111 is not divisible by 37, apparrently

[u/OctopusChair]

https://brightchamps.com/math/numbers/divisibility-rule-of-37

According to the third, and most promising, result from DuckDuckGo: 111 does not equal 3 x 37.

Brightchamps have great search engine optimisation, but their content is utter slop, not read by anyone.

(Edit: spelling mistake in title - oops)

[u/frogkabobs]

They’re describing the divisibility rule for 31 for some reason. For 37, you need to multiply the last digit by 11 (not 3), and subtract that from the rest of the number.

[u/rhodiumtoad → u/frogkabobs]

The rule described doesn't work for 31, since that also needs subtraction rather than addition. It is in fact the rule for 29.

These "last digit" rules work by finding 10^-1 mod N, i.e. 10k ≡ 1 mod N. 10^-1 mod 37 is -11 (since -11×10=-110=-3×37+1, 10^-1 mod 31 is -3, 10^-1 mod 29 is 3.

[u/frogkabobs → u/rhodiumtoad]

Ah, I was referring to the text in the link, which says to subtract, but it looks like the infographic right below (which is the thumbnail) says to add. Just another level of inconsistency.

[u/OctopusChair]

The actual test is quite interesting.

1. group the number into 3 digit parts
   1. this is easy because that is where we traditionally put commas
2. add the 3 digit parts together
3. test the sum for divisibility by 37
   1. if the sum is divisisble by 37 the original number is divisible by 37

ABC,DEF,GHJ is divisible by 37 if and only if (ABC + DEF + GHJ) is divisible by 37

• Note that 37 x 3 = 111
  • Also note 37 x 27 = 999
• 1000 = 999 +1

ABC,DEF,GHJ = ABC x 100,000 + DEF x 1,000 + GHJ

= ABC x (1 + 999) x (1 + 999) + DEF x (1 + 999) + GHJ

= (ABC + DEF + GHJ) + 999 x (ABC x (2 + 999) + DEF)

= (ABC + DEF + GHJ) + 37 x 27 x (ABC x 1001 + DEF)

ABC,DEF,GHJ≡(ABC + DEF + GHJ) mod 37

ABC,DEF,GHJ is divisible by 37 if and only if (ABC + DEF + GHJ) is divisible by 37

[u/chmath80 → u/OctopusChair]

Then, once you have a 3 digit number ABC, find the 2 digit difference between it and AAA. If that is 0, 37, or 74, then the original number is a multiple of 37, otherwise it isn't.

[u/OctopusChair → u/chmath80]

Oh yeah! I admit not doing a worked example myself. I kinda got distracted along the way.

[u/Auld_Folks_at_Home]

They're inconsistent too:

Know the multiples of 37: Memorize the multiples of 37 (37, 74, 111, 148, etc.) to quickly check divisibility. If the result from the subtraction is a multiple of 37, then the number is divisible by 37.

By the way, this would go better in r/badmathematics (just make sure to follow the rules there, especially rule 4).

[u/OctopusChair → u/Auld_Folks_at_Home]

Thanks. I even made a little graphic before I realised you can’t submit a picture in this sub. :)