running the full long division. To find a remainder quickly, match the divisor to its shortcut: use the last digits for 2, 5 and 10, the digit sum for 9, and the alternating digit sum for 11. The remainder of 1234 ÷ 9 is 1, found by adding 1 + 2 + 3 + 4 = 10 and reducing 10 modulo 9. The Remainder Calculator confirms any shortcut in one step, and each trick below is checked against the standard method for finding a remainder.
These shortcuts deliver 3 gains: they cut a multi-digit division to a single mental step, they reveal what a remainder of zero means before any work begins, and they scale to exponents that no calculator can expand. Ruling out small divisors quickly this way also speeds up testing a number for primality. The tricks cover dividing by 2, 5, 9, 10, 11, powers of 2, and repeated powers through the cycle method.
Divide by 2: Read the Last Digit
The remainder of any division by 2 is 0 for even numbers and 1 for odd numbers. Only the final digit matters. The remainder of 5473 ÷ 2 is 1, because 3 is odd. The remainder of 918 ÷ 2 is 0, because 8 is even.
This is the fastest of all remainder tricks, since it ignores every digit except the last.
Divide by 5: Check the Last Digit
The remainder of a division by 5 depends only on the last digit of the dividend. A final digit of 0 or 5 leaves 0, and every other final digit maps to a fixed remainder.
| Last digit | Remainder ÷ 5 |
|---|---|
| 0 or 5 | 0 |
| 1 or 6 | 1 |
| 2 or 7 | 2 |
| 3 or 8 | 3 |
| 4 or 9 | 4 |
The remainder of 5473 ÷ 5 is 3, since the last digit 3 falls in the third row. No division is needed.
Divide by 10, 100 or 1000: Split the Digits
The remainder of a division by a power of 10 equals the trailing digits of the dividend. Dividing by 10 leaves the last digit, dividing by 100 leaves the last 2 digits, and dividing by 1000 leaves the last 3 digits.
The remainder of 45678 ÷ 100 is 78. The remainder of 45678 ÷ 1000 is 678. The remainder of 45678 ÷ 10 is 8. This trick turns a division into a simple read of the number’s tail.
Divide by 9: Add the Digits
The remainder of a division by 9 equals the digit sum reduced modulo 9. Add every digit, and if the total reaches 9 or more, add its digits again until a single value below 9 remains.
Find 1234 ÷ 9: 1 + 2 + 3 + 4 = 10, then 1 + 0 = 1. The remainder is 1.
Find 8657 ÷ 9: 8 + 6 + 5 + 7 = 26, then 2 + 6 = 8. The remainder is 8.
This trick works because 10, 100 and 1000 all leave a remainder of 1 when divided by 9, so every digit contributes only its own value. A digit sum that reduces to 9 means the remainder is 0 and the number divides evenly.
Divide by 3: Add the Digits
The remainder of a division by 3 equals the digit sum reduced modulo 3. The digit sum of 1234 is 10, and 10 ÷ 3 leaves 1, so 1234 ÷ 3 leaves 1. The digit sum method for 3 and 9 is one of the divisibility rules these shortcuts rely on.
Divide by 11: Alternate the Signs
The remainder of a division by 11 equals the alternating digit sum, taken from right to left. Add the last digit, subtract the next, add the next, and continue.
Find 1234 ÷ 11: 4 − 3 + 2 − 1 = 2. The remainder is 2.
Find 90728 ÷ 11: 8 − 2 + 7 − 0 + 9 = 22, then 22 leaves 0 when divided by 11. The remainder is 0, so 11 divides 90728 exactly.
A negative alternating sum adds 11 until the result lands between 0 and 10. An alternating sum of −3 becomes −3 + 11 = 8. This same shift keeps the remainder of a negative number non-negative.
