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Q: Is 432 divisible by 2 3 5 6?

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To be divisible by 2 the number must be even, that is its last digit must be 2, 4, 6, 8, or 0; the last digit of 432 is 2 which is one of {2, 4, 6, 8, 0} so 432 is even and divisible by 2. To be divisible by 3 sum the digits of the number; if this sum is divisible by 3 then the original number is divisible by 3. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 3, 6, or 9 then the original number is divisible by 3; For 432: 4 + 3 + 2 = 9 which is one of {3, 6, 9} so 432 is divisible by 3. To be divisible by 5 the last digit must be a 0 or 5; the last digit of 432 is a 2 which is not 0 nor 5, so 432 is not divisible by 5. To be divisible by 6, the number must be divisible by both 2 and 3; these have been tested above and found to be true, so 432 is divisible by 6. To be divisible by 9 sum the digits of the number; if this sum is divisible by 9 then the original number is divisible by 9. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 9 then the original number is divisible by 9; For 432: 4 + 3 + 2 = 9 which is 9 so 432 is divisible by 9 To be divisible by 10 the last digit must be a 0; the last digit of 432 is a 2 which is not 0, so 432 is not divisible by 10. 432 is divisible by 2, 3, 6 and 9, but not divisible by 5 nor 10.

Yes, it is divisible by 2, 3 and 9, but is not exactly divisible by 5 or 10.

All except 5

No.

No, 432 divided by 5 is 86.4

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, you can tell using the divisibility rules. The answers are yes for all but 5 and 10.

Is 5225 divisible by 3? A number is divisible by 3, if the sum of its digits is evenly divisible by 3. 5 + 2 + 2 + 5 = 14. Since 14 is not divisible by 3, neither is 5225.

It is divisible by 2 and 3. It isn't divisible by 5 and 10.

30 is divisible by, and is the least common multiple of, 2, 3, and 5.

It is divisible by 2, 5 and 10, but not 3 and 9.

One of the infinitely many integers divisible by 2, 3 and 5 is 3000.

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.

It is divisible by 3.

20 is divisible by 2, 5 and 10. It is not exactly divisible by 3 or 9.

29,178 is divisible by 2, 3 and 9, not by 5 or 10.

It is divisible by 2 & 3. It is not divisible by 5, 9 & 10. 534 is even → divisible by 2 5 + 3 + 4 = 12 → divisible by 3 534 does not end in 0 or 5 → not divisible by 5 5 + 3 + 4 = 12 → not divisible by 9 534 does not end in 0 → not divisible by 10

It is divisible by 2 and 5

It is evenly divisible by 2 and by 3.

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

475 is evenly divisible by 5, but not by 2, 3, 4, or 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.

It is divisible by 2, 5 and 10 but not the rest.

It is only divisible by 3, and not 2, 4, 5, 6, or 10.

325 is divisible by 5 leaving no remainder