The rule for divisibility is carried out by:
- Look at the last 3 digits (right-most),
- Check if they are divisible by 8,
- If yes, then the original number is divisible by 8.
This concept is similar to that of the divisibility by 4.
Concept:
Taking 1 four-digit number abcd. Re-writting it ==> 1000a + 100b + 10c + d
We see that the 1000a digit is of no concern to the divisibility test as it is definitely divisible by 8 since 1000 / 8 = 125. Even if it is 2000, 30000,...it is still divisible.
This leave us with the only the last 3 digits to decide whether the original number is divisible by 8. This concept is surely less complex than the Divisibility by 7, right?
For more information about the concept for Divisibility by 7, please click here.
To view concept for Divisibility by 9, please click here.
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Tuesday, 5 August 2008
Divisibility by 7 | Concept Explained
Divisibility rule is applied when the number is may be too big for computation using a calculator. But it is a good exercise to know this rule as it analyze the in-depth "story" of the integration between division and other mathematical operations. Each divisibility tells a different concept or "story".
Divisibility Rule for 7:
Why it works? The concept is explained below.
Let the number be "ab". It can be re-written as 10a + b.
We doubled the last digit"b" ==> 2b. For what?
Let's deviate a bit.
If a number is divisible by 7, doubling it will also be divisible by 7.
Example: 21 /7 = 3, doubling 21 ==> 42 / 7 = 6 (still divisible after 21 is doubled)
Therefore 10a + b ==> doubling it ==> becomes 20a + 2b.
By the rule stated at the top, we need to subtract "2b" from "a" (the rest of the digit besides the first digit), that is, a - 2b, for divisibility checking.
This means the 20a + 2b need to be negated and to be added to 21a to achieve the targetted a - 2b.
Result become, -(20a + 2b) + 21a ==> -20a -2b + 21a ==> a -2b.
Note now, the added 21a is definitely divisible by 7, therefore, leaving -(20a + 2b) as the deciding factor for divisibility consideration.
If -(20a + 2b) is divisible by 7, then it implies that the a - 2b is also divisible by 7!
Why go through such hassle of doing a - 2b when we can just double the original number, that is, (20a + 2b) of above and get the answer?
The reason is that by doubling the original number, we will get an even larger number that the calculator cannot handle or display. Therefore the rule using a - 2b has to be used since the number of digits is reduced from original. Clear?
:)
For Concept on Divisibility by:
5 and 6 Click here
3 and 4 Click here
2 Click here
11 Click here
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Divisibility Rule for 7:
- Take the last (rightmost) digit,
- double it,
- Subtract it from the rest (or vice versa),
- check whether the result is divisible by 7,
- if yes, the original is divisible by 7. (This may have to be done a few rounds)
Why it works? The concept is explained below.
Let the number be "ab". It can be re-written as 10a + b.
We doubled the last digit"b" ==> 2b. For what?
Let's deviate a bit.
If a number is divisible by 7, doubling it will also be divisible by 7.
Example: 21 /7 = 3, doubling 21 ==> 42 / 7 = 6 (still divisible after 21 is doubled)
Therefore 10a + b ==> doubling it ==> becomes 20a + 2b.
By the rule stated at the top, we need to subtract "2b" from "a" (the rest of the digit besides the first digit), that is, a - 2b, for divisibility checking.
This means the 20a + 2b need to be negated and to be added to 21a to achieve the targetted a - 2b.
Result become, -(20a + 2b) + 21a ==> -20a -2b + 21a ==> a -2b.
Note now, the added 21a is definitely divisible by 7, therefore, leaving -(20a + 2b) as the deciding factor for divisibility consideration.
If -(20a + 2b) is divisible by 7, then it implies that the a - 2b is also divisible by 7!
Why go through such hassle of doing a - 2b when we can just double the original number, that is, (20a + 2b) of above and get the answer?
The reason is that by doubling the original number, we will get an even larger number that the calculator cannot handle or display. Therefore the rule using a - 2b has to be used since the number of digits is reduced from original. Clear?
:)
For Concept on Divisibility by:
5 and 6 Click here
3 and 4 Click here
2 Click here
11 Click here
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Divisibility by 5, 6 | Concept Explained
Divisibility by 5
The Rule states: Any number ending with a "5" or "0" can always be divided by 5.
(This is so since multiples of "5" end with either a "0"or "5".)
This is one of the simplest rule in the Divisibility checking.
Concept: 5 times any integer always gives another number ending with a "0" or "5". Therefore by just simply looking at the last digit, we can determine whether the original number can be divided by 5.
Divisibility by 6
Rule: If any integer can be divided by 2 and also 3, then the number can be divided by 6.
Concept:
Since 2 and 3 are prime numbers that multiply to give a 6, these 2 numbers suffice to determine whether the original integer can be divided by 6.
For Concept on Divisibility of 7, click here.
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The Rule states: Any number ending with a "5" or "0" can always be divided by 5.
(This is so since multiples of "5" end with either a "0"or "5".)
This is one of the simplest rule in the Divisibility checking.
Concept: 5 times any integer always gives another number ending with a "0" or "5". Therefore by just simply looking at the last digit, we can determine whether the original number can be divided by 5.
Divisibility by 6
Rule: If any integer can be divided by 2 and also 3, then the number can be divided by 6.
Concept:
Since 2 and 3 are prime numbers that multiply to give a 6, these 2 numbers suffice to determine whether the original integer can be divided by 6.
For Concept on Divisibility of 7, click here.
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Divisibility by 3, 4 | Concept explained
Divisibility by 3
A number can be checked for divisibility by 3 using the following 3 steps:
1) Add up all the digits in the number.
