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

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Q: Which divisibility rule would you apply to tell whether a number is divisible by 3?

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Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule repeatedly as necessary. Here is an example. You want to know if 50661 is divisible by 13. Add 4 times the last digit to the remaining numbers you have after removing the last digit. So we add 1x4=4 to the number we have after taking the 1 after 50661. This is 5066+4=5070. Now do this again. 4x0 is 0 and add this to the number you get by taking 5070 and truncating the number.. ie drop the last 0. So we have 507+0=507. Now add 7x4=28 so 50. Of course 50 comes from 507 after dropping the 7. You have 78. Now since 78 is 6x13, the original number must have been divisible by 13 too. This is one of the more complicated divisibility rules.

For any practical purpose, it is easier to simply divide, instead of looking for fancy divisibility rules. However, you can apply the divisibility rules for 3 and for 7. This works because (a) their product is 21, and (b) these numbers are relatively prime.

If a number is divisible by six, then it must also be divisible by both two and three. To check it then, you simply apply both rules for each of those: Is it an even number? Is the sum of it's digits divisible by three? If both answers are yes, then the number is divisible by six. To find out if a number is divisible by seven take it's last digit, double it, and subtract it from the remaining digits. If the result is divisible by seven, then the original number is as well. For example is three 343 divisible by 7? Let's find out: 3 * 2 = 6, and 34 - 6 = 28. Is 28 divisible by seven? Yes, but if you're not sure, you can repeat the process. 8 * 2 = 16, 2 - 16 = -14. 14 is of course divisible by 7. If the last three digits of a number are divisible by 8, then the entire number is. For example, I know that 10923485710234985723908471859256 is divisible by eight, because I know that the last three digits, 256, form a number that's divisible by eight. Not sure about those last three digits? Simply divide them by two, three times in a row. If the result is a whole number, then it's divisible by eight. 256 / 2 = 128, 128 / 2 = 64, 64 / 2 = 32, so 256 is divisible by eight, and therefore 10923485710234985723908471859256 is also. If the sum of the digits is divisible by 9, then the number itself is. For example, is 8936523 divisible by 9? Well, 8 + 9 + 3 + 6 + 5 + 2 + 3 = 36. Is 36 divisible by 9? 3 + 6 = 9. 9 is obviously divisible by 9, so yes, 8936523 is also. If the last digit is a zero, then the number is divisible by 10. For example, 12340 is divisible by ten, but 12345 is not.

Check the sum of the digits, if the sum divisible by 3 then the number is divisible by 3. Example 910: the sum of digits= 9 + 1 + 0 = 10, but 10 is not divisible by 3, so the number 910 is not divisible by 3. Example 2154: the sum of digits = 2 + 1 + 5 + 4 = 12, this sum is divisible by 3, so the number 2154 is divisible by 3. if the sum is long you can check the sum of the sum and apply the same rule. Example 52498731: the sum of digits = 5 + 2 + 4 + 9 + 8 + 7 + 3 + 1 = 39, the sum of the digits for 39 = 3 + 9 = 12, so the original number, i.e. 52498731 is divisible by 3.

Apply for the rules of 3 and 8. Numbers are divisible by 24 only if they are divisible by both 3 and 8.

Related questions

Since 1992 is divisible by 4, it is a leap year - meaning it has a February 29.(About divisibility by 4, special rules apply at the end of each century.)Since 1992 is divisible by 4, it is a leap year - meaning it has a February 29.(About divisibility by 4, special rules apply at the end of each century.)Since 1992 is divisible by 4, it is a leap year - meaning it has a February 29.(About divisibility by 4, special rules apply at the end of each century.)Since 1992 is divisible by 4, it is a leap year - meaning it has a February 29.(About divisibility by 4, special rules apply at the end of each century.)

yes because you have 3 divided by 56

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again. Source~http://mathforum.org/k12/mathtips/division.tips.html But with incrediby large numbers you're just better off just dividing to see.

14 is difficult as it is 2 x 7 and 7 is difficult.There are few, if any, divisibility rules which are quicker than dividing by 14, but one I can offer:If the number is odd, it is not divisible by 14, otherwise apply the following test for divisibility by 7:split the number into blocks of three digits from the right hand end (like normally writing a large number) - if a block has less than 3 digits, put leading zeros to make it 3 digitsStarting from the right hand end alternatively subtract and add the first digits of each block together, the second digits of each block together and the third digits of each block together.Add twice the above sum of first digits to three times the sum of the second digits to the sum of the third digitsIf the result of step 3 is divisible by 7, the original number is divisible by 7.If the number is both even (divisible by 2) and divisible by 7 it will be divisible by 14.Examples:563361Odd (not divisible by 2), so not divisible by 14. 74024562936Even (divisible by 2), so apply 7 test: 74024562936 → 074 024 562 936→ 2 x (9 - 5 + 0 - 0) + 3 x (3 - 6 + 2 - 7) + (6 - 2 + 4 - 4) = 2 x 4 + 3 x -8 + 4 = -12: not divisible by 7⇒ 74024562936 is not divisible by 14123789456Even (divisible by 2), so apply 7 test: 123789456 → 123 789 456→ 2 x (4 - 7 + 1) + 3 x (5 - 8 + 2) + (6 - 9 + 3) = 2 x -2 + 3 x -1 + 0 = -7: divisible by 7⇒ 123789456 is divisible by 14

