6,142 = 'n' × 3

When the two numbers are divided, there is a remainder.

Scroll down for the 2nd method...

6,142 is not a prime number but a composite one.

* A composite number is a natural number that has at least one other factor than 1 and itself.

The number 6,142 is not divisible by 3.

When the two numbers are divided, there is a remainder.

Is 6,142 divisible by 3? | Mar 25 13:56 UTC (GMT) |

Is 731 divisible by 17? | Mar 25 13:56 UTC (GMT) |

Is 3,550 divisible by 655? | Mar 25 13:56 UTC (GMT) |

Is 846 divisible by 7? | Mar 25 13:56 UTC (GMT) |

Is 23,343 divisible by 3? | Mar 25 13:56 UTC (GMT) |

The list of all the pairs of numbers that were checked on whether they are divisible or not |

### 1. Divisibility:

**One natural number is said to be divisible by another natural number if after dividing the two numbers, the remainder of the operation is zero.****Example:**Let's divide two different numbers: 12 and 15, by 4.- When dividing 12 by 4, the quotient is 3 and the remainder of the operation is zero.
- But when we divide 15 by 4, the quotient is 3 and the operation leaves a remainder of 3.
- We say that the number 12 is divisible by 4 and 15 is not divisible by 4.
- We also say that 4 is a divisor, or a factor, of 12, but not a factor (divisor) of 15.

**We say that the number "a" is divisible by "b"**, if there is an integer number "n", such that:- a = n × b.
- The number "b" is called a divisor (or a factor) of "a" ("n" is also a divisor, or a factor, of "a").

### 2. Some divisibility rules:

- 0 is divisible by any number other than itself.
- 1 is a divisor (a factor) of every number.
**Improper factors:**Any number "a", different of zero, is divisible at least by 1 and itself. In this case the number itself, "a", is called**an improper factor**(or an**improper divisor**). Some also consider 1 as an improper factor (divisor).**Prime numbers:**A number which is divisible only by 1 and itself is also called**a prime number**.**Coprime numbers:**If the greatest common factor of two numbers, "m" and "n", the GCF (m; n) = 1 - then it means that the two numbers are coprime, in other words they have no divisor other than 1. If a number "a" is divisible by these two coprime numbers, "m" and "n", then "a" is also divisible by their product, (m × n).**Example:**- The number 84 is divisible by 4 and 3 and is also divisible by 4 × 3 = 12.
- This is true because the two divisors, 3 and 4, are coprime.

### 3. Calculating the divisors (factors):

**Calculating the divisors (factors)**of a number is very useful when simplifying fractions (reducing fractions to lower terms).- The established rules for finding factors (divisors) are based on the fact that the numbers are written in the decimal system:
- Multiples of 10 are divisible by 2 and 5, because 10 is divisible by 2 and 5
- Multiples of 100 are divisible by 4 and 25, because 100 is divisible by 4 and 25
- Multiples of 1,000 are divisible by 8, because 1,000 is divisible by 8.
- All the powers of 10, when divided by 3, or 9, have a remainder equal to 1.

- Due to the rules of operations with remainders, we have the following remainders when dividing numbers by 9:
- 600 leaves a remainder equal to 6 = 1 × 6 (1 for every 100)
- 240 = 2 × 100 + 4 × 10, then the remainder will be equal to 2 × 1 + 4 × 1 = 6

- When a number is divided by 3 or 9, the remainder is equal to what you get by dividing the sum of the digits of that number by 3 or 9:
- 7,309 has the sum of its digits: 7 + 3 + 0 + 9 = 19, which is divided with a remainder by either 3 or 9. So 7,309 is divisible by neither 3 nor by 9.

- All the even powers of 10, such as 10
^{2}= 100, 10^{4}= 10,000, 10^{6}= 1,000,000, and so on, when divided by 11 have a remainder equal to 1. - All the odd powers of 10, such as 10
^{1}= 10, 10^{3}= 1,000, 10^{5}= 100,000, 10^{7}= 10,000,000, and so on, when divided by 11 have a remainder equal to 10. In this case,**the alternating sum of the digits**of the number has the same remainder as the number itself when divided by 11. **How is the alternating sum of the digits**being calculated - it is shown in the example below.

- For instance, for the number: 85,976: 6 + 9 + 8 = 23, 7 + 5 = 12, the alternating sum of the digits: 23 - 12 = 11. So 85,976 is divisible by 11.

### 4. Quick ways to determine whether a number is divisible by another one or not:

- 2, if the last digit is divisible by 2. If the last digit of a number is 0, 2, 4, 6 or 8, then the number is divisible by 2. For example, the number 20: 0 is divisible by 2, so then 20 must be divisible by 2 (indeed: 20 = 2 × 10).
- 3, if the sum of the digits of the number is divisible by 3. For example, the number 126: the sum of the digits is 1 + 2 + 6 = 9, which is divisible by 3. Then the number 126 must also be divisible by 3 (indeed: 126 = 3 × 42).
- 4, if the last two digits of the number make up a number that is divisible by 4. For example 124: 24 is divisible by 4 (24 = 4 × 6), so 124 is also divisible by 4 (indeed: 124 = 4 × 31).
- 5, if the last digit is divisible by 5 (the last digit is 0 or 5). For example 100: the last digit, 0, is divisible by 5, then the number 100 must be divisible by 5 (indeed: 100 = 5 × 20).
- 6, if the number is divisible by both 2 and 3. For example, the number 24 is divisible by 2 (24 = 2 × 12) and is also divisible by 3 (24 = 3 × 8), then it must be divisible by 6. Indeed, 24 = 6 × 4.
- 7, if the last digit of the number (the unit digit), doubled, subtracted from the number made up of the rest of the digits gives a number that is divisible by 7. The process can be repeated until a smaller number is obtained. For example, is the number 294 divisible by 7? We apply the algorithm: 29 - (2 × 4) = 29 - 8 = 21. 21 is divisible by 7. 21 = 7 × 3. But we could have applied the algorithm again, this time on the number 21: 2 - (2 × 1) = 2 - 2 = 0. Zero is divisible by 7, so 21 must be divisible by 7. If 21 is divisible by 7, then 294 must be divisible by 7.
- 8, if the last three digits of the number are making up a number that is divisible by 8. For example, the number 2,120: 120 is divisible by 8 since 120 = 8 × 15. Then 2,120 must also be divisible by 8. Proof: if we divide the numbers, 2,120 = 8 × 265.
- 9, if the sum of the digits of the number is divisible by 9. For example, the number 270 has the sum of the digits equal to 2 + 7 + 0 = 9, which is divisible by 9. Then 270 must also be divisible by 9. Indeed: 270 = 9 × 30.
- 10, if the last digit of the number is 0. Example, 140 is divisible by 10, since 140 = 10 × 14.
- 11 if the alternating sum of the digits is divisible by 11. For example, the number 2,915 has the alternating sum of the digits equal to: (5 + 9) - (1 + 2) = 14 - 3 = 11, which is divisible by 11. Then the number 2,915 must also be divisible by 11: 2,915 = 11 × 265.
- 25, if the last two digits of the number are making up a number that is divisible by 25. For example, the number made up by the last two digits of the number 275 is 75, which is divisible by 25, since 75 = 25 × 3. Then 275 must also be divisible by 25: 275 = 25 × 11.