None is zero

Relevant wiki: Cryptogram

$\large \begin{array} { l l l l l } & &D & O & G\\ +& &C & A & T \\ \hline & 1 & 0 & 0 & 0 \\ \end{array}$

Consider the units column: $G+T$ ends in a $0$ which implies that $G+T=0$ or $G+T=10$ and $G+T$ can not be $20$ or more.

Since none of the digits is $0 \implies$ $G+T=0+0=0$ is false. Thus, $G+T=10$ which means we need to carry $1$ to the tens column.

Since the result in the tens column is $0$ and there is a carried $1$ into this column, this implies that $O+A$ ends in a $9 \implies$ $O+A=9$ and we can not have $O+A$ as $19$ or more. In the tens digit we must have $O+A=9$ plus the carried $1$ which results in a $0$ in the tens digit.

Same analysis for the hundreds column, we must have $D+C=9$ plus the carried one $\implies D+C= 9$ .

Thus $( D+O+G+C+A+T)=(G+T)+(O+A)+(D+C) = 10 + 9 + 9= \boxed{28}$ .

Let's say that G and T both add up to 10. After you've added them together, you'd have to carry the tens digit over the O and A, which have to add up to 9 since we've already carried the one. Once again we follow this procedure with the tens digit carried over the D and C, and that means that D and C must also add up to 9.

9 + 9 + 10 = $\boxed{28}$

G + T = 10 , O + A = 9 and D + C = 9 Hence G +T + O + A + D + C = 28.

I realized that dog+cat=1000 could also be translated as 1000-dog=cat or vice versa. So my simple solution was to do 1000 minus random 3 digit numbers without the same digit until I find a combo without the digits 0 or 5, as those require another 5 or 0 to equal a digit ending in 0. My combo was 126 and 874, but that's one way to do it.

The sum by replacing variables with constants is 351+ 649 . So D is 3, O is 5, G is 1, C is 6, A is 4 and T is 9 Sum of the digits is 28.

G+T must give 0 at the end so example: 876+ 124= 1000 So 8+7+6+1+2+4=28

Relevant wiki: Cryptogram

Clearly

G+T=10 ,

O+A=9 & (Since O+A+1=10)

D+C=9 (Since D+C+1=10)

Therefore, (G+T)+(O+A)+(D+C)=28

=>D+O+G+C+A+T=28

We need 6 unique numbers to assign the letters. 0 is an illegal number. There are 9 possible unique numbers (1 through 9) There are 2 sets of numbers that need to sum 1000. In that case, we can use the following pairs of unique numbers to reach the solution. I listed the lowest and highest possible numbers, and some in the middle. 231 + 769, ..., 521 + 479, 532 + 468, 568 + 432, 579 + 421, 587 + 413, ..., 876 + 124,

In every case the unique single digits add up to ((D+C)+(O+A)+(G+T)) = (9 + 9 + 10) = ||28|

9*3 + 1 = 28 Note all numbers and variables later explained are positive integers so I don't have to copy and paste everything. where m is a number that is a sum of two numbers that have the same number of digits, with m also being 10^n, the solution is 9n + 1 as if you subtract 1 from m, you'd get 99...99 for an n number of 9s. Therefore, the sum of all numbers as a ones place, not necessarily distinct, from the two earlier numbers that were explained to have the same number of digits is 9n+1, as you need 1 more to change the 99...99 for an n number of 9s to be turned into an n number of 0s with 1 as the leading digit.