Q: Is 5890 divisible by 9

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is not

yes!

No.

5890 ÷ 3 = 1963.33 recurring.

The trick to determining whether a number is divisible by 9 is to add all the individual numbers of the large number together and see if the result is divisible by 9. Let's add 5 + 8 + 9 + 0 = 22. Since 22 is not neatly divisible by 9, then we know that the larger 5890 is also not divisible by 9. Of course, you can divide it by 9, but there will be something left over.

No.

NO

Any number ending is zero is divisible by 2. 5890 / 2 = 2945

yes

If the one's place is a 0, then it is divisible by 10.

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.

yes the answer is 1472.5

Yes; 1178.

Yes, both.

No, none of them.

5,890 is divisible by: 1, 2, 5, 10, 19, 31, 38, 62, 95, 155, 190, 310, 589, 1178, 2945 and 5890.

5,890 is divisible by: 1, 2, 5, 10, 19, 31, 38, 62, 95, 155, 190, 310, 589, 1178, 2945, 5890.

A trick here is to use something called the digital root (take the sum of all the digits until you have a single digit.) 5 + 8 + 9 + 0 = 22; 2 + 2 = 4 Anything divisible by 3 will have a digital root of 3, 6, or 9. Therefore, 5890 is not divisible by 3 [without a remainder.]

No because: 5+8+9+0=22 22 is not divisible by 6.

All numbers divisible by 3 are NOT divisible by 9. As an example, 6, which is divisible by 3, is not divisible by 9. However, all numbers divisible by 9 are also divisible by 3 because 9 is divisible by 3.

No - any number divisible by 9 will have the sum of its digits divisible by 9.

NO. 1110 is not divisible by 9. A number is divisible by 9 if the sum of its digits is divisible by 9. 1110 = 1 + 1 + 1 + 0 = 3 (3 is not divisible by 9, thus, 1110 is not divisible by 9.)

0.0085

81: 8 + 1 = 9 which is divisible by 9, so 81 is divisible by 9 162: 1 + 6 + 2 = 9 which is divisible by 9, so 162 is divisible by 9 199: 1 + 9 + 9 = 19 → 1 + 9 = 10 → 1 + 0 = 1 which is not divisible by 9, so 199 is not divisible by 9. 1125: 1 + 1 + 2 + 5 = 9 which is divisible by 9, so 1125 is divisible by 9. So 199 is the only one not divisible by 9.

9 is divisible by 1, 3 and 9.

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