Which is a counterexample to the conjecture?
any number that is divisible by 3 is divisible by 9.
a.
63
b.
69
c.
18
d.
27

Answer B

If the sum of the digits in the number is a multiple of the number 9 is divisible by 9
A 63
digits : 6,3
sum 6 + 3 = 9
9 : 9 = 1

The number of 63 is divisible by 9

B 69
digits 6,9
sum 6 + 9 = 15
     1,(6)

     15  : 9
   -  9
 
       60 
    -  54
   
         60
the number 69 is not a multiple of 9 and divided by 9 do not receive natural number
C 18
digits 1,8
sum 1+8 = 9

9:9 = 1

The number of 63 is divisible by 9

D 27
digits 2,7
sum  2 + 7 = 9

9 : 9 = 1
The number of 63 is divisible by 9

difficult

621450450
digits 6,2,1,4,5,0,4,5,0
1 sum 6 + 2 + 1 + 4 + 5 + 0 + 4 + 5 + 0 = 27
2 sum 2 + 7 = 9 
9 : 9 = 1
The number of 63 is divisible by 9

621450450 :9 = 69050050

Just look outside and then you’ll see where the awnser is