Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Divisbility by 7 [#permalink]
15 Apr 2013, 13:10

1

This post received KUDOS

686 Subtract twice the last digit (2*6) from the # formed by the remaining digits(68) 68-12=56 If the diff is divisible by 7, the original # is a multiple of 7. 686-56=630 630 is divisible by 7 => 686 is divisible by 7.

I have never heard of this rule before. Hope I got it right (it should be right I tesed other numbers) _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Divisbility by 7 [#permalink]
15 Apr 2013, 13:19

2

This post received KUDOS

Hi

Lets break down the process in steps for easier understanding and throw couple of examples

1. Take the units digits or last digit in the number. Lets use 441 last digit is 1 and for 354 it is 4 ) 2. Double the last digits which will be 2 and 8 3. Subtract the number you doubled from rest of the digits which means ( 44-2 = 42 and 35 - 8 = 27) 4. Check if the result in Step 3 is divisible by 7. ( 42 divisbile by 7 - Yes (7 X 6 =42 then 441 divisible by 7. Similarly 27 not divisible by 7 so 354 is NOT divisible by 7) 5. Fist pump if you got the concept

Re: Divisbility by 7 [#permalink]
15 Apr 2013, 13:26

1

This post received KUDOS

Expert's post

Zarrolou wrote:

686 Subtract twice the last digit (2*6) from the # formed by the remaining digits(68) 68-12=56 If the diff is divisible by 7, the original # is a multiple of 7. 686-56=630 630 is divisible by 7 => 686 is divisible by 7.

I have never heard of this rule before. Hope I got it right (it should be right I tesed other numbers)

This is the correct illustration of the "divisible by 7 shortcut". Nicely done.

Of course, it's hard to remember many obscure rules that will likely never become relevant on the GMAT, so take these rules with a grain of salt. The best strategy is still to logically and systematically drop multiples until you get to the right answer. For 686, you might be able to see that it's 14 away from 100, so this is 7x98. Otherwise drop multiples of 7, 70, 700, etc. 700 is too big, so we can drop 70x9=630, leaving us with 56 (or 7x8).

This technique is good because it works for all numbers quickly. The only ones you should definitely know are 2, 3, 5 and 10. All others may have simple shortcuts but being able to drop multiples quickly to determine divisibility is the most flexible strategy.

Re: Divisbility by 7 [#permalink]
15 Apr 2013, 13:29

Zarrolou wrote:

686 Subtract twice the last digit (2*6) from the # formed by the remaining digits(68) 68-12=56 If the diff is divisible by 7, the original # is a multiple of 7. 686-56=630 630 is divisible by 7 => 686 is divisible by 7.

I have never heard of this rule before. Hope I got it right (it should be right I tesed other numbers)

HulkSmashGMAT wrote:

Hi

Lets break down the process in steps for easier understanding and throw couple of examples

1. Take the units digits or last digit in the number. Lets use 441 last digit is 1 and for 354 it is 4 ) 2. Double the last digits which will be 2 and 8 3. Subtract the number you doubled from rest of the digits which means ( 44-2 = 42 and 35 - 8 = 27) 4. Check if the result in Step 3 is divisible by 7. ( 42 divisbile by 7 - Yes (7 X 6 =42 then 441 divisible by 7. Similarly 27 not divisible by 7 so 354 is NOT divisible by 7) 5. Fist pump if you got the concept

Cheers!

- Hulk

First of all thank you for the kind cooperation. In fact , I tried the rule with 63 and 56 it did not work , but it worked on trying it with 49 . This what making me puzzled _________________

Re: Divisbility by 7 [#permalink]
15 Apr 2013, 13:37

VeritasPrepRon wrote:

Zarrolou wrote:

686 Subtract twice the last digit (2*6) from the # formed by the remaining digits(68) 68-12=56 If the diff is divisible by 7, the original # is a multiple of 7. 686-56=630 630 is divisible by 7 => 686 is divisible by 7.

I have never heard of this rule before. Hope I got it right (it should be right I tesed other numbers)

This is the correct illustration of the "divisible by 7 shortcut". Nicely done.

Of course, it's hard to remember many obscure rules that will likely never become relevant on the GMAT, so take these rules with a grain of salt. The best strategy is still to logically and systematically drop multiples until you get to the right answer. For 686, you might be able to see that it's 14 away from 100, so this is 7x98. Otherwise drop multiples of 7, 70, 700, etc. 700 is too big, so we can drop 70x9=630, leaving us with 56 (or 7x8).

This technique is good because it works for all numbers quickly. The only ones you should definitely know are 2, 3, 5 and 10. All others may have simple shortcuts but being able to drop multiples quickly to determine divisibility is the most flexible strategy.

Hope this helps! -Ron

Kudos for the great explanation .what about 56 , and 63 ? this rule did not work _________________

Re: Divisbility by 7 [#permalink]
15 Apr 2013, 13:43

1

This post received KUDOS

Expert's post

Zarrolou wrote:

56 6*2=12 5-12=-7 -\frac{7}{7}=-1

63 3*2=6 6-6=0 \frac{0}{7}=0

It works, both are divisible by 7

Exactly. They still work if you consider negative numbers or 0. I guess the base case would be 7. We'd get 0 - (2x7) = -14, which is in fact divisible by 7.. but it would be a little silly to use this rule to check if 7 is divisible by 7

I'd say any 2 digit number should be pretty easy to do. The multiplication table takes you to 7x12 = 84. After that there's 91 and 98 and then triple digits. Feel free to use this for triple digits upwards, although again it's never the only way to check for divisibility.

To stay away from MBA application forums for a while. At least until I hear from Stanford. I’m not certain I can handle the constant anxiety that comes...

As all my good friends know, patience is not one of my strong points. And Stanford is stretching beyond what I thought was my capacity but I am handling...

Today, I participated in the Business Technology Group competition sponsored by Microsoft. With a case released just two days ago, our group was under intense time schedule to complete the...

Undoubtedly, Durham is an amazing and beautiful place. Just this morning I took a picture at one of the paths that take you from the parking lot to school. You...