No, 75 is not evenly divisible by nine.
75 is divisible by: 1 3 5 15 25 75.
Yes. 75 / 3 = 25
The simplest for of 75/100 is very simple to figure out, here are the basic steps: 1. 75 25 3 ------ ÷ ------ = ------ 100 25 4 You do that step because 25 is divisible by both 75 & 100. Your final fraction should be 3/4.
There are eight pairs of numbers divisible by three that sum to 150, and 75 is also divisible by three, so: sum = 17 * 75 = 750 + 525 = 1275
80 is divisible by 5 and 4 and is greater than 75
No, 75 is not evenly divisible by nine.
No. 75 is not evenly divisible by ten.
75 is divisible by: 1 3 5 15 25 75.
yes, because if you divide 300 by 4 you get 75 with a remaider of 0.
75 is divisible by: 1 3 5 15 25 75
87
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.
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.
Neither is divisible by 8.
No.
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