Divisibility test for 7, 11, etc.

I ran across this divisibility test for 7 in an old pre-algebra textbook.  After figuring out how it works, I was able to generalize it.  Since then, I've seen some other discussions of the generalization on the web, but it deserves to be better known, I think.  The version for 11 is especially easy, but few people seem to know it.

The test for 7 goes like this (using 37499 as the number to be tested):

1. Break off the last digit from the rest of the number to create two numbers:  3749 and 9
2. Double the broken off last digit, and subtract that from the other new number:  3749-18=3731
3. If the result is divisible by 7, then the original number is.  Of course, you can test the result using the same procedure:  the numbers are 373 and 1, so 373-2=371 .... and again:  37-2=35, which is clearly divisble by 7, so the original number was.

It works like this:  the first of the two numbers created by breaking off the last digit (3749 in the example) is effectively the number in the tens column, so call it t, and call the units digit u.  So the original number is 10t+u.  The number created in step 2 is t-2u.  So the claim is that 7|(t-2u) implies 7|(10t+u).  I'm using the vertical bar to mean "divides".

Noting that multiplying a number by a non-multiple of 7 doesn't change divisibility by 7, nor does adding multiples of 7, we get:

7|(t-2u) iff
7|10(t-2u) = 10t-20u iff
7|10t-20u+21u = 10t+u which was what we wanted.

The reason you need a 2 in the process is that you need to add a multiple of 7 with a final digit of 1 in the last step of the proof.  Since that multiple is 21, you need to multiply by 2 (it's fairly obvious if you follow the 2 through the proof).  The generalization uses the same pattern for other primes with multiples ending in 1 whose first digits are manageable:

For 7, the multiple is 21, so you multiply the final digit by 2 and subtract.
For 11, the multiple is 11, so you multiply the final digit by 1 and subtract (easy!!).
For 17, the multiple is 51, so you multiply the final digit by 5 and subtract.

You can also repeat the proof with addition and subtraction reversed, requiring multiples with a final digit of 9 (and using the next higher first digit to multiply--repeating the proof makes it clear why):

For 13, the multiple is 39, so you multiply the final digit by 4 and add.
For 19, the multiple is 19, so you multiply the final digit by 2 and add.
For 29, the multiple is 29, so you multiply the final digit by 3 and add.

Other tests are possible, but the arithmetic gets unwieldy.  In these tests, with a paper and pencil to keep track of successive steps, the calculations are almost immediate, especially for 11:

testing 351,718:

351,718 (35171-8)
35,163 (3516-3)
3513 (351-3)
348 (34-8)
26....not divisible.

Ever since I've found this, I've been preaching the gospel of easy divisibility tests for awkward primes.

Comments

[Vito Prosciutto]

Very cool. The 11 test is actually a bit simpler than the version I knew (take the sum of every other digit and subtract the sum of the remaining digits and if that number is divisible by 11 then you're done.

[J.D. Fisher]

Nice generalization. Very nice work.

[Geoff]

Another quick divisibility test for 11:
Any palindromic number (i.e. a number that's written the same forwards or backwards, like 2112 or 12344321) which has an even number of digits is divisible by 11.

The converse isn't true, of course.

[Cedric]

thanks for the divisiblity of 7 been tryin to find it =)

[Jonathan]

hmm, I don't know if you look this far back in your archives. Your divisibility test for 7 is clunky for longer numbers.

Try this instead:

break the number into groups of three digits (breaks go where the commas go). Add the even groups, subtract the odd ones (or vice versa). If the result is divisible by 7, so is the number.

eg: 82,342,509,668

509 + 82 = 591
748 + 668 =1416
1416 - 591 = 825

Now is 825 divisible by 7? Apply your test, which is new to me: 82 - 2(5) = 72. Nope

[Jonathan]

Oh, and the same test works for 13.

(why? because 1001 = 7*11*13)

[Nici]

If you tests for 11 are true, it would make both 777 AND 767 divisible by 11. Me no thinkee so.

[Polymath]

Nici: i don't think your examples disprove my test.

for 777:

77 - 14 = 63, which is divisible by 7.

for 767:

76 - 14 = 62, which isn't.

[danielle]

heyy lol i fink its wierd 2 bcuz i red da comment n its false
767 dont work m8 dus it lol
rite listen, wot it is yeh, is cuz...767 take 7 x2 and yh we so u got 76 - 14 well dat makes 62 n it dont work !!! ur fake m8 i neeed da real answer ne1 got ne help ?

[Dumbassatic poker]

Umm danielle I hope to god you are joking because you just proved yourself an atard by contradiction... If the answer you get is right you must be wrong? wtf. Oh btw Carlos Mencia just nominated you for DEE DEE DEE of the year congrats!

[hd]

yepz whateva

[Nabi Bux Abro]

I want to get the knowledge about the number divisible by 13.

[peter]

i am a student strugling on a question the question is when 823519 is divided by a number which is bigger than one the remainder is three times the remainder obtained by dividing 274658 by the same number find the divisor?

[mukesh]

most easy test

[andrew]

in answer to Nabi Bux Abro's question:

1} 823519 = Ax + 3R
2} 274658 = Cx + R
A, C, R, x are integer unknowns, and 'x' is the divisor that you're looking for.

Subtract 3 times {2} from {1} to get
455 = (A-3C)x +3R - 3R
or
5*91 = (A-3C)x.
'x' is either 5 or 91. 5 does not work. 91 does, since 823519/91 has remainder 60, and 274658/91 has remainder 20.

[Joe Salami]

You stupid ham prosciutto, it's the same test, c-(b-a)=c-b+a!

[rahul]

The divisibility test of 7 is good & cool.

[toni,bit sindri,barhiya]

it is very intelligent work.

[emergency response]

I ran this divisibility test by myself and get the same results. It's amazing!

[Mad Teddy]

Check out my web page:

http://www.madteddy.com/divisibl.htm

where I present a document which explains how it is possible to generate divisibilty tests for any divisor ending in 1, 3, 7, or 9 (i.e. relatively prime to 10). Starting from the "break off the last digit, double it, and subtract" test for 7, I've presented a table for every number of this type up to 109. It would be a straightforward matter to extend it as far as you like.

(Note that the usual tests for 3, 9, and 11 are actually examples of this type of test, thinly disguised.)

[Daniel Ngila]

I must say congrats to whoever figured out the divisibility test of 7.Just for the record i did it when i was 16.No offense.

[Emma]

too hard !!!!!!!

[greta gonzales]

great!! it helps a lot to math majors students

[Silvio Moura Velho]

I created a rule for divisibility by seven, eleven and thirteen whose algorithm for divisibility by seven is this:
N = a,bcd; a' ≣ ( − cd mod 7 + a ) mod 7; cd is eliminated and if 7|a'b then 7|N. The procedure is applied from right to left repetitively till the leftmost pair of digits is reached. If the leftmost pair is incomplete consider a = 0.
Example: N = 382,536, using simple language:
36 to 42 = 6; 6 + 2 − 7 = 1 → 15; 15 to 21 = 6; 6 + 3 − 7 = 2 → 28; 7|28 and 7|N.
This rule is mentioned in my unpublished (officially registered) book: "Divisibility by 7, the end of a myth?".
This rule is applied in seconds to large numbers entirely through mental calculation.