305 can only be [exactly] divided by 5 and not 2, 3, 6, 9 nor 10.Divisibility tests:2: NoNumber must be even (last digit divisible by 2). Last digit is 5 which is not even, so the whole number (305) not divisible by 2.3: NoAdd up the digits of the number; if this sum is divisible by 3, then the original number is divisible by 3: 3 + 0 + 5 = 8 which is not divisible by 3, so the original number (305) is not divisible by 3.5: YesLast digit is 0 or 5. Last digit is 5, so 305 is divisible by 5. 6: No6 = 2 x 3, so number must pass both tests for divisibility of 2 and 3; 305 fails tests for (both) 2 and 3 (above), so is not divisible by 6. 9: NoAdd up the digits of the number; if this sum is divisible by 9, then the original number is divisible by 9: 3 + 0 + 5 = 8 which is not divisible by 9, so the original number (305) is not divisible by 9.10: NoThe last digit of the number must be 0. Last digit is 5, so 305 is not divisible by 10.
If, when divided by 3, it leaves no remainder. Also, if you add the individual numbers up and that number is divisible by three, the larger one is as well. Example: 12345 adds up to 15. 15 is divisible by three. So is 12345.
Add the digits. If the total is divisible by three, the original number is divisible by three. It's not and it's not.
If a number is divisible by both three and four, it's divisible by twelve.
Take the number 3336. You know it's divisible by 1 because everything is. You know it's divisible by 2 because it's even. You know it's divisible by 3 because the digits add up to a multiple of 3 and you know it's divisible by 4 because the last two digits are divisible by 4. So you've found at least four factors: 1,2,3 and 4.
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.
a 3 digit number that is divisible by on is a three digit number that is a multiple of one.
108 is the smallest 3-digit number divisible by 2,3,4 & 6
300
999.
There are 300 three digit numbers that are divisible by neither 2 nor 3. There are 999 - 100 + 1 = 900 three digit numbers. 100 ÷ 2 = 50 → first three digit number divisible by 2 is 50 × 2 =100 999 ÷ 2 = 499 r 1 → last three digit number divisible by 2 is 499 × 2 = 998 → there are 499 - 50 + 1 = 450 three digit numbers divisible by 2. 100 ÷ 3 = 33 r 1 → first three digit number divisible by 3 is 34 × 3 = 102 999 ÷ 3 = 333 → last three digit number divisible by 3 is 333 × 3 = 999 → there are 333 - 34 + 1 = 300 three digit numbers divisible by 3. The lowest common multiple of 2 and 3 is 6, so these have been counted in both those divisible by 2 and those divisible by 3. 100 ÷ 6 = 16 r 4 → first three digit number divisible by 6 is 17 × 6 = 102 999 ÷ 6 = 166 r 3 → last three digit number divisible by 6 is 166 × 6 = 996 → there are 166 - 17 + 1 = 150 → there are 450 + 300 - 150 = 600 three digit numbers that are divisible by either 2 or 3 (or both). → there are 900 - 600 = 300 three digit numbers that are divisible by neither 2 nor 3.
The highest six-digit number divisible by 3, 4, and 5 is 999960.
120
030
300
300
-10