2) Check whether the sum of all the diigts can be divided by 3
3) If the answer is Yes, then the number can be divided by 3.
Concept of the Divisibility by 3 Rule:
Let the number be represented by ABC (with "A", "B" and "C" being the digit of the number).
Therefore ABC = A(100) + B(10) + C.
We can also write the above maths expression of ABC as
ABC = A(99+1) + B(9+1) + C which leads us to
ABC = 99A+A + 9B+ B + C.
Rearranging, ABC = 99A + 9B + A + B + C.
Look at the last 3 terms. It is the addition of the individual digits "A", "B" and "C".
Look also at the leftmost 2 terms. 99A and 9B are definitely divisible by 3 since 9 = 3 x 3 and 99 = 11 x 3.
From the above observations, to know whether number ABC can be divided by 3, we only need to divide the summation of the digits (A + B+ C) by 3 to decide the outcome.
Divisibility by 4
A number can be checked for divisibility by 4 using the following steps:
1) Look at the last 2 digits (least significant) of the number,
2) Check to see if the number formed (last 2 digits) can be divided by 4, and
3) If the answer is Yes, then the original number can be divided by 4.
Concept of the Divisibility by 4 Rule:
Let the number be DEF.
And DEF = D(100) + EF.
The rule states to look at the digits EF only. How about the digit D?
D is the hundred unit. Therefore it has to be 100, 200, 300, ......
100, 200, 300 ... is definitely divisible by 4. So it does not become a concern or a factor for consideration in the divisibility rule. Therefore only the last 2 digits (EF) decides the outcome.
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A number can be checked for divisibility by 3 using the following 3 steps:
1) Add up all the digits in the number.
2) Check whether the sum of all the diigts can be divided by 3
3) If the answer is Yes, then the number can be divided by 3.
Concept of the Divisibility by 3 Rule:
Let the number be represented by ABC (with "A", "B" and "C" being the digit of the number).
Therefore ABC = A(100) + B(10) + C.
We can also write the above maths expression of ABC as
ABC = A(99+1) + B(9+1) + C which leads us to
ABC = 99A+A + 9B+ B + C.
Rearranging, ABC = 99A + 9B + A + B + C.
Look at the last 3 terms. It is the addition of the individual digits "A", "B" and "C".
Look also at the leftmost 2 terms. 99A and 9B are definitely divisible by 3 since 9 = 3 x 3 and 99 = 11 x 3.
From the above observations, to know whether number ABC can be divided by 3, we only need to divide the summation of the digits (A + B+ C) by 3 to decide the outcome.
Divisibility by 4
A number can be checked for divisibility by 4 using the following steps:
1) Look at the last 2 digits (least significant) of the number,
2) Check to see if the number formed (last 2 digits) can be divided by 4, and
3) If the answer is Yes, then the original number can be divided by 4.
Concept of the Divisibility by 4 Rule:
Let the number be DEF.
And DEF = D(100) + EF.
The rule states to look at the digits EF only. How about the digit D?
D is the hundred unit. Therefore it has to be 100, 200, 300, ......
100, 200, 300 ... is definitely divisible by 4. So it does not become a concern or a factor for consideration in the divisibility rule. Therefore only the last 2 digits (EF) decides the outcome.
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Divisibility by 2
We may come across a number and wonder if it can be divided by 2.
Checking concept:
If a number is even, then it can be divided by 2 as "2" is an even number.
For a review of the mathematical relation between odd and even number, you may visit this link for information.
If the number is odd, a remainder will occur and thus not divisible by 2.
This is the simplest among the Divisibility Rule.
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Checking concept:
If a number is even, then it can be divided by 2 as "2" is an even number.
For a review of the mathematical relation between odd and even number, you may visit this link for information.
If the number is odd, a remainder will occur and thus not divisible by 2.
This is the simplest among the Divisibility Rule.
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Can It Be Divided By 11?
To be able to decide whether a number can be divided by 11, we need to understand this multiply by 11 concept first.
Click here to review the x 11 mental concept.
It is simply to do the reverse process of the mental multiplication by 11.
Example: Can the number 352 be divided by 11?
Looking at the centre digit, we can deduce that it is the addition of the 2 extreme digits.
As such the number 352 can be divided by 11.
Example: How about 3751 ?
Again, looking at the digit 3 and 7, we can see that their subtraction produces 7-3 = 4.
For the other 2 digits of 5 and 1, their subtraction also results in 5- 1 = 4.
This means that the number 3751 can be divided by 11 (based on the mental multiplication by 11).
Example: How about 542 ?
The centre digit should be 5 - 2 = 3 to be divisible by 11.
Therefore the number 542 can be quickly deduced not able to be divided by 11.
Check other divisibility concept by clicking the link here.
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Click here to review the x 11 mental concept.
It is simply to do the reverse process of the mental multiplication by 11.
Example: Can the number 352 be divided by 11?
Looking at the centre digit, we can deduce that it is the addition of the 2 extreme digits.
As such the number 352 can be divided by 11.
Example: How about 3751 ?
Again, looking at the digit 3 and 7, we can see that their subtraction produces 7-3 = 4.
For the other 2 digits of 5 and 1, their subtraction also results in 5- 1 = 4.
This means that the number 3751 can be divided by 11 (based on the mental multiplication by 11).
Example: How about 542 ?
The centre digit should be 5 - 2 = 3 to be divisible by 11.
Therefore the number 542 can be quickly deduced not able to be divided by 11.
Check other divisibility concept by clicking the link here.
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