Q: If every digit of a number is divisible by 3 then the number itself is divisible by 3?

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Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.

A prime number is one that has only two factors: one and itself.You can tell if a number is prime quickly by using the Laws of Divisibility:Numbers that end in 0, 2, 4, 6, and 8 are divisible by 2.If the sum of a number's digit is divisible by 3, the number itself is divisible by 3.If the last two digits of a number are divisible by 4, the number itself is divisible by 4.If a number ends in 0 or 5, it is divisible by 5.If a number is divisible by both 2 and 3, it is divisible by 6.If you double the last digit and subtract it from the rest of the number and the answer is 0 or divisible by 7, it is divisible by 7. For example, 4364: 4 doubled is 8. 436 - 8 = 424. You still cannot readily tell if the number is divisible by 7, so you can do it again: 4 doubled is 8. 42 - 8 = 34. 34 is not divisible by 7 so 436 is not divisible by 7If the last three digits of a number are divisible by 8, the number itself is divisible by 8.If the sum of the digits of a number is divisible by 9, the number itself is divisible by 9.If a number ends with 0, it is divisible by 10.If you add every other digit and then subtract the rest of the digits and the answer is 0 or is divisible by 11, the number itself is divisible by 11. For example: 1364: 1+ 6 = 7 and 7 - 3 - 4 = 0. The number is divisible by 11.If the number is divisible by 3 and 4, it is divisible by 12.There are more divisibility rules that can be used for large numbers.

every alternate number from 100 to 998

Every sixth number from 1,002 to 9,996 is divisible by 3 and 6. There are 1,500 of them.

Yes. Every non-zero number is divisible by itself.

I'm not quite sure about that theory, but if all the numbers in the number add up to a number that's dividable by 3, then the number itself would then be divisible by 3. Ex. 141 1+4+1= 6 6 is divisible by 3, so 141 is too. 141/3= 47.

One and itself.

Yes. Every number is divisible by itself or 1.

1795 is a composite number because it is divisible by 5. Every number (except 5) ending with digit 5 or 0 is divisible by 5.

There is no such number. Once you have one n-digit number which is divisible by 4 then 10 times that number will be an n+1 digit number which is divisible by 4. And this process can continue ad infinitum.

Every three-digit number that ends with a zero or a 5 is divisible by 5.It doesn't matter what the first 2 digits are.

A prime number is divisible only by itself - 102 is divisible by 2 and 51 so it is not a prime.A prime number is divisible only by 1 and itself.102 is a composite number because it is divisible by 2, 17, and 51 as well as by 1 and itself.Every even number (except 2) is composite.

135. Every increase of 30 is also. Such as 165, 195, 225.

14

Every number is divisible by any non-zero number.421 is a prime so its factors are only 1 and itself.

Yes, every number that's divisible by 9 is divisible by 3

The smallest 3 digit number full stop is 100, as every smaller number has 2 digits. 100 divides by 4, thus the answer is 100.

To find out if 1859 is a prime number, use the rules of division:1859 is not an even number, so it is not divisible by 2.The digits of 1859 add up to a number that is divisible by 3, so it is not divisible by 3.The last two digits (59) is not divisible by 4, so neither is 1859.The last digit is not 5, so it is not divisible by 5.1859 is not divisible by 2 and it is not divisible by 3, so it is not divisible by 6.When you double the last digit (9 doubled is 18), and subtract that number from the rest of the number (859 - 18 = 841), if the answer is 0 or divisible by 7, the number itself is divisible by 7. If you can't readily tell if the number is divisible by 7, you can do the steps again. Starting with 841, doubling the 1 is still 1, so take 1 away from 84 to get 83. 83 is not divisible by 7, so neither is 1859.The last three digits (859) are not divisible by 8, so 1859 is not divisible by 8.If the sum of the digits is divisible by 9, the number is divisible by 9. The sum of the digits is 23, which is not divisible by 9.1859 does not end with a 0, so it is not divisible by 10.If the sum of every second digit (8 + 9) minus the sum of the other digits (1 + 5) equals 0 or is divisible by 11, the number itself is divisible by 11. 17 - 6 = 11, so 1859 is divisible by 11.Since 1859 is divisible by 11, it is a composite number (a number that has at least three factors: 1, the number itself, and at least one oher factor).

It isn't. 438 is divisible by 6 because 438 is a multiple of 6.

Yes, because if it had zero as the final digit, it'd be divisible by 2 and 5, and thus not prime.

11

The number 1. Every other number is divisible by 1 and itself - so at least two numbers.

Every number? No. It isn't divisible by 61, for instance.

Every number is divisible by any non-zero number.Any element of the set of numbers of the form 12*k, where k is an integer in the range [84, 833], is evenly divisible.

Yes Every even number, which is a whole number whose least significant digit (the one at the right) is 0, 2, 4, 6 or 8, is evenly divisible by 2.