standarddeviationcalculator.net

Updated Free · runs in your browser

Math

Divisibility calculator

Enter a whole number to see which of 2 to 12 divide it, with the digit rule for each one worked through on your number. Switch to a single divisor to test any other number.

Any whole number, as long as you like.

Divisors from 2 to 12 that go into 7,560 10 of 11
Divisible by2, 3, 4, 5, 6, 7, 8, 9, 10, 12
Not divisible by11
Digit sum18
÷ResultRemainder
2✓ divisible0
3✓ divisible0
4✓ divisible0
5✓ divisible0
6✓ divisible0
7✓ divisible0
8✓ divisible0
9✓ divisible0
10✓ divisible0
11✗ no3
12✓ divisible0
Show the working, step by step

  1. 2: The last digit is even (0, 2, 4, 6 or 8).

    last digit 0 → even → divisible

  2. 3: The sum of the digits is divisible by 3.

    7 + 5 + 6 + 0 = 18; 18 ÷ 3 = 6 → divisible

  3. 4: The number formed by the last two digits is divisible by 4.

    60 ÷ 4 = 15 → divisible

  4. 5: The last digit is 0 or 5.

    last digit 0 → divisible

  5. 6: Divisible by both 2 and 3.

    by 2: ✓ (last digit 0); by 3: ✓ (digit sum 18) → divisible

  6. 7: Double the last digit and subtract it from the rest of the number; repeat. The number is divisible by 7 exactly when the result is.

    756 − 2 × 0 = 756; 75 − 2 × 6 = 63; 63 = 7 × 9 → divisible

  7. 8: The number formed by the last three digits is divisible by 8.

    560 ÷ 8 = 70 → divisible

  8. 9: The sum of the digits is divisible by 9.

    7 + 5 + 6 + 0 = 18; 18 ÷ 9 = 2 → divisible

  9. 10: The last digit is 0.

    last digit 0 → divisible

  10. 11: Starting from the last digit, alternately add and subtract the digits. The result must be divisible by 11 (0 counts).

    0 − 6 + 5 − 7 = −8; not a multiple of 11 → not divisible

  11. 12: Divisible by both 3 and 4.

    by 3: ✓ (digit sum 18); by 4: ✓ (last two digits 60) → divisible

The rules

DivisorRuleWhy it works
2Last digit even10 is divisible by 2
3Digit sum divisible by 310 leaves remainder 1 mod 3
4Last two digits divisible by 4100 is divisible by 4
5Last digit 0 or 510 is divisible by 5
6Divisible by 2 and by 36 = 2 × 3, coprime factors
7Rest minus twice the last digit divisible by 710a + b ≡ 0 ⇔ a − 2b ≡ 0 (mod 7)
8Last three digits divisible by 81,000 is divisible by 8
9Digit sum divisible by 910 leaves remainder 1 mod 9
10Last digit 0Place value
11Alternating digit sum divisible by 1110 leaves remainder −1 mod 11
12Divisible by 3 and by 412 = 3 × 4, coprime factors

A worked example: 7,560

  • 2, 5, 10: the last digit is 0, so all three divide it.
  • 3 and 9: 7 + 5 + 6 + 0 = 18, which is divisible by both.
  • 4: the last two digits, 60, are 4 × 15.
  • 8: the last three digits, 560, are 8 × 70.
  • 6 and 12: follow from 2, 3 and 4.
  • 7: 756 − 2 × 0 = 756; 75 − 2 × 6 = 63 = 7 × 9.
  • 11: 0 − 6 + 5 − 7 = −8, not a multiple of 11. The remainder is 3.

So 7,560 is divisible by 10 of the 11 numbers from 2 to 12, everything except 11. That fits its prime factorisation, 7,560 = 2³ × 3³ × 5 × 7, which has no 11.

Where the rules come from

Every rule is modular arithmetic on place value. Write the number as a sum of digits times powers of ten, and replace each power of ten by its remainder on division by d. For 3 and 9 every power of ten is 1, so the remainder is the digit sum. For 11 the powers of ten alternate between 1 and −1, which gives the alternating sum. For 2, 4, 8, 5 and 25, high enough powers of ten are 0, so only the last few digits count.

Common mistakes

  • Combining factors that are not coprime. Divisible by 2 and 4 does not mean divisible by 8 (take 4).
  • Starting the 11 test from the left on an even-length number. It only flips the sign, so the verdict is the same, but the remainder is not.
  • Using the digit sum for 6. 15 has digit sum 6 but is odd, so it is not divisible by 6.

Common questions

Why does the digit-sum rule work for 3 and 9?

Because 10 leaves remainder 1 when divided by 3 or 9, so do 100, 1,000 and every power of ten. A number such as 7,560 = 7 × 1,000 + 5 × 100 + 6 × 10 + 0 therefore leaves the same remainder as 7 + 5 + 6 + 0 = 18. Since 18 is divisible by 9, so is 7,560.

What is the divisibility rule for 7?

Double the last digit and subtract it from the rest of the number; repeat until the number is small. 7,560 → 756 − 0 = 756 → 75 − 12 = 63 = 7 × 9, so 7,560 is divisible by 7. For long numbers it is quicker to split into groups of three digits from the right and alternately add and subtract them, because 1,001 = 7 × 11 × 13.

How does the rule for 11 work?

Alternately add and subtract the digits starting from the right. For 7,560 that is 0 − 6 + 5 − 7 = −8, which is not a multiple of 11, so 7,560 is not divisible by 11. For 1,331 it is 1 − 3 + 3 − 1 = 0, and 0 counts as a multiple, so 1,331 = 11³ is.

Why is divisibility by 6 checked with 2 and 3?

6 = 2 × 3, and 2 and 3 share no factor. A number divisible by both must be divisible by their product. The same works for 12 = 3 × 4, but not with 2 and 6: 6 is divisible by 2 and 6 yet not by 12, because 2 and 6 share the factor 2.

Is there a rule for 4, 8 and 16?

Yes, look only at the end of the number. 100 is divisible by 4, so only the last two digits matter for 4; 1,000 is divisible by 8, so the last three matter for 8; for 16 it is the last four. 7,560 ends in 60 (divisible by 4) and 560 (560 = 8 × 70), so it is divisible by both 4 and 8.

Written and reviewed by our editorial team. Last updated . Method and sources: how these numbers are computed.