Modular Arithmetic Clock

Choose modulus n and operation. See result on a modular clock and inspect addition/multiplication tables.

Worked example — Modular Arithmetic Clock

Expression	7 x 5 (mod 12)
Result	11

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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A prime modulus lets you divide again 🖖

Modular arithmetic deals with integers under a cyclic modulus n, written as a ≡ b (mod n). This algebraic system forms a commutative ring Z/nZ. If the modulus p is prime, the ring becomes a field, meaning every non-zero element has a multiplicative inverse. This structure underlies modern cryptography, cyclic codes, and clock arithmetic, demonstrating how numbers behave under periodic boundaries.

Why it is called clock arithmetic 🖖

On a 12-hour clock, 4 hours after 11 o'clock is 3, not 15 — the numbers wrap back to the start once they pass 12. That is exactly modular arithmetic: a ≡ b (mod n) means a and b leave the same remainder when divided by n. So 15 ≡ 3 (mod 12). The core takeaway: modular arithmetic keeps only the remainder after division, and this tool's clock lets you watch the numbers loop around.

The nine-divisibility trick is hidden modular arithmetic 🖖

The old rule "a number is divisible by 9 if its digits sum to a multiple of 9" is pure modular arithmetic. Because 10 ≡ 1 (mod 9), every power of 10 is also ≡ 1, so any number is congruent to the sum of its digits (mod 9). The same fact powers "casting out nines", a centuries-old way to check arithmetic by hand.

MODULAR ARITHMETIC — WHEN CAN YOU DIVIDE, AND WHEN CAN YOU NOT?

Which Modular Arithmetic Case Are You In?

Working mod n means keeping only the remainder, and addition, subtraction and multiplication all survive that unharmed. Division is the one that does not. Whether you can divide by a number — whether it has an inverse — depends entirely on whether it shares a factor with n, which is why a prime modulus behaves so differently from a composite one. Which case you are in depends on the operation and on what kind of number n is.

Addition — always well behaved, whatever n is 13 + 5 ≡ 6 (mod 12)

A prime modulus — every nonzero element can be divided by φ(7) = 6 ⇒ gcd(a, 7) = 1 ∀a ≠ 0

A composite modulus — most elements cannot be divided by 2 × 3 ≡ 0, φ(6) = 2

Repeated powers — they cycle, and Fermat says where a^p−1 ≡ 1 (mod p)

Addition — always well behaved, whatever n is

What you know: Any modulus at all. Adding, subtracting and multiplying mod n are always defined, and every element has an additive opposite.

What to check: 13 + 5 ≡ 6 (mod 12)

Worked example: A 12-hour clock: 13 + 5 = 18, and 18 mod 12 = 6. Five hours after one o'clock is six o'clock.

Open this case: clock mod 12

$Addition — always well behaved, whatever n is. The addition table mod 12: every row is the previous one shifted along by one, and every value appears exactly once. Any modulus at all. Adding, subtracting and multiplying mod n are always defined, and every element has an additive opposite.$

The addition table mod 12: every row is the previous one shifted along by one, and every value appears exactly once.

A prime modulus — every nonzero element can be divided by

What you know: n is prime. Then no nonzero element shares a factor with n, so every one of them has a multiplicative inverse.

What to check: φ(7) = 6 ⇒ gcd(a, 7) = 1 ∀a ≠ 0

Worked example: Mod 7: 3 × 4 = 12 ≡ 5. All six nonzero values 1…6 are invertible, so the multiplication table has no zeros away from the top row.

Open this case: prime mod 7

A prime modulus — every nonzero element can be divided by. The multiplication table mod 7: no zeros below the first row, and each row a permutation of 1 to 6. n is prime. Then no nonzero element shares a factor with n, so every one of them has a multiplicative inverse. — The multiplication table mod 7: no zeros below the first row, and each row a permutation of 1 to 6.

A composite modulus — most elements cannot be divided by

What you know: n is composite. Only the values coprime to n have inverses; the rest are zero divisors, and dividing by them is meaningless.

What to check: 2 × 3 ≡ 0, φ(6) = 2

Worked example: Mod 6: 2 × 4 = 8 ≡ 2. Neither 2 nor 4 is invertible, since gcd(2,6) = 2 and gcd(4,6) = 2. Only 1 and 5 are units — two out of six.

Open this case: composite mod 6

A composite modulus — most elements cannot be divided by. The multiplication table mod 6, with zeros scattered through it and only two rows that are permutations. n is composite. Only the values coprime to n have inverses; the rest are zero divisors, and dividing by them is meaningless. — The multiplication table mod 6, with zeros scattered through it and only two rows that are permutations.

