hi
 
i learnt in primary school : if the sum of all digits of a number
is divisible by 3
then this number is divisible by 3
for example : 24111 is divisible by 3
because 2+4+1+1+1 =9 , 9 divisible by 3
 
 
 
 
there is also others theories called
9 proof for verify an addition
 
french joke impossible to translate
with the Fermat's small theorem
u can see in this chanel
 
 
 
i can explain this ...
 
with magics play cards as a pretidigitator
 
 
here is the sequence (n+1)(n+1)(n+1)
 
if i develop this expression
 
the first term is n powered by 3
 
 
all the middle terms are multiple of n
 
then divisible by n
 
 
but the last term (not factor of n)
 
will be 1*1*1 = 1
this last term is THEMOSTIMPORTANT
because not multiple of n
 
 
if i multiply this sequence
 
by 5
 
 
then this LAST TERM will be now multiple of 5
 
 
 
 
because 10 = 9+1
 
 
100 = (9+1)(9+1)
 
100 = (9+1)(9+1)(9+1)
if i develop
all the terms will be divisible by 9
 
 
exept the LASTONE
 
 
so 10,100,1000 not divisible by 3 or 9
 
 
 
see bellow...
 
 
 
 
then 100 not divisible by 3
 
 
 
 
as the Fermat's small theorem
ONLY THE LASTONE not multiple of n
 
 
if i multiply 100 by 5
 
 
 
the last one is now = 1*1*1*5 = 5
 
 
 
 
 
 
 
all the terms are divisible by 3 or 9 exeplt the last one
divisible by 5
 
 
 
if now i take 3
 
 
3 *100
 
the last one =3
so
all this nomber ,300 is now divisible
by 3
 
because the last one is also divisible by 3
 
 
 
now we take 1452
 
 
 
 
 
 
what are the last indices not nultiple of n
 
 
 
 
 
 
 
 
 
 
 
