Q: How can you check if a number is divisible by 9 when usiing the divisibility rule for 9?

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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.

You could combine the tests for divisibility by 3 and 4. To test for divisibility by three, add all the digits together and see if they're divisible by three. If necessary, you can keep repeating the addition until you come up with a single-digit number. To test for divisibility by four, take the last two digits. If that two-digit number is divisible by four, then the whole number is. This is because any multiple of 100 is divisible by 4, so only the last two digits matter. Combined, these two tests will allow you to quickly check for divisibility by 12.

No. To check divisibility by 3 add the digits of the number together and if the sum is divisible by 3, then so is the original number. 36226 → 3 + 6 + 2 + 2 + 6 = 19 19 is not divisible by 3, so 36226 is not divisible by 3.

The divisibility rule for 3 is add up the digits and check for divisibility. 7+2+4 = 13.13 is not evenly divisible by 3 so 724 is not divisibleby 3.

Do the division on a calculator. If you get a whole number - no digits after the decimal point - it is divisible. Otherwise it isn't.Note: Specifically for divisibility by 5, you can also check whether the last digit is 5 or 0.

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There are two ways of answering this.Check the number for divisibility by 2.Check the quotient for divisibility by 2.Check the quotient for divisibility by 2.Check the quotient for divisibility by 2.Check the quotient for divisibility by 2.Check the quotient for divisibility by 2.For large numbers, the check can be restricted to the number formed by the last six digits.

Multiply the last digit by 7. Subtract that number from the remaining digits. If that number is divisible by 23, then the original number is divisible by 23.

No, it isn't. Check with the divisibility rule for 3 which states that if a number is divisible by 3, the sum of its digits must be divisible by 3...

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.

To check for divisibility by 6 you need to check for divisibility by 2 and by 3. Divisibility by 2: If the number ends with a 0, 2, 4, 6 or 8 it is divisible by 2. If it is not then it is not divisible by 2 and so cannot be divisible by 6. Divisibility by 6: If the digital root is divisible by 3 then the number is divisible by 3. If it is not then it is not divisble by 3 and so cannot be divisible by 6. The digital root is simply the sum of all the digits in the number. And, if the answer has more than 2 digits, you can repeat the process as many times as you like. For example, 7987 7+9+8+7 = 31 3+1 = 4 So the digital root of 7987 is 4. That is not divisible by 3 and so neither is 7987.

No. To check divisibility by nine, sum the digits. If they add up to 9 (or a multiple of 9) then the number is divisible by 9. This trick works for 9 divisibility and 3 divisibility. In this problem: 3 + 0 + 9 + 1 = 13, which is not divisible by 9. Suppose the number was 3591: 3+5+9+1=18, which is divisible by 9 (you can check 1+8=9). You can keep summing the digits of the new numbers till you get to a number which is either 9 or some other number. (If it's 3 or 6, then the number is divisible by 3)

There is no easy rule for divisibility by 34.

Not evenly if you look at the last number and it is an even number such as 2,4,6,8 then it would be evenly divisible by 2.

Yes.To check for divisibility by 6, check that the number passes the tests for divisibility by 2 and 3 (since 2 x 3 = 6), namely is the number even and when the digits are added together are they a multiple of 3.For 462:it is even4 + 6 + 2 = 12 which is divisible by 3 (3 x 4 = 12)So it is divisible by 6.462 / 6 = 77

To check for divisibility by 9 sum the digits of the number and if this sum is divisible by 9 then so is the original number. For 32643: 3 + 2 + 6 + 4 + 3 = 18 which is divisible by 9 so 32643 is divisible by 9. As 9 = 3 × 3, any number divisible by 9 is also divisible by 3, thus as 32643 is divisible by 9 it is also divisible by 3. However, for completeness: to check for divisibility by 3 sum the digits of the number and if this sum is divisible by 3 then so is the original number. For 32643: 3 + 2 + 6 + 4 + 3 = 18 which is divisible by 3 so 32643 is divisible by 3.

876 is divisible by 3. As an easy check for the divisibility by 3 for any number, simply add up the digits (8 + 7 + 6 for 876) and if the sum is a multiple of 3, then the number is divisible by 3.

You could combine the tests for divisibility by 3 and 4. To test for divisibility by three, add all the digits together and see if they're divisible by three. If necessary, you can keep repeating the addition until you come up with a single-digit number. To test for divisibility by four, take the last two digits. If that two-digit number is divisible by four, then the whole number is. This is because any multiple of 100 is divisible by 4, so only the last two digits matter. Combined, these two tests will allow you to quickly check for divisibility by 12.