180 is divisible by 2, 3, 4, 5, 9, and 10.
Your question is impossible to answer. Any number that is divisible by both 2 and 5 will also be divisible by 10. 30 and 60 are not divisible by 9.
To be divisible by 2 the number must be even, that is its last digit must be 2, 4, 6, 8, or 0; the last digit of 432 is 2 which is one of {2, 4, 6, 8, 0} so 432 is even and divisible by 2. To be divisible by 3 sum the digits of the number; if this sum is divisible by 3 then the original number is divisible by 3. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 3, 6, or 9 then the original number is divisible by 3; For 432: 4 + 3 + 2 = 9 which is one of {3, 6, 9} so 432 is divisible by 3. To be divisible by 5 the last digit must be a 0 or 5; the last digit of 432 is a 2 which is not 0 nor 5, so 432 is not divisible by 5. To be divisible by 6, the number must be divisible by both 2 and 3; these have been tested above and found to be true, so 432 is divisible by 6. To be divisible by 9 sum the digits of the number; if this sum is divisible by 9 then the original number is divisible by 9. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 9 then the original number is divisible by 9; For 432: 4 + 3 + 2 = 9 which is 9 so 432 is divisible by 9 To be divisible by 10 the last digit must be a 0; the last digit of 432 is a 2 which is not 0, so 432 is not divisible by 10. 432 is divisible by 2, 3, 6 and 9, but not divisible by 5 nor 10.
To determine if 927 is divisible by a number, we need to check if it can be evenly divided by that number without leaving a remainder. 927 is divisible by 3 because the sum of its digits (9 + 2 + 7 = 18) is divisible by 3, which means 927 is also divisible by 3. However, 927 is not divisible by 2, 5, 9, or 10 because it does not end in 0 or 5 for 2 and 5, and it does not have all the factors of 9 or 10.
Since 1845 is an odd number, it can't be divisible by 2, 6 or 10. It is divisible by 3, 5 and 9.
90
There are no numbers that satisfy this. If a number is divisible by both 2 and 5, then it must also be divisible by 10.
180 is divisible by 2, 3, 4, 5, 9, and 10.
Your question is impossible to answer. Any number that is divisible by both 2 and 5 will also be divisible by 10. 30 and 60 are not divisible by 9.
The least number divisible by 2, 3, 4, 5, 6, 9, and 10 is 180.
All numbers divisible by 90 are also divisible by 9 and 10
90
No. 189 is only evenly divisible by 3 and 9 (from the set provided). Using the following rules of divisibility on the number 189: Divisible by 2? No - the number is not even Divisible by 3? Yes - the sum of the digits (1 + 8 + 9 = 18) is divisible by 3 Divisible by 4? No - the last two digits are not evenly divisible by 4 Divisible by 5? No - the last digit is not a 0 or a 5 Divisible by 6? No - the number is not even Divisible by 9? Yes - the sum of the digits is divisible by 9 Divisible by 10? No - the number is not divisible by 2 or 5
To be divisible by 2 the number must be even, that is its last digit must be 2, 4, 6, 8, or 0; the last digit of 432 is 2 which is one of {2, 4, 6, 8, 0} so 432 is even and divisible by 2. To be divisible by 3 sum the digits of the number; if this sum is divisible by 3 then the original number is divisible by 3. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 3, 6, or 9 then the original number is divisible by 3; For 432: 4 + 3 + 2 = 9 which is one of {3, 6, 9} so 432 is divisible by 3. To be divisible by 5 the last digit must be a 0 or 5; the last digit of 432 is a 2 which is not 0 nor 5, so 432 is not divisible by 5. To be divisible by 6, the number must be divisible by both 2 and 3; these have been tested above and found to be true, so 432 is divisible by 6. To be divisible by 9 sum the digits of the number; if this sum is divisible by 9 then the original number is divisible by 9. The test can be repeated on the sum until a single digit remains, in which case if this single digit is 9 then the original number is divisible by 9; For 432: 4 + 3 + 2 = 9 which is 9 so 432 is divisible by 9 To be divisible by 10 the last digit must be a 0; the last digit of 432 is a 2 which is not 0, so 432 is not divisible by 10. 432 is divisible by 2, 3, 6 and 9, but not divisible by 5 nor 10.
To determine if 927 is divisible by a number, we need to check if it can be evenly divided by that number without leaving a remainder. 927 is divisible by 3 because the sum of its digits (9 + 2 + 7 = 18) is divisible by 3, which means 927 is also divisible by 3. However, 927 is not divisible by 2, 5, 9, or 10 because it does not end in 0 or 5 for 2 and 5, and it does not have all the factors of 9 or 10.
No, a number can only be divisible by a number smaller than it
Since 1845 is an odd number, it can't be divisible by 2, 6 or 10. It is divisible by 3, 5 and 9.