A prank I recommend to readers is to use the number 91 when a group situation calls for a random prime number. If done subtly enough, a decent portion of mathematicians will believe you. Granted, its not a particularly *funny* prank…

So why 91? Look at the main ways to quickly check primality of two digit numbers:

For these, 91 is the only composite number that all of them miss (being 7 times 13). Most mathematicians run those checks in their head, or (more likely) they are pretty good at spotting immediately numbers which fail the above checks. Anything that passes these checks at least ‘smells’ prime.

This fact is also useful for something other than lame pranks. It means that you can know for sure which two digit numbers are prime, simply by adding one last check to the list:

Its not super useful, but it does cut down on the amount of time it takes to check numbers like 83. It smells prime, and its not 91, so its prime! Its like the quickest Sieve method imaginable.

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February 11, 2008 at 11:51 am |

Check for 7s, A little known by fairly useful divisibility test: Double the 1s digit and subtract the result from the previous digits: 9- 2(1)=7. If the result is divisible by 7, then the number is divisible by 7.

Two more examples, to check 343 (yeah I know it is 7x7x7), 34-6-28. Meanwhile, 2- 2(8)=-14.

2401 = (7x7x7x7):

240-2(1)=238 = 7×34.

23 -2(8)=23-16=7.

For big (between 100-9800) that end in 1s 3s or 7s,

when other divisibility tests immediately fail, check for numbers ending in 3, 7, or 9 that are near the square roots. Notice this works for 91 as well.

February 11, 2008 at 11:53 am |

GRRH typo: 34-6=28.

February 11, 2008 at 3:18 pm |

@ carter

do you have a proof for that? if so, its a fairly interesting divisibility property, and I wonder if it could be generalized.

February 11, 2008 at 4:58 pm |

Great post! I, too, have enjoyed the fact that 91 “smells” prime, but isn’t. However, I’ve never considered using that fact for the underhanded motives of fooling others. Now I will.

A couple comments mentioned divisibility tests. I’d like to point out a paper of mine, “Stupid Divisibility Tricks,” available at my website. I hope you enjoy!

February 11, 2008 at 5:39 pm |

Joe-

Think of it instead as subtracting 21 times the units digit from the original number. The resulting number is a multiple of 10, so you can divide it by ten to get the final number. Neither of the two operations changed divisibilty by 7.

This is also how one can think of the divisibility by 3 and 11 tests (checking the sum of the digits, and checking the alternating sum of the digits), since these amount to a series of operations which don’t change divisibility. This all can be said a bit slicker in modular arithmetic, but there’s no need.

February 15, 2008 at 12:17 am |

You know about 57, the so-called Grothendieck prime? (You can tell by adding the digits that it is a multiple of 3). A quick google on the phrase found several descriptions of the incident, as described here: http://www.ams.org/notices/200410/fea-grothendieck-part2.pdf

April 29, 2010 at 8:55 am |

@carter I know I’m a little late to the party, but I only just found this page. Your idea is a nice one. Maybe I’m just not understanding what your saying but to me, this fails as early as the number 21 = 7* 3. Using your method to check if it’s a multiple of 7 we have 2 – 2(1) = 0. As I understand it, your method should give a difference which is some multiple of 7. So does getting a 0 mean a contradiction, or can you just view it as 7*0 ?

May 9, 2010 at 10:33 am |

Actually if you perform all the other tests first the only numbers that needs to be checked for square number is 49=7^2, as for numbers between 100 and 200 you get 119,133 and 161

that behaves like 91.

October 21, 2010 at 9:53 am |

@joe fredette the proof seems simple enough….

let 10a+b be divisible by 7

10a+b=7k

:.a=(7k-b)/10

*note that a is already an integer :.a-2b is an integer

a-2b= (7k-b)/10 – 2b = (7k-21b)/10

which has to be divisible by 7 (the /10 doesnt make a difference coz we already know it i an integer)

October 21, 2010 at 9:54 am |

sorry i am a li’l late too………

November 6, 2011 at 10:11 am |

βαπτιστικα ροουχα…[...]My Favorite Prime Number with Four Divisors « The Everything Seminar[...]…

March 11, 2012 at 12:53 am |

Damn autocorrects…[...]My Favorite Prime Number with Four Divisors « The Everything Seminar[...]…

November 6, 2012 at 9:30 am |

Thank you Sarthak for the info. One piece of information that might help anyone stumbling onto this and wondering about the “/10″:

Picking up from Sarthak, we know that (7k-21b)/10 is cleanly divisible by 7, by factoring: 7 * (k-3b) / 10. We are left with the question if (k-3b)/10 is always an integer. Using the same logic as Sarthak, we can do the following: k=(10a+b)/7; k-3b=(10a+b)/7 – 3b = (10a-20b)/7.

Therefore, we can also definitively say that (k-3b)/10 is also an integer. No surprise for the intuition, but I still like seeing everything worked out to the end.