How to disproof with logic: If the digit sum is dividable by 3 the number is dividable by 12 for ℕ ∩ [12, 100)

Question

Assumption:

If the digit sum of $x ∈ ℕ ∩ [12, 100)$ is dividable by 3, it is also dividable by 12.

Examples:

• $12 → 1+2 = 3$;
  $3 \mod 3 = 0 \Rightarrow 12 \mod 12 = 0$
• $13 → 1+3 = 4$;
  $4 \mod 3 = 1 \Rightarrow 13 \mod 12 ≠ 0$
• $24 → 2+4 = 6$;
  $6 \mod 3 = 0 \Rightarrow 24 \mod 12 = 0$

Assumption (more formal):

• Let $x ∈ ℕ ∩ [12, 100)$
• Notation used for x:
  • $a_0$ describes the first digit from the right
  • $a_1$ describes the seconds digit from the right
  • ...
  • Example: x = 32: $a_1 = 3$, $a_0 = 2$
• $(a_1 + a_0) \mod 3 = 0 \Rightarrow (a_1 \times 10 + a_0) \mod 12 = 0$

Question

How would I start from here? I know I could pick $15$ or other numbers to disprove by example. But I'd like to imagine (for the sole purpose of exercise) finding a specific example is too difficult.

Answer 1

Always try out enough examples to see whether or not the statement is actually provable.

$x=15$ is a counterexample: its digits sum to $6$, which is divisible by $3$, but $15$ isn't divisible by $12$.

Answer 2

To disprove something you need only one example that does not apply. if you want numbers >12 , 15 is the simplest . any other like 27 or 33 or 333 will do. if you want to improve your skills better prove that if the digit sum of n is dividable by 3 the number ist dividable by three .

Answer 3

Proving it for three write the number $a_2a_1a_0$ as $100a_2+10a_1+a_0$ and consider 10, 100, 1000 mod 3=1 so what remainder has $a_n*10^n$

Answer 4

This is how I approached the problem in the end.

Assumption:
$x∈N∩[12,100): (a_2 + a_1 + a_0) \mod3=0⇒(a_1×10+a_0)\mod12=0$

Possible values for $a_1$ and $a_0$ are:
$a_1 ∈ \{0,1,2,3,4,5,6,7,8,9\}$
$a_0 ∈ \{0,1,2,3,4,5,6,7,8,9\}$

From the assumption we know the following is valid
$(a_1 + a_0)\mod3 = 0$

This allows only the following permutations of $a_2$ to $a_0$:
$\{03,06,09,12,...,99\}$

From here we have a clear set we could to check (what other people already wrote, but I feel this explanation here has more structure.)
From here we could also try to figure out how to find a better algorithm to check all values.

Out of scope
In case we would have more digits we would have a formula like this:

$ (∑a_k \times 10^k ) \mod 12 = 0$

= $ (∑ (a_k \mod 12 ) \times (10^k \mod 12) ) \mod 12 = 0$

= $ (∑ a_k \times (10^k \mod 12) ) \mod 12 = 0$

Here $(10^k \mod 12)$ would amount to

• $k=0 ⇒ (10^k \mod 12) = 1$
• $k=1 ⇒ (10^k \mod 12) = 10$
• $k=2 ⇒ (10^k \mod 12) = 4$
• $k=3 ⇒ (10^k \mod 12) = 4$

So for testing we could use these simplified values.