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Q: Is 75 divisible by 2

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No. 75 is not evenly divisible by two.

no

80 is divisible by 5 and 4 and is greater than 75

It could be 76 - not divisible by 5. It could be 75 - not divisible by 2.

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

No. 75 is not evenly divisible by ten.

No, 75 is not evenly divisible by nine.

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

No it is not. Since 75 has an odd number at the end of it, it is not divisible by 4.

Every number is divisible by any non-zero number. However, 75 is not evenly divisible by 8.

Both composite 75 is divisible by 5 112 is divisible by 2 ... and other numbers

Yes, not evenly, quotient is 75, remainder is 1

No. 375 is only divisible by: 1, 3, 5, 15, 25, 75, 125, 375.

The first number is 80

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

75 is divisible by 1, 3, 5, 15, 25, and 75 Answer: Any number is divisible by any other number, 75 is no exception. The answer is not always an integer value. Examples: 75/2=37.5, 75/750=0.1Division by zero is undefined and division by an imaginary number (j or (-1)0.5 ) may be done by multiplying top and bottom by the conjugate of the complex number first

Neither is divisible by 8.

No. 75 is not evenly divisible by 67.

-75

no

no.. take an example.. 150 is divided by both 75 and 3..but not by 225..

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

900 is one possible answer to this confusingly worded question.

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

Prime Numbers are those which are either divisible by themselves or by 1. Going according to the simple definition, 5 is a prime number while 75 is not as 75 is divisible by 75,15,5,3,1.