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

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Q: Is 75 divisible by 4
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Related questions

What number is greater than 75 and is divisible by 2 and 5?

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


Is 9 divisible by 75?

No, 75 is not evenly divisible by nine.


Is 75 divisible by 10?

No. 75 is not evenly divisible by ten.


What are the divisible of 75?

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


Is 300 divisible by 4?

yes, because if you divide 300 by 4 you get 75 with a remaider of 0.


What numbers are divisible by 75?

75 is divisible by: 1 3 5 15 25 75


What is 348 divided by 4 and how did you get that answer?

87


How do you get 75 percent to three fourths?

75% is 75/100. Both 75 and 100 are divisible by 25 so you are allowed to divide 25 by each value which gets you to 3/4. 75/25= 3, 100/25= 4.


What is the rule to determine if a number is divisible by 12?

That the number is divisible by 4* and the sum of its digits is a multiple of 3. *If the number has three of more digits then it is only necessary to look at the tens and units to determine if it is divisible by 4, as 4 is a factor of 100 and therefore of any multiple of 100. Examples : 75 : is not divisible by 4 although its digits total 12 which is a multiple of 3. 132 : is divisible by 4 as 32 is divisible by 4, and its digits total 6 which is divisible by 3, then 132 is divisible by 12.


Is 75 and 124 divisible by 8?

Neither is divisible by 8.


Is 75 divisible by 2?

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