Not always adorable!

I claim that all numbers above $10$ are adorable.

Let $a=a'+2$ , $b=b'+3$ , and $c=4$ .

Then, $m=a'+b'+9$ where $a',b'\ge 0$ and $2\mid a'$ , $3\mid b'$

Now by Chicken McNugget theorem, the smallest number not representable by $a'+b'$ is $2\times 3-2-3=1$ . Thus, $a'+b'$ can equal $2$ or any number greater than $2$ .

Thus, $a'+b'+9\ge 11$ so $m\ge 11$ is not adorable.

It is a simple exercise now to check that all $m \le 10$ is adorable except $m=9$ , where $m=2+3+4$ .

Thus, the answer is $1+2+3+\cdots +7+8+10=\boxed{46}$

Let $a=2p, b=3q$ and $c=4r$ . Now for $p,q,r\geq 2$ ,we will show that if $m$ is adorable, then $m+1$ is also adorable. This can be proved as follows. $$ If $m = 2p+3q+4r$ then $m+1 = 2(p-1)+3(q+1) + 4r$ $$ So we see that all $m \geq 18 = 2*2+2*3+2*4$ are adorable. Also the least adorable number is $2+3+4=9$ . And since the least increment possible is of $2$ , $10$ is not adorable. All numbers $9\leq m\leq18$ can also be show to be adorable easily). $$ So we have our list of non-adorable numbers as : $1,2,3,4,5,6,7,8,10$ which sum to $\boxed{46}$

For odd $m \geq 9$ , $m = (m-7) + 3 + 4$ . Since $m$ is odd, $m -7$ is even. Also, $m \geq 9 \implies m-7 \geq 2$ . Therefore all odd $m \geq 9$ are adorable.

For even $m \geq 12$ , we have $m = (m-10) + 6 + 4$ . Since $m$ is even, $m -10$ is also even and also, $m \geq 12 \implies m - 10 \geq 2$ . Therefore all even $m \geq 12$ are adorable.

Now, $a \geq 2, b \geq 3 , c \geq 4 \implies m = a + b + c \geq 9$ . Therefore $[1,2,3,..,6,7,8]$ are not adorable

The only number left to check is $10$ . Since the smallest adorable number is $9$ , and the smallest increment possible is $2$ , $10$ is also not adorable.

Adding all these numbers, we get $\boxed{46}$ .

$\bullet$ First of all, note that if $k$ is adorable, then any multiple of $k$ is adorable.

$\bullet$ $2 \mid a\quad ; 3\mid b\quad ; 4\mid c \implies a+b+c \geq 2+3+4 = 9$ . Thus all the numbers less than 9 are NOT adorable. (1 to 8)

$\bullet$ Note that if $p>7$ is a prime, then $p = (p-7) +3 +4$ and this tells us that $p$ is adorable. ( $2$ always divides $p-7$ as $p$ will be odd).

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Hence if any number is adorable, we conclude that it must be composed of only the primes $\leq 7$ .

$\therefore m = 2^x 3^y 5^z 7^w$

Now note that

$2^4= 16 = 6 + 6 +4$

$3^2 = 9 = 2+3+4$

$5^2 = 25= 2+3 +20$

$7^2 = 49= 2+3+44$

This tells us $x,y,z \leq 1$ and $x \leq 3$ .

So if a number is adorable, it has very few possibilities, all will be divisors of $2^3\times 3\times 5\times 7$ .

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$2^3 \times 3\times 5\times 7$ has $(3+1)(1+1)(1+1)(1+1) = 32$ divisors.

We now see

$3\times 5 = 15 = 2+ 9 + 4\\ 5\times 7 = 35 = 2 + 9 + 24 \\ 3\times 7 = 21 = 2 + 3 + 16$

Hence $15,21,35$ and all their multiples are Adorable, so out of the 32 factors, only those of the form $2^k \times p$ are considered, which will be factors of

$2^3 \times 3 ,\quad$ $2^3 \times 5, \quad 2^3 \times 7$ .

Now note that

$2^2 \times 3 = 12 = 2+6+4$ ... This cancels 12 and 24

$2^2 \times 5 = 20 = 2+6+12$ ... This cancels 20 and 40

$2\times 7 = 14 = 4+6+4$ ... This cancels 14,28 and 56

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$\\ \text{Number}-- \text{Divisors}---- \text{Adorable} --\text{Not Adorable}$

$2^3\times 3 --1,2,3,4,6,8,12,24-- 12,24 -- 1,2,3,4,6,8\\ 2^3\times 5 -- 1,2,4,5,8,10,20,40 -- 20,40 -- 1,2,4,5,8,10\\ 2^3\times 7 -- 1,2,4,7,8,14,28,56 -- 14,28,56 -- 1,2,4,7,8 \\$

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hence the only numbers NOT adorable are $1,2,3,4,5,6,7,8$ and $10$ .

Their sum is $\boxed{46}$

First of all let's see that if a number n is adorable, then the number n+2 is also adorable.

Proof.

Let's take an adorable number n and write it like this: n = 2x + 3y + 4z

We can write the number n+2 like this:

n+2 = 2x + 3y + 4z + 2 = 2(x+1) + 3y + 4z

Which is also adorable. CVD.

So, since we know that 9 is the first adorable number ( 9 = 2 + 3 + 4) and since it is odd, we can deduce that every odd number over 9 is also adorable.

In the same way we prove that since 12 is the first adorable even number (12 = 2 + 3×2 + 4), every even number over 12 is also adorable.

Every other number is not adorable, so our list of non-adorable numbers is: 1,2,3,4,5,6,7,8,10.

The sum of these nine numbers is 46.

Let's say $a$ is adorable. Which means $a=2k+3d+4s$ where $k, d, s$ are three integer numbers.

Now the next adorable number will be $b=2(k+1)+3d+4s$ . This implies that adorable numbers are distributed at a distance of $2$

The first adorable number is $9$ so $11, 13, 15, 17, ...$ would also be adorable. That means that all odd numbers greater that nine are adorable.

Also since $12$ is adorable (which can be easily verify), all even numbers greater that twelve are also adorable.

Finally it's easy to check that $1, 2, 3, 4, 5, 6, 7, 8$ and $10$ are not adorable