Carminati–McLenaghan invariants

The base conversion divisibility test is a process that can be used to determine whether or not a certain (positive) natural number a can be divided evenly into a larger natural number b. It is the general case for the well-known test for divisibility by nine. For other divisors, applying this test is generally harder than figuring it out by normal division.

Example

Is 312 divisible by 13?

• a=13
• b=312
• x=a+1=14
• y=b (base-14)=184 (312 in base x)
• z=1+8+4=13
• z/a=13/13=1, a natural number

312 is divisible by 13.

Example

this can be solved by another method. a=3 b=1 c=2 now, 10a=30 b=1 4c=8; thus 10a+b+4c=30+1+8=39,which is divisible by 13. For 2-digit numbers:- if a+4b is divisible by 13 then 10a+b is divisible by 13.

Example

is 91 divisible by 13? a=9 b=1 therefore a+4b=9+4*1=13,which is divisible by 13 thus 91(13*7=91) is divisible by 13.

Dividing by nine

The trick for determining if a number is divisible by nine is well-known: If the sum of the digits of a number is divisible by nine, then the number itself is as well. This is a special case of the general rule, made easy because no base conversion is necessary since 9 + 1 = 10, and we already use base 10.

Example: Is 2,340 divisible by 9?

• a=9
• b=2,340
• x=a+1=10
• y=b (base-10)=2,340
• z=2+3+4+0=9
• z/a=9/9=1, a natural number

2,340 is divisible by 9.

Proof

Any number can be expressed as

$numbe{r}_{(base)}=\sum _{i=0}^{n}digit{s}_i\times bas{e}^i$

We know that under Modulo Arithmetic, $base{\equiv }_{(base-1)}1$

Thus $number{\equiv }_{(base-1)}\sum _{i=0}^{n}digit{s}_i\times 1$