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Shortcut: Divisible by 7 ? [#permalink]
21 May 2010, 09:17

7

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We all know the rules for 2,3,4,5,6,8,9 and 10. But what is less talked about is the number 7. Here is the shortcut to see whether the number is divisible by 7 or not.

Step1: Take a number and multiply each of the digit beginning on the right hand side by 1,3,2,6,4,5 (repeat the pattern if the number is large enough)

Step 2: Add the product of the numbers. If the sum is divisible by 7 -- so is the number.

e.g. Check for number 133

3(1) + 3(3) + 1(2) = 14 --> which is divisible by 7. So, 133 is divisible by 7.

Another example: check for no. 2016

6(1) + 1(3) + 0(2) + 2(6) = 21 --> which is also divisible by 7. So, the no. 2016 is divisible by 7. _________________

Re: Shortcut: Divisible by 7 ? [#permalink]
21 May 2010, 10:16

That is totally weird. I can't be bothered to check if it's true (I assume it is)... I'd love to see the proof as to why it works. Not because I'm that big of a math geek, but because I'm sure it would boggle my mind even further. _________________

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Re: Shortcut: Divisible by 7 ? [#permalink]
21 May 2010, 10:24

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This post was BOOKMARKED

Here's another one:

1. Lop off the last digit of the number in question. For 2016, that would be 6.

2. Double that number. Double the 6 to get 12.

3. Subtract that number from the remaining digits of the number in question, as if they were a stand-alone number. So you don't subtract 12 from 2016, or from 2010. You subtract 12 from 201 to get 189.

4. Repeat steps 1-3 until you get to a number you recognize as either a multiple of 7 or not. If you get a multiple of 7 after step 3, the original number was a multiple of 7. If you get a non-multiple of 7 after step 3, the original number was not a multiple of 7.

Take 189, lop of the 9, double to get 18, and subtract 18 from the 18 we got when we took 9 off the end of 189. Result 0 is divisible by 7, so 2016 also was.

Another example: 13,587

1. Separate: 1358 and 7 2. Double the 7: 14 3. Subtract: 1358-14 = 1344 1. Separate: 134 and 4 2. Double the 4: 8 3. Subtract: 134-8 = 126 Recognize 126 as 14 less than 140, i.e. a multiple of 7.

Conclusion: 13,587 is a multiple of 7. _________________

Emily Sledge | Manhattan GMAT Instructor | St. Louis

Re: Shortcut: Divisible by 7 ? [#permalink]
21 May 2010, 10:32

dalmba wrote:

That is totally weird. I can't be bothered to check if it's true (I assume it is)... I'd love to see the proof as to why it works. Not because I'm that big of a math geek, but because I'm sure it would boggle my mind even further.

I totally agree! Both of these methods are so wacky, unlike the more intuitive tests for divisibility by other single digit numbers.

I have to wonder, wouldn't simple long division by 7 be fastest? _________________

Emily Sledge | Manhattan GMAT Instructor | St. Louis

Re: Shortcut: Divisible by 7 ? [#permalink]
21 May 2010, 10:43

I have book for my kids that gives the rules for divisibility by 7 as: Multiply the final digit by 5 and then add the answer to the number preceding it. If the answer is divisible by 7 then the whole number is divisible by 7. (the book is called Speed Math for Kids by Bill Handley)

So for your example 13587. 1358 +35=1393. 1393 is divisible by 7 (=199.) so the larger number is.

Ultimately I do not see this speeding you up at all. It is absolutely slower AND more error prone for me. I would just divide it out. Not thinking this will be very relevant on the GMAT anyway.

If you found my comments helpful, please give kudos. ( only need a few more) Thanks, Skip

Re: Shortcut: Divisible by 7 ? [#permalink]
07 Jan 2015, 02:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Shortcut: Divisible by 7 ? [#permalink]
11 Jan 2015, 09:23

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Yes, no need to learn the divisibility rule for 7 as far as the GMAT is concerned. For every single computation I have seen on the GMAT, it would be faster to just divide by 7 than resorting to these rules.

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