6. To check for divisibility by 9, add the digits of the number together and if the sum is divisible by 9, then the original number is divisible by 9. If the test is repeated on the sum(s) until a single digit remains, then this is the remainder when the original number is divided by 9. Subtracting this remainder from 9 will give the smallest number that needs to be added to to the original number to make it divisible by 9. For 75: 7 + 5 = 12 1 + 2 = 3 so 75 ÷ 9 has a remainder of 3, therefore add 9 - 3 = 6 to 75 to make it divisible by 9. (75 + 6 = 81 = 9 x 9).
75 is divisible by: 1 3 5 15 25 75.
No it is not. Since 75 has an odd number at the end of it, it is not divisible by 4.
81: 8 + 1 = 9 which is divisible by 9, so 81 is divisible by 9 162: 1 + 6 + 2 = 9 which is divisible by 9, so 162 is divisible by 9 199: 1 + 9 + 9 = 19 → 1 + 9 = 10 → 1 + 0 = 1 which is not divisible by 9, so 199 is not divisible by 9. 1125: 1 + 1 + 2 + 5 = 9 which is divisible by 9, so 1125 is divisible by 9. So 199 is the only one not divisible by 9.
Yes. 75 / 3 = 25
6. To check for divisibility by 9, add the digits of the number together and if the sum is divisible by 9, then the original number is divisible by 9. If the test is repeated on the sum(s) until a single digit remains, then this is the remainder when the original number is divided by 9. Subtracting this remainder from 9 will give the smallest number that needs to be added to to the original number to make it divisible by 9. For 75: 7 + 5 = 12 1 + 2 = 3 so 75 ÷ 9 has a remainder of 3, therefore add 9 - 3 = 6 to 75 to make it divisible by 9. (75 + 6 = 81 = 9 x 9).
No. 375 is divisible by 1, 3, 5, 15, 25, 75, 125, 375.
No. 75 is not evenly divisible by ten.
75 is divisible by: 1 3 5 15 25 75.
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
75 is divisible by: 1 3 5 15 25 75
No it is not. Since 75 has an odd number at the end of it, it is not divisible by 4.
1, 3, 5, 9, 15, 25, 45, 75, 225
3 and 5, yes. The rest, no.
The prime factors of 675 are 3, 3, 3, 5, and 5, so 675 is divisible by 3 and 5. It is also divisible by the product of any two or more of those factors so, for instance, 675 is divisible by 9, 15, 25, 27, 75, etc.
Neither is divisible by 8.
No.