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1, 2, 0, 3 and 8 are not divisible by 4 and/or 9.

Neither is 12038.

Q: Is 1 2 0 3 8 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.

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.

yes, 0 is divisible by everything.

Check your divisibility rules: 2: One's digit is an even number YES 3: The sum of the digits is divisible by 3 NO: 7 + 1 + 0 = 8 which is not divisible by 3 5: One's digit is a 5 or a 0 YES 9: The sum of the digits is divisible by 9 NO: 7 + 1 + 0 = 8 which is not divisible by 9 10: One's digit is a 0 YES

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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.

not divisible by 9.but it is divisible by 4.

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

Check your divisibility rules: 2: One's digit is an even number YES 3: The sum of the digits is divisible by 3 NO: 7 + 1 + 0 = 8 which is not divisible by 3 5: One's digit is a 5 or a 0 YES 9: The sum of the digits is divisible by 9 NO: 7 + 1 + 0 = 8 which is not divisible by 9 10: One's digit is a 0 YES

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

Yes, the number 1720 is divisible by 3 because the sum of its digits (1 + 7 + 2 + 0) is equal to 10, which is divisible by 3.

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.

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.