Repeated powers — they cycle, and Fermat says where

What you know: Raising to a power mod n. For prime p and a not divisible by p, the exponents wrap with period dividing p − 1.

What to check: a^p−1 ≡ 1 (mod p)

Worked example: Mod 13: 7¹² ≡ 1. Fermat's little theorem guarantees it for every base from 1 to 12, without computing a single large power.

Open this case: fermat mod 13

Repeated powers — they cycle, and Fermat says where. The powers of each base mod 13, with the column at exponent 12 coming back to 1 for every unit. Raising to a power mod n. For prime p and a not divisible by p, the exponents wrap with period dividing p − 1. — The powers of each base mod 13, with the column at exponent 12 coming back to 1 for every unit.

Problem solved in full

Evaluating 7 12 mod 13 on the full grid 5 steps
Evaluate 7¹² mod 13 without ever writing 7¹² down. This is the a^k operation with n = 13, a = 7 and k = 12, and the grid is the full table of a^k mod 13.
1. 7¹² is 13 841 287 201 — eleven digits for an answer that has to come out between 0 and 12. Reduction survives multiplication: the residue of a product depends only on the residues of its factors, so you may reduce after every step rather than at the end.
  
  $7^{12} = 13841287201, x y mod n = ((x mod n) (y mod n)) mod n$
2. Fermat's little theorem settles the answer before any arithmetic happens. For a prime p and an a that p does not divide, a^p−1 ≡ 1; here p = 13 and k is exactly p − 1, so the result is 1 and everything below is a check on the theorem rather than a search for the answer.
  
  $p = 13 prime, 13 ∤ 7 ⟹ 7^{12} = 7^{p - 1} \equiv 1 (mod 13)$
3. Squaring doubles the exponent, so three squarings reach 7⁸. Each line reduces before the next one starts, which is why no number in the whole calculation exceeds 100.
  
  $7^{2} = 49 \equiv 10, 7^{4} \equiv 1 0^{2} = 100 \equiv 9, 7^{8} \equiv 9^{2} = 81 \equiv 3 (mod 13)$
4. 12 is 8 + 4, and both of those powers were computed on the way up, so one more multiplication finishes it. Four multiplications in total, against the eleven that a left-to-right product of twelve sevens would need — and the saving grows with the exponent, not with the modulus.
  
  $12 = 110 0_{2} = 8 + 4 ⟹ 7^{12} \equiv 7^{8} \cdot 7^{4} \equiv 3 \times 9 = 27 \equiv 1 (mod 13)$
5. The order of 7, meaning the least k with 7^k ≡ 1, must divide 12, so it is one of 1, 2, 3, 4, 6 and 12. Checking the five proper divisors rules every one of them out, so the order is exactly 12.
  
  $7^{1} \equiv 7, 7^{2} \equiv 10, 7^{3} \equiv 5, 7^{4} \equiv 9, 7^{6} \equiv 12 ⟹ ord_{13} (7) = 12$
Answer

The three squarings and the closing multiplication reproduce the tool's row for a = 7 without an error: 10 at k = 2, 9 at k = 4, 3 at k = 8, and 1 at k = 12 where Fermat said it would be. Because the order is 12 and not a proper divisor of it — the order test needed 12 at k = 6, which is not 1 — that row is a permutation of 1 through 12, hitting every non-zero residue exactly once, which makes 7 a primitive root mod 13. Now read the row in the other direction: given 11, find k. There is no square-and-multiply for that, only a search, and that asymmetry between easy forwards and hard backwards is the whole basis of Diffie–Hellman key exchange. The algorithm scales in a way the table cannot: a 2048-bit exponent costs at most about 4 000 modular multiplications, while the power it names would take roughly 2.7 × 10⁶¹⁶ digits to write out.

$ord_{13} (7) = 12 ⟹ {7^{k} mod 13}_{k = 1}^{12} = {1, 2, \dots, 12}$

Learning path

When two things land on the same value

Leads to Birthday paradox

References (1)

The congruence sign this tool prints, and the book that introduced it: C. F. Gauss, Disquisitiones arithmeticae. Leipzig, 1801 — Section I defines a ≡ b (mod n); Section II is where a prime modulus starts behaving like a field. Digitised by the Smithsonian Libraries.

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

clock mod 12 - Clock arithmetic: 13 ≡ 1 (mod 12), so 13+5 wraps to 6.
prime mod 7 - Prime modulus: nonzero multiplication rows behave as permutations.
composite mod 6 - Composite modulus shows repeating rows where gcd(row,n) > 1.
fermat mod 13 - Fermat-style pattern: a^(p-1) ≡ 1 mod p for prime p and gcd(a,p)=1.