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Simple and quick divisibility test for 7 [#permalink]

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06 Feb 2013, 00:27

Here is something that I found while scratching my head and looking for the rules of divisibility for the number -7.

Let's do it by example a) 2695 -Multiply the number of hundreds the number has by 2. 2695 = 2600 + 95 So, 26*2 = 52

-Add the remaining two digit number. Here the remaining two digit number is 95 Hence, 52+95 = 147

- Check whether the resulting sum is divisible by 7. if yes, then the original number is divisible by 7. 147/7 = integer Therefore, 2695 is divisible by 7

Another example: b) 26957

Step 1 => 26900 + 57 Therefore, 269*2 = 538

Step 2 => 538 + 57 = 595

Step 3 => Check if 595/7 is an integer. Indeed, 595/7 = 85 Thus, 26957 is divisible by 7.

This in my opinion is the fastest way to check the divisibility of a number by 7.

I'd be glad to see some appreciation in the form of Kudos. _________________

Actually looks a little tedious. 269*2, 595/7 , 538+57 one of these may be the exact point where things may go wrong.

I follow a similar stratgey but with simple calculations involving 0s. It may or may not help others. Take out huge chunks from calculations that are multiple of 7. get help of 0s as much as possible

Such as, 28957 28000 is multiple of 7, so need to check only for 957 700 is multiple of 7, so need to check only for 257 210 is multiple of 7, so need to check only for 47 and we know the ans.

Actually looks a little tedious. 269*2, 595/7 , 538+57 one of these may be the exact point where things may go wrong.

I follow a similar stratgey but with simple calculations involving 0s. It may or may not help others. Take out huge chunks from calculations that are multiple of 7. get help of 0s as much as possible

Such as, 28957 28000 is multiple of 7, so need to check only for 957 700 is multiple of 7, so need to check only for 257 210 is multiple of 7, so need to check only for 47 and we know the ans.

Hope it helps

You changed the number from 26957 to 28957, was it intentional? When the thousandth digits are multiple of 7 then, I'd go with your suggestion. I like it. However, when its not, as is the case here (26957), What will you do? 21000+5957 ? => 5600+357 => 350 + 7 => 7...!

Hmmm, it may take same amount of time. I guess its a matter of preference. However, I like the suggestion. Will keep in mind.
_________________

Re: Simple and quick divisibility test for 7 [#permalink]

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06 Feb 2013, 01:02

Vishwa25 wrote:

You changed the number from 26957 to 28957, was it intentional? When the thousandth digits are multiple of 7 then, I'd go with your suggestion. I like it. However, when its not, as is the case here (26957), What will you do? 21000+5957 ? => 5600+357 => 350 + 7 => 7...!

Hmmm, it may take same amount of time. I guess its a matter of preference. However, I like the suggestion. Will keep in mind.

Intentional, to show the calculations.

Quote:

However, when its not, as is the case here (26957), What will you do?

you can take your number: 26957 21000 is multiple of 7, so need to check only for 5957 5600 is multiple of 7, so need to check only for 357 350 is multiple of 7, so need to check only for 7 and we know the ans.
_________________

Re: Simple and quick divisibility test for 7 [#permalink]

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31 Aug 2013, 11:06

Actually all the above methods were cool but lenthy(ingleesh prowblem ) i derived something which is really easy and sort if you memorize given numbers....

I found a method which if you keep in mind it will make little faster..... keep in mind these number right digit->left digit 1->2(i.e double) 2->4(i.e double) 3->6(i.e double) 4->1(i.e 1 greater than 3 so 4-3=1) 5->2(i.e 2 greater than 3 so 5-3=2) 6->5(i just muggup this number )

how you going to memorize these is up to you now if the unit digit is one of them the right handside digit should be according to that else subtract it. here is the example 1) to check 7203 the right hand digit is 3 so the left hand side digit must be 6 but here it is not so subtract 6 from 0 the remaining number(i.e 714) is divisible by 7 let's check it the last digit is now 4 so the previous digit should be 1 and it is now remains 7 which is already divisible by 7

2) to check 509796 the right hand digit is 6 to the left hand side digit must be 5 but it is 9 (oh i forgot to say if the digit is greater than 7 you can also subtract 7 from it so 9 became 2) now subtract 5 from it, it remains 4. now the left hand side number is divisible by 7, let's check now th number(which is actually 50974) has it's last digit 4 so the previous digit should be 1 but here it's not so subtract 1 from 7 it became 6 now the previous digit should be 5 but it is 9 so subtract 5 from 9 now the number(which is actually 504) has it's last digit 4 so the previous digit should be 1 but here it's again not so subtract 1 from 0 which gives 49 which is divisible by 7

Actually it's very easy if you remember the Process/Method

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