For example you have two numbers 4 and 5 and we add them the result is 9 which is also divisible by 3 if we divide it by 3 the result will be.
So it means that total/addition of all the numbers which after addition can be divisible by 3.
I am a 3 digit number divisible by 7 but not 2 the sum of my digits is 4 what number am I
No, because the sum of its digit is not divisible by 3
It is divisible by 3 but not divisible by 9. To test for divisibility by 3, sum the digits and if the sum is divisible by 3 then so is the original number; the test can be repeated ion the sum, so keep summing until a single digit remains and if this single digit is 3, 6 or 9, then the original number is divisible by 3: 5673 → 5 + 6 + 7 + 3 = 21 → 2 + 1 = 3 which is one of {3, 6, 9} so 5673 is divisible by 3. To test for divisibility by 9, sum the digits and if the sum is divisible by 9 then so is the original number; the test can be repeated ion the sum, so keep summing until a single digit remains(this single digit is known as the digital root of the number) and if this single digit is 9, then the original number is divisible by 9: 5673 → 5 + 6 + 7 + 3 = 21 → 2 + 1 = 3 which is not 9 so 5673 is not divisible by 9.
Add the digits of the number together and if the sum is divisible by 3 then the original number is divisible by 3. The test can be applied to the sum and so the summation can be repeated until a single digit remains; if this digit is 3, 6 or 9 then the original number is divisible by 3.
yes 552 is divisible by 6 and the result is 92... to find out if a number is divisible by 6 then that number must be divisible by 2 and 3.. 1.If a number is divisble by three then the last digit must be divisible by 2 (so that the last digit must be 0,2,4,6,8) in this number the last digit is 2 so that it is divisible by 2 2.If a number is divisble by 3 then the sum of its digit must be divisible by 3. In this case the sum of the digits of 552 is 5+5+2= 12 and 12 is divisible by 3.... If this two rules fit in then that number is divisible by 6
I am a 3 digit number divisible by 7 but not 2 the sum of my digits is 4 what number am I
No, because the sum of its digit is not divisible by 3
3 and 9. 93 has a digit sum of 12, initially, which is divisible by 3, but not by 9. So 93 is divisible by 3, but not by 9. 99 has a digit sum of 18, initially, which is divisible by 3 and 9. So 99 is divisible by both 3 and 9.
301, if you mean that it's divisible by 7 and its DIGITS add up to 4
Yes, because the sum of its digit is divisible by 3
It is divisible by 3 but not divisible by 9. To test for divisibility by 3, sum the digits and if the sum is divisible by 3 then so is the original number; the test can be repeated ion the sum, so keep summing until a single digit remains and if this single digit is 3, 6 or 9, then the original number is divisible by 3: 5673 → 5 + 6 + 7 + 3 = 21 → 2 + 1 = 3 which is one of {3, 6, 9} so 5673 is divisible by 3. To test for divisibility by 9, sum the digits and if the sum is divisible by 9 then so is the original number; the test can be repeated ion the sum, so keep summing until a single digit remains(this single digit is known as the digital root of the number) and if this single digit is 9, then the original number is divisible by 9: 5673 → 5 + 6 + 7 + 3 = 21 → 2 + 1 = 3 which is not 9 so 5673 is 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 26 is 6 which is one of {2, 4, 6, 8, 0} so 26 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 26: 2 + 6 = 8 which is not one of {3, 6, 9} so 26 is not divisible by 3. To be divisible by 5 the last digit must be a 0 or 5; the last digit of 26 is a 6 which is not 0 nor 5, so 26 is not divisible by 5. 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 26: 2 + 6 = 8 which is not 9 so 26 is not divisible by 9. 26 is divisible by 2 but not divisible by 3, 5 nor 9.
Add the digits of the number together and if the sum is divisible by 3 then the original number is divisible by 3. The test can be applied to the sum and so the summation can be repeated until a single digit remains; if this digit is 3, 6 or 9 then the original number is divisible by 3.
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.
129.129.129.129.
No. To check divisibility by 3 add the digits together and if the sum is divisible by 3, then so is the original number. If the digits of the sum are summed and this is repeated until a single digit remains, then only if the digit is 3, 6 or 9 is the original number divisible by 3. 341 → 3 + 4 + 1 = 8 which is not divisible by 3 (not one of 3, 6 or 9), so 341 is not divisible by 3.
721