To find out, if a number is divisible by 7, take the last digit, double it, and subtract it from the rest of the number. If you get an answer divisible by 7 (including zero), then the original number is divisible by 7.If you don't know the new number's divisibility, you can apply the rule again.Eg: Take the number 343The units digit is 3.Doubling this 3,we get 6.Subtracting 3 from 34(that is the remaining 2 digits in the number), we get 34-6=28.As 28 is divisible by 7, The whole number, i.e., 343 is divisible by 7.Proof of Divisibilty Rule for 7Let 'D' ( > 10 ) be the dividend.Let D1 be the units' digitand D2 be the rest of the number of D.i.e. D = D1 + 10D2We have to prove(i) if D2 - 2D1 is divisible by 7,then D is also divisible by 7and (ii) if D is divisible by 7,then D2 - 2D1 is also divisible by 7.Proof of (i) :D2 - 2D1 is divisible by 7 ⇒ D2 - 2D1 = 7k where k is any natural number.Multiplying both sides by 10, we get10D2 - 20D1 = 70kAdding D1 to both sides, we get(10D2 + D1) - 20D1 = 70k + D1⇒ (10D2 + D1) = 70k + D1 + 20D1⇒ D = 70k + 21D1 = 7(10k + 3D1) = a multiple of 7.⇒ D is divisible by 7. (proved.)Proof of (ii) :D is divisible by 7 ⇒ D1 + 10D2 is divisible by 7 ⇒ D1 + 10D2 = 7k where k is any natural number.Subtracting 21D1 from both sides, we get10D2 - 20D1 = 7k - 21D1⇒ 10(D2 - 2D1) = 7(k - 3D1)⇒ 10(D2 - 2D1) is divisible by 7Since 10 is not divisible by 7,(D2 - 2D1) is divisible by 7. (proved.)

28 is divisible by...1, because everything is.2, because it's even.4, because the last two digits are divisible by 4.Divide them into 28.All Factors of 28:1, 2, 4, 7, 14, 28

Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule repeatedly as necessary. Here is an example. You want to know if 50661 is divisible by 13. Add 4 times the last digit to the remaining numbers you have after removing the last digit. So we add 1x4=4 to the number we have after taking the 1 after 50661. This is 5066+4=5070. Now do this again. 4x0 is 0 and add this to the number you get by taking 5070 and truncating the number.. ie drop the last 0. So we have 507+0=507. Now add 7x4=28 so 50. Of course 50 comes from 507 after dropping the 7. You have 78. Now since 78 is 6x13, the original number must have been divisible by 13 too. This is one of the more complicated divisibility rules.

For any practical purpose, it is easier to simply divide, instead of looking for fancy divisibility rules. However, you can apply the divisibility rules for 3 and for 7. This works because (a) their product is 21, and (b) these numbers are relatively prime.

If a number is divisible by six, then it must also be divisible by both two and three. To check it then, you simply apply both rules for each of those: Is it an even number? Is the sum of it's digits divisible by three? If both answers are yes, then the number is divisible by six. To find out if a number is divisible by seven take it's last digit, double it, and subtract it from the remaining digits. If the result is divisible by seven, then the original number is as well. For example is three 343 divisible by 7? Let's find out: 3 * 2 = 6, and 34 - 6 = 28. Is 28 divisible by seven? Yes, but if you're not sure, you can repeat the process. 8 * 2 = 16, 2 - 16 = -14. 14 is of course divisible by 7. If the last three digits of a number are divisible by 8, then the entire number is. For example, I know that 10923485710234985723908471859256 is divisible by eight, because I know that the last three digits, 256, form a number that's divisible by eight. Not sure about those last three digits? Simply divide them by two, three times in a row. If the result is a whole number, then it's divisible by eight. 256 / 2 = 128, 128 / 2 = 64, 64 / 2 = 32, so 256 is divisible by eight, and therefore 10923485710234985723908471859256 is also. If the sum of the digits is divisible by 9, then the number itself is. For example, is 8936523 divisible by 9? Well, 8 + 9 + 3 + 6 + 5 + 2 + 3 = 36. Is 36 divisible by 9? 3 + 6 = 9. 9 is obviously divisible by 9, so yes, 8936523 is also. If the last digit is a zero, then the number is divisible by 10. For example, 12340 is divisible by ten, but 12345 is not.

Of course you can apply. Whether or not it will be granted depends on a number of factors.

Check the sum of the digits, if the sum divisible by 3 then the number is divisible by 3. Example 910: the sum of digits= 9 + 1 + 0 = 10, but 10 is not divisible by 3, so the number 910 is not divisible by 3. Example 2154: the sum of digits = 2 + 1 + 5 + 4 = 12, this sum is divisible by 3, so the number 2154 is divisible by 3. if the sum is long you can check the sum of the sum and apply the same rule. Example 52498731: the sum of digits = 5 + 2 + 4 + 9 + 8 + 7 + 3 + 1 = 39, the sum of the digits for 39 = 3 + 9 = 12, so the original number, i.e. 52498731 is divisible by 3.

The same rules of divisibility apply for large numbers as well as small ones. Divide by two or three a couple of times if you're able, and the number might become more manageable.