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That means the number itself is divisible by three. the sum of the digits in 144, for example, is 9. 144 divided by 3 is 48. The sum of the digits in 21 is 3. 21 divided by 3 is 7.If the number is an even number and the sum of the digits is divisible by 3, the number is also divisible by 6. 144 is even and the sum of the digits is divisible by 3, so 144 is also divisible by 6. 144/6 = 24. And finally, if the sum is divisible by 9, the number itself is also divisible by 9. 27 is an example of this.

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I am a 3 digit number divisible by 7 but not 2 the sum of my digits is 4 what number am I

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.

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 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.

Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.

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.

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

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.

301

721

A number is divisible by 3 if the sum of its digits is divisible by 3.

a 3 digit number that is divisible by on is a three digit number that is a multiple of one.

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

If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

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 516 this gives: → 5 + 1 + 6 = 12 → 1 + 2 = 3 3 is not 9 so 516 is not divisible by 9. (This single digit is known as the digital root of the number.)

No. To be divisible by 6 the number must be divisible by both 2 and 3. To be divisible by 2, the last digit of the number must be even (ie one of {0, 2, 4, 6, 8}). The last digit of 35 is 5 which is not even and so 35 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 so is the original number. As the test can be applied to the sum, repeatedly summing the digits of the sums until a single digit remains, then the original number is divisible by 3 only if this single digit is one of {3, 6, 9}. For 35 3 + 5 = 8 which is not divisible by 3 (nor is is one of {3, 6, 9}, thus 35 is not divisible by 3. As 35 is not divisible by both 2 and 3 (in fact it is divisible by neither 2 nor 3) it is not divisible by 6.

301

The way to know if a number is divisible by 3 is if its digit sum adds up to a number that is divisible by 3 (3, 6 or 9). The way to know if a number is divisible by 2 is if that number is even. That is, a number is divisible by two if the units are 0, 2, 4, 6 or 8. A number is divisible by 6 if it meets both these rules. In this case, 402 has the digit sum 6, and the units value is 2. Therefore, we can say that 402 is divisible by 6.

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.

Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.

If the number is even, it is a multiple of 2 If the sum of the digits make a number divisible by 3, the number is a multiple of 3 If the number ends in 5 or 0, the number is a multiple of 5 If the number is divisible by 2 and 3, the number is a multiple of 6 If the sum of the digits make a number divisible by 9, the number is a multiple of 9

the number 3 is divisible by 3

By 3 only.It is not divisible by 2, 4, 5, 6, 9 & 10.Division tests:2:If the number is even, ie if the last digit is even (0, 2, 4, 6, 8) it is divisible by 2. 633 is odd (last digit is 3 which is not an even digit), so 633 is not divisible by 2.3:Sum the digits; if the sum is divisible by 3, the original number is divisible by 3. (Can repeat the summing until a single digit remains; if this digit is 3, 6 or 9 (ie divisible by 3) then the original number is divisible by 3.)6 + 6 + 3 = 151 + 5 = 66 is divisible by 3, so 663 is divisible by 3.4:Add the last (units) digit to twice the previous (tens) digit; if this sum is divisible by 4, so is the original number. (Can repeat summing until a single digit remains; if this digit is 4 or 8 (ie divisible by 4) then the original number is divisible by 4.)6 x 2 + 3 = 151 x 2 + 5 = 77 is not divisible by 4, so 663 is not divisible by 4.Note: all multiples of 4 are even; 633 is odd so it cannot be divisible by 4 (the above test does not need to be done).5:The last digit of the number must be 0 or 5. 3 is not 0 nor 5, so 663 is not divisible by 5.6:Number must pass the 2 and 3 tests (above). Fails 2 test (above), so 663 is not divisible by 6.9:Sum the digits; if the sum is divisible by 9, the original number is divisible by 9 (Can repeat the summing until a single digit remains; if this digit is 9 (ie divisible by 9) then the original number is divisible by 9)6 + 6 + 3 = 151 + 5 = 66 is not 9, so 663 is not divisible by 9.10:Last digit must be 0. 3 is not 0, so 663 is not divisible by10.

To be divisible by 6, the number must be divisible by both 2 and 3:To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};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.If the number is not divisible by 2 or 3 (or both) then the number is not divisible by 6.examples:126Last digit is even so it is divisible by 2 1 + 2 + 6 = 9 which is divisible by 3, so it is divisible by 3â†’ 126 is divisible by both 2 and 3, so it is divisible by 6124Last digit is even so it is divisible by 2 1 + 2 + 4 = 7 which is not divisible by 3, so it is not divisible by 3â†’ 126 is divisible by 2 but not divisible by 3, so it is not divisible by 6123Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 3 = 6 which is divisible by 3, so it is divisible by 3â†’ 123 is divisible by 3 but not divisible by 2, so it is not divisible by 6121Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 1 = 4 which is not divisible by 3, so it is not divisible by 3â†’ 121 is not divisible by either 2 or 3, so it is not divisible by 6

531 is one of them and that any 3 digit number whose digital sum is 9 is also divisible by 9