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A number is divisible by 6 if the number is divisible by 2 AND 3.

Q: What is the test of divisibility for 6?

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Test of divisibility by 2:If a number is even then the number can be evenly divided by 2.5890 is an even number so, it is divisible by 2.Test of divisibility by 3:A number is divisible by 3 if the sum of digits of the number is a multiple of 3.Sum of digits = 5+8+9+0 = 22, which is not a multiple of 3.So, 5890 is not divisible by 3.Test of divisibility by 6:In order to check if a number is divisible by 6, we have to check if it is divisible by both 2 and 3 because 6 = 2x3.As we have seen above that 5890 is not divisible by 3 so, 5890 fails to pass the divisibility test by 6.Test of divisibility by 9:If the sum of digits of a number is divisible by 9 then the number is divisible by 9.Sum of digits = 5+8+9+0 = 22, which is not a multiple of 9.So, 5890 is not divisible by 9.Test of divisibility by 5:If the last digit of a number is 0 or 5, then it is divisible by 5.It is clear that 5890 is divisible by 5.Test of divisibility by 10:If the last digit of a number is 0, then the number is divisible by 10.It is clear that 5890 is divisible by 10 as the last digit is 0.

Edward Chavez

Every number has a test for divisibility. The issue is that the tests get more complicated as the divisor increases. For primes up to 50, see either of the attached links.

all even numbers

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If the number is also divisible by 2 and 3

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To test divisibility for 20, you need to use the tests for divisibility by 4 and 5.The test for divisibility by 4 is that the last 2 digits of the number, given as a 2-digit number, are divisible by 4.Example for 4:We are testing the number 11042.42/4 = 10.5 which is not a whole number. Therefore 11042 is not divisible by 4.The test for divisibility by 5 is that the last digit of the number is either 5 or 0.

It is divisibility by 3 and divisibility by 5.Divisibility by 3: the digital root of an integer is obtained by adding together all the digits in the integer, with the process repeated if required. If the final result is 3, 6 or 9, then the integer is divisible by 3.Divisibility by 5: the integer ends in 0 or 5.

Test of divisibility by 2:If a number is even then the number can be evenly divided by 2.5890 is an even number so, it is divisible by 2.Test of divisibility by 3:A number is divisible by 3 if the sum of digits of the number is a multiple of 3.Sum of digits = 5+8+9+0 = 22, which is not a multiple of 3.So, 5890 is not divisible by 3.Test of divisibility by 6:In order to check if a number is divisible by 6, we have to check if it is divisible by both 2 and 3 because 6 = 2x3.As we have seen above that 5890 is not divisible by 3 so, 5890 fails to pass the divisibility test by 6.Test of divisibility by 9:If the sum of digits of a number is divisible by 9 then the number is divisible by 9.Sum of digits = 5+8+9+0 = 22, which is not a multiple of 9.So, 5890 is not divisible by 9.Test of divisibility by 5:If the last digit of a number is 0 or 5, then it is divisible by 5.It is clear that 5890 is divisible by 5.Test of divisibility by 10:If the last digit of a number is 0, then the number is divisible by 10.It is clear that 5890 is divisible by 10 as the last digit is 0.

By tautology. If it did not work, it would not be a divisibility rule!

Edward Chavez

Every number has a test for divisibility. The issue is that the tests get more complicated as the divisor increases. For primes up to 50, see either of the attached links.

all even numbers

7623 is divisible by 3.Test of divisibility by 3:Sum of digits of 7623 = 7+6+2+3 = 18, which is a multiple of 3, so the number is divisible by 3.If sum of the digits of a number is a multiple of 9 then it is divisible by 9.So, 7623 is also divisible by 9.Therefore, test of divisibility can help a lot in determining whether a number is divisible by any other number.

Yes

6 = 2 x 3So it must satisfy the divisibility tests for both 2 and 3, namely:Any even number for which the sum of its digits is divisible by 3.