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26 Jul 2010, 00:34
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Divisibility of Integers By Certain Integers
Here by divisibility we mean that the answer of the division is an integer with 0 remainder.
2: A number is divisible by 2 if it is an even number. Any number with 0, 2, 4, 6, 8 in the unit place is an even number.
3: A number is divisible by 3 if the sum of it's digits is divisible by 3. For example, 132 is divisible by 3 because "1+3+2=6" is divisible by 3.
4: A number is divisible by 4 if it is at least twice divisible by 2 or it's last two digits are divisible by 4. For example, 124 is divisible by 4 because 24 is divisible by 4.
5: A number whose unit digit is 0 or 5 is divisible by 5.
6: A number which is divisible by both 2 (see the point above) and 3 ( see the point above) is divisible by 6.
7: To check the divisibility of a number with 7, we need to do long division. There is no short cut as such.
8: A number is divisible by 8 if it is at least three times divisible by 2 or it's last 3 digits (in case of big numbers) are divisible by 8. 5856 is divisible by 8 because "856" is divisible by 8.
9: A number is divisible by 9 if the sum of the digits is divisible by 9. For example, 8361 is divisible by 9 because "8+3+6+1 = 18" is divisible by 9.
10: A number with zero in the digit place is divisible by 10.
I hope this helps.
Regards
Farukh
Here by divisibility we mean that the answer of the division is an integer with 0 remainder.
2: A number is divisible by 2 if it is an even number. Any number with 0, 2, 4, 6, 8 in the unit place is an even number.
3: A number is divisible by 3 if the sum of it's digits is divisible by 3. For example, 132 is divisible by 3 because "1+3+2=6" is divisible by 3.
4: A number is divisible by 4 if it is at least twice divisible by 2 or it's last two digits are divisible by 4. For example, 124 is divisible by 4 because 24 is divisible by 4.
5: A number whose unit digit is 0 or 5 is divisible by 5.
6: A number which is divisible by both 2 (see the point above) and 3 ( see the point above) is divisible by 6.
7: To check the divisibility of a number with 7, we need to do long division. There is no short cut as such.
8: A number is divisible by 8 if it is at least three times divisible by 2 or it's last 3 digits (in case of big numbers) are divisible by 8. 5856 is divisible by 8 because "856" is divisible by 8.
9: A number is divisible by 9 if the sum of the digits is divisible by 9. For example, 8361 is divisible by 9 because "8+3+6+1 = 18" is divisible by 9.
10: A number with zero in the digit place is divisible by 10.
I hope this helps.
Regards
Farukh
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26 Jul 2010, 01:41
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Hi,
There is a way to check the divisibility by 7.
Check the below:
Divisibility by 7, 11, and 13
Let a number be ....kjihgfedcba where a, b, c, d, are respectively units digits, tens digits, hundreds digits, thousands digits and so on. Starting from right to left, we make groups of three digit numbers successively and continue till the end. It is not necessary that the leftmost group has three digits.
Grouping of the above number in groups of three, from right to left, is done in the following manner kj,ihg,fed,cba
We add the alternate groups (1 st , 3 rd , 5th etc.. and 2nd , 4th , 6th , etc..) to obtain two sets of numbers, N 1 and N 2 .
In the above example, N 1 = cba + ihg and N 2 = fed + kj
Let D be difference of two numbers, N 1 and N 2 i.e. D = N 1 - N 2 .
- If D is divisible by 7, then the original number is divisible by 7.
- If D is divisible by 11, then the original number is divisible by 11
- If D is divisible by 13 then the original number is divisible by 13.
Regards,
Jack
There is a way to check the divisibility by 7.
Check the below:
Divisibility by 7, 11, and 13
Let a number be ....kjihgfedcba where a, b, c, d, are respectively units digits, tens digits, hundreds digits, thousands digits and so on. Starting from right to left, we make groups of three digit numbers successively and continue till the end. It is not necessary that the leftmost group has three digits.
Grouping of the above number in groups of three, from right to left, is done in the following manner kj,ihg,fed,cba
We add the alternate groups (1 st , 3 rd , 5th etc.. and 2nd , 4th , 6th , etc..) to obtain two sets of numbers, N 1 and N 2 .
In the above example, N 1 = cba + ihg and N 2 = fed + kj
Let D be difference of two numbers, N 1 and N 2 i.e. D = N 1 - N 2 .
- If D is divisible by 7, then the original number is divisible by 7.
- If D is divisible by 11, then the original number is divisible by 11
- If D is divisible by 13 then the original number is divisible by 13.
Regards,
Jack
26 Jul 2010, 01:55
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Thanks for the additional comments Jack!
It is an excellent add on!
It is an excellent add on!
