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3 and 5, yes.

The rest, no.

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Q: Is 75 divisible by 2 or 3 or 5 or 6 or 9 or 10?
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Is 672 divisible by 5 10 2 3?

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


Is 75 divisible by 2 3 4 5 6 7 8 9 or 10?

3 and 5, yes. The rest, no.


Are 9 and 10 divisible by 2 3 4 5 6 10?

9 is divisible by 3 10 is divisible by 2, 5 and 10. Everything else, no.


Is the number 75 divisible by 2 3 4 5 9 10?

To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8}; The last digit of 75 is 5, which is not one of these so it is not divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3. For this gives: 75→7 + 5 = 12 12→1 + 2 = 3 3 is one of {3, 6,9} so it is divisible by 3. To be divisible by 4, add the last (ones) digit to twice the previous (tens) digit; if this sum is divisible by 4, then so is the original number. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {4, 8} is the original number divisible by 4. For this gives: 75→5 + 2×7 = 19 19→9 + 2×1 = 11 11→1 + 2×1 = 3 3 is not one of {4, 8} so it is not divisible by 4. To be divisible by 5, the last digit must be one of {0, 5}. The last digit of is 5 which is one of {0, 5} so it is divisible by 5. To be divisible by 9, sum the digits of the number and if this sum is divisible by 9, then the original number is divisible by 9. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is 9 is the original number divisible by 9. For this gives: 75→7 + 5 = 12 12→1 + 2 = 3 3 is not 9 so it is not divisible by 9. To be divisible by 10, the last digit must be 0 The last digit is 5 which is not 0, so it is not divisible by 10. → 75 is divisible by 3 and 5 75 is not divisible by 2, 4, 9, 10


Is 244 divisible by 3?

No since only numbers in which its sum can be divisible by 3 is divisible by 3. 2 + 4 + 4 is 10, and 10 is not divisible by 3.

Related questions

Name a number between 100 and 200 that is divisible by 2 3 and 10?

All of them but I assume you mean evenly divisible. 150. 150/2 is 75 150/3 is 50 and 150/10 is 15


Is 672 divisible by 5 10 2 3?

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


Is 75 divisible by 2 3 4 5 6 7 8 9 or 10?

3 and 5, yes. The rest, no.


What numbers IS 150 divisible by?

150 is divisible by 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.


Is 7580 divisible by 2 3 9 10?

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


Is 432 divisible by 2 3 5 9 or 10?

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


Are 9 and 10 divisible by 2 3 4 5 6 10?

9 is divisible by 3 10 is divisible by 2, 5 and 10. Everything else, no.


Is 20 divisible by 2 3 5 9 or 10?

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


Is the number 75 divisible by 2 3 4 5 9 10?

To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8}; The last digit of 75 is 5, which is not one of these so it is not divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3. For this gives: 75→7 + 5 = 12 12→1 + 2 = 3 3 is one of {3, 6,9} so it is divisible by 3. To be divisible by 4, add the last (ones) digit to twice the previous (tens) digit; if this sum is divisible by 4, then so is the original number. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {4, 8} is the original number divisible by 4. For this gives: 75→5 + 2×7 = 19 19→9 + 2×1 = 11 11→1 + 2×1 = 3 3 is not one of {4, 8} so it is not divisible by 4. To be divisible by 5, the last digit must be one of {0, 5}. The last digit of is 5 which is one of {0, 5} so it is divisible by 5. To be divisible by 9, sum the digits of the number and if this sum is divisible by 9, then the original number is divisible by 9. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is 9 is the original number divisible by 9. For this gives: 75→7 + 5 = 12 12→1 + 2 = 3 3 is not 9 so it is not divisible by 9. To be divisible by 10, the last digit must be 0 The last digit is 5 which is not 0, so it is not divisible by 10. → 75 is divisible by 3 and 5 75 is not divisible by 2, 4, 9, 10


Is 800 divisible by 2 3 5 9 and 10?

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


What are the divisible of 75?

75 is divisible by: 1 3 5 15 25 75.


Is 75 divisible by 2 prime numbers?

Yes. The prime factorization of 75 is 3 x 5 x 5. So 75 is divisible by the two prime numbers 3 and 5.