Powers of 2: Use the Bitwise Trick
The remainder of a division by a power of 2 equals the value of the low bits, found with the bitwise AND expression n & (2ᵏ − 1). Dividing by 8 keeps the last 3 bits, dividing by 16 keeps the last 4 bits.
Find 77 mod 8: the expression 77 & 7 compares 77 (1001101 in binary) with 7 (0000111) and keeps the low 3 bits, 101, which is 5. The remainder is 5.
This trick runs in a single processor instruction, which is why hashing and buffer code prefer power-of-2 sizes. It is one piece of the modular arithmetic behind the tricks, and the % operator applies it across programming languages.
The Power Cycle for Exponents
The remainder of a large power follows a repeating cycle, so the exponent is reduced to its position in that cycle rather than expanded.
Find 2¹⁰⁰ mod 7. The powers of 2 leave a repeating pattern of remainders when divided by 7:
| Power | Value | Remainder ÷ 7 |
|---|---|---|
| 2¹ | 2 | 2 |
| 2² | 4 | 4 |
| 2³ | 8 | 1 |
| 2⁴ | 16 | 2 |
The cycle 2, 4, 1 repeats every 3 powers. Since 100 leaves a remainder of 1 when divided by 3, the answer matches the first position in the cycle, so 2¹⁰⁰ mod 7 = 2.
This trick solves problems that would otherwise need a 31-digit number. The cycle length is the key: find where the remainders start repeating, divide the exponent by that length, and read the remainder off the cycle position.
Break the Dividend Into Parts
A large dividend splits into friendly pieces whose remainders add. The remainder of 1234 ÷ 7 splits into 1200 ÷ 7 and 34 ÷ 7. The first leaves 3, the second leaves 6, and 3 + 6 = 9, which reduces to 2 after subtracting one more 7. The remainder is 2.
This method suits any divisor and rewards choosing pieces that divide cleanly. Rounding a chunk to the nearest multiple of the divisor shrinks the arithmetic at every step.
Shortcut Summary
| Divisor | Shortcut | Example |
|---|---|---|
| 2 | Last digit odd or even | 5473 → 1 |
| 5 | Map the last digit | 5473 → 3 |
| 10, 100, 1000 | Read the trailing digits | 45678 ÷ 100 → 78 |
| 3, 9 | Reduce the digit sum | 1234 ÷ 9 → 1 |
| 11 | Alternating digit sum | 1234 → 2 |
| Power of 2 | Bitwise AND with 2ᵏ − 1 | 77 mod 8 → 5 |
| Exponents | Power cycle position | 2¹⁰⁰ mod 7 → 2 |
Frequently Asked Questions
What is the fastest way to find a remainder?
The fastest way depends on the divisor. Division by 2, 5 or 10 reads a single digit, division by 9 sums the digits, and division by a power of 2 uses one bitwise operation. Matching the divisor to its shortcut beats long division every time.
Do these shortcuts work for every divisor?
No, each shortcut targets a specific divisor. Divisors without a rule, such as 7 or 13, need the split-into-parts method or long division with remainders.
How do you find the remainder of a large power?
Find the remainder of a large power with the cycle method. List the remainders of the first few powers until they repeat, divide the exponent by the cycle length, and read the answer from the matching cycle position.
Why does adding the digits give the remainder for 9?
Adding the digits works for 9 because 10, 100 and every higher power of 10 leaves a remainder of 1 when divided by 9. Each digit therefore contributes only its face value to the total remainder.
Conclusion
Remainder shortcuts replace long division with a single step for common divisors: the last digit handles 2, 5 and 10, the digit sum handles 3 and 9, the alternating digit sum handles 11, and a bitwise AND handles powers of 2. The power cycle extends the same thinking to exponents that no calculator can expand.
These 3 gains, a shorter calculation, an instant zero-remainder test, and scaling to huge powers, make the tricks the practical layer on top of the full method. Enter any dividend and divisor into the Remainder Calculator to verify a shortcut, and reach for long division whenever the divisor has no rule.