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Can you divide 2 3 5 6 9 10 into 305?
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.
What are the divisible by 3?
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.
How do you know if 436 is divisible by 3?
Add the digits. If the total is divisible by three, the original number is divisible by three. It's not and it's not.
How can you tell if a number is divisible by 12?
If a number is divisible by both three and four, it's divisible by twelve.
How do you use divisibility rules to find at least for factors of a number?
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.
