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Q: Is 1 2 0 3 6 divisible by 4 and 9?

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If you multiply 417 by any integer (0, 1, 2, 3, -1, -2, -3, etc.), you will get a number that is divisible by 417.If you multiply 417 by any integer (0, 1, 2, 3, -1, -2, -3, etc.), you will get a number that is divisible by 417.If you multiply 417 by any integer (0, 1, 2, 3, -1, -2, -3, etc.), you will get a number that is divisible by 417.If you multiply 417 by any integer (0, 1, 2, 3, -1, -2, -3, etc.), you will get a number that is divisible by 417.

1, 2, 0, 3 and 8 are not divisible by 4 and/or 9. Neither is 12038.

The sum of its digits is 1+2+0 = 3 which is divisible by 3. So 120 is divisible by 3. The number 120 ends in a 0 and so it is divisible by 5.

2390 is not divisible by 3 but is divisible by 5. 2 + 3 + 9 + 0 = 14; 1 + 4 = 5: 5 is not divisible by 3, so 14 is not divisible by 3, so 2390 is not divisible by 3. 2390 ends with a 0, so 2390 is divisible by 5.

A number is divisible by 6 if it is divisible both by 2 and 3 1240 is even then is divisible by 2 1+2+4+0 = 7 which is not divisible by 3 then 1240 is not divisible by 3 Thus 1240 is not divisible by 6

yes, 0 is divisible by everything.

Last digit (0) is even, so it is divisible by 2 4 + 3 + 2 + 0 = 9 which is divisible by 3, so it is divisible by 3 last digit is 0 or 5, so it is divisible by 5 4 + 3 + 2 + 0 = 9 which is divisible by 9, so it is divisible by 9 last digit is 0, so it is divisible by 10 → 4320 is divisible by all the numbers 2, 3, 5, 9, 10

21 is not divisible by 2 nor 5, but it is divisible by 3. To be divisible by 2 the last digit must be even (one of {0, 2, 4, 6, 8}; the last digit 1 is not one of these, thus 21 is not divisible by 2 To be divisible by 3 sum the digits and if this total is divisible by 3, then so is the original number; the test can be repeated on the sum, so by repeatedly summing the digits of the totals until a single digit remains only if that single digit is 3, 6 or 9 is the original number divisible by 3. 21 → 2 + 1 = 3 which is divisible by 3 so 21 is divisible by 3. To be divisible by 5, the last digit must be 0 or 5. The last digit of 21 is 1 which is not 0 nor 5, thus 21 is not divisible by 5.

No. To be divisible by 6, the number must be divisible by both 2 and 3. To be divisible by 2 the number must be even, ie its last digit must be one of {0, 2, 4, 6, 8}; 105's last digit is 5 which is odd so 105 is not divisible by 2. To be divisible by 3, sum the digits of the number and if the result is divisible by 3, then so is the original number. For 1-5: 1 + 0 + 5 = 6 which is divisible by 3 therefore 105 is divisible by 3. Although 105 is divisible by 3 it is not divisible by 2, thus it is not divisible by 6.

yep it is because 2 is divisible by 2 and 3 is divisible by 3 and 0 is divisible by 5 have fun suckers!

0, 1, 2, 3, 6, 9, and 18...

There is no 4-digit number that is divisible by zero.

It is divisible only by 3; It is not divisible by 2, 4, 5, 6, 9, 10. 17211 is odd, so not divisible by 2, 4, 6 nor 10. 1 + 7 + 2 + 1 + 1 = 12 which is divisible by 3, so 17211 is divisible by 3, but 12 is not divisible by 9, so 17211 is not divisible by 9. 17211 does not end in 5 or 0 so not divisible by 5

All odd whole numbers are not divisible by 2.The term "whole number" has three distinct definitions, in my experience.- Positive Integers ( 1, 2, 3, ... )- Non-Negative Integers ( 0, 1, 2, 3, ... )- Integers ( ..., -3, -2, -1, 0, 1, 2, 3, ... )The largest whole number that is not divisible by two would be the largest odd number, which is not defined.The smallest Positive Integer which is not divisible by two would be 1.The smallest Non-Negative Integer which is not divisible by two would be 0.The smallest Integer which is not divisible by two would be the most negative odd number, which is not defined.

A prime number has only 2 factors which are 1 and itself. Composite numbers are everything else except 1 and 0. 1 and 0 are neither prime, nor composite. A prime number is not divisible by 3.

you add the numbers together and if that is divisible by three then so is that number for example: the number 111, you would do 1+1+1=3 so 111 is dividable by 3; or the number 1,620, 1+6+2+0= 9 (which is divisible by 3) so 1,620 is divisible by 3

All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.

Yes, it is divisible by 2 because it ends with 2, it is also divisible bye 3 because 1+3+2=6. 6 is divisible by 3, therefor 132 is divisible by 3.

Yes, it is divisible by both. 120 is even, so it is a multiple of 2; 1 + 2 + 0 = 3 which is one of 3, 6, 9, so it is a multiple of 3.

To determine if a number is divisible by 6, it must be divisible by both 2 and 3. To determine if a number is divisible by 2, it should be even - in other words, it should end with 0, 2, 4, 6, or 8. To determine if a number is divisible by 3, the sum of its digits should be divisible by 3. 54,132 is an even number, so it is divisible by 2. 5 + 4 + 1 + 3 + 2 = 15, which is divisible by 3, so 54,132 is divisible by 3. Since 54,132 is divisible by both 2 and 3, it is divisible by 6.

A number is divisible by 2 if that number is an even number, which means it ends in 0, 2, 4, 6, or 8. A number is divisible by 3 if the sum of its digits is evenly divisible by 3.

Because 410 ends in 0, it is divisble by 10. To determine if it's divisible by 6, it needs to be divisible by 2 and 3. it is divisible by 2, as it is even, but is not divisible by 3;

57 divided by 3 is 19. (Any number in which all the digits add up to equal a number divisible by 3 is divisible by 3. 5+7=12 which is divisible by 3.) And since 57 is divisible by more numbers than just 1 and itself it is a composite number.

Yes, 120 is divisible by 3. A good rule of thumb for the multiples of 3 is to add up the digits in the number. If the number that comes out is divisible by 3, then it is a multiple of 3. For example, in 120, 1 + 2 + 0 = 3. Therefore, the number is divisible by 3. The process can be repeated for larger numbers. For example, 2688 is divisible by 3 because 2 + 6 + 8 + 8 = 24, and 2 + 4 = 6.

If the sum of the digits equals 3, the number is evenly divisible by 3. 2 + 0 + 1 = 3, so 201 is evenly divisible by 3. 1, 3, 67, 201