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# Divisbility by 7

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15 Apr 2013, 14:00
1
"Subtract twice the last digit from the # formed by the remaining digits. If the diff is divisible by 7, the original # is a multiple of 7."

I read this on Kaplan 's twitter account , and I remember reading something similar in GMAT Club's book ... could anybody please clarify more ????

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15 Apr 2013, 14:10
1
$$686$$
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)
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15 Apr 2013, 14:19
2
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
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15 Apr 2013, 14:26
1
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
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15 Apr 2013, 14:29
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
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15 Apr 2013, 14:37
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
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15 Apr 2013, 14:39
1
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
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15 Apr 2013, 14:43
1
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.

Thanks!
-Ron
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04 Jan 2018, 05:04
Wow, very interesting shortcut, I've never heard of it before. Although I think this would only save time if the original number had 3 digits. Otherwise, a simple division might be quicker.
Re: Divisbility by 7 &nbs [#permalink] 04 Jan 2018, 05:04
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