0
$\begingroup$

I would like to know the divisibility tests for 13, 16, 17, 19. I also would appreciate the proof for the divisibility test done. Please oblige!

Rgds Jayanth

$\endgroup$
2
  • $\begingroup$ The only useful test is the one for 13 (which also covers 7 and 11): form the alternating sum of blocks of 3 from right to left: 2911272 -> 2-911+272=-637 and 637 is 13 times 49. $\endgroup$
    – almagest
    Apr 6, 2016 at 11:15
  • $\begingroup$ See math.stackexchange.com/questions/328562/… $\endgroup$ Apr 6, 2016 at 11:18

1 Answer 1

0
$\begingroup$

Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary. Example: $50661\rightarrow5066+4=5070\rightarrow507+0=507\rightarrow50+28=78$ and $78$ is $6\times13$, so $50661$ is divisible by $13$.

Test for divisibility by 17. Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary. Example: $3978\rightarrow397-5\times8=357\rightarrow35-5\times7=0.$ So $3978$ is divisible by $17$.

Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary. EG: $101156\rightarrow 10115+2\times6=10127\rightarrow1012+2\times7=1026\rightarrow102+2\times6=114$ and $114=6\times19$, so $101156$ is divisible by 19.

NOTE: I found it from some blog.

$\endgroup$
3
  • $\begingroup$ Some error with the math $\endgroup$ Apr 6, 2016 at 11:27
  • $\begingroup$ Last paragraph. $\endgroup$ Apr 6, 2016 at 11:27
  • $\begingroup$ I mean syntax for latex $\endgroup$ Apr 6, 2016 at 11:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .