My first cryptogram

Only solution where C+C+C=C is possible is if C=5.

Since the final answer is CCC i.e., 555, and they are 3 equal numbers which add to it, then just divide 555 by 3:

555÷3=185

$3(100a+10b+c) = 111c$ , or simply $10a+b=\frac{18}{5}c$ . Here we will notice that between 1 to 9, the only possible solution is that $(a,b,c) = (1,8,5)$ , thus, $\overline{abc} = 185.$

• Finding C

3C =XC : Only solution is, C=5 ( $3\times5 = 15$ )

Reminder, X = 1

• Finding B

3B+1(reminder) = X5

i.e 3B = X4 Only solution is, B 8 ( $3\times8 = 24$ )

Reminder, X = 2

• Finding C

3A+2(reminder) = 5

i.e 3C = 3. Only solution is, A=1

Hence $\overline{ABC} = 185.$

As A = B = C = 0 is not where we should cease particularly when we know that it is usual to have different digits being meant by different alphabets' representation, we should look for others.

Among 1 to 9, only 5 added itself consecutively for twice such as multiplied by 3 which shall produce digit at $10^0$ of its own, therefore, C = 5.

With C = 5, it carries forward a 1 to add onto another digit multiplied by 3. Only 8 $\times$ 3 + 1 shall give $10^1$ of 5 among 0 to 9, therefore, B = 8.

With C = 5 and B = 8, it carries forward a 2 onto another digit (different) multiplied by 3. Only 1 $\times$ 3 + 2 give $10^2$ of 5 among 1 to 9 (first digit is usually meant for non-zero), therefore, A = 1.

850 $\times$ 3 = 2550 is obviously not the answer, so, $10^0$ or C not considered for 0 is correct.

Therefore, 185 is the only answer.

Answer: $\boxed{185}$

i copied from the internet

Assume A, B, and C are single digit integers. The highest number one of these can be is 9. From the first column: 3C = n(10)+C. Because 3(9)=27, n can't be greater than 2. Therefore, there are three options: $3C=C$ $3C=10+C$ $3C=20+C$ Solving each, one gets: $C=0$ $C=5$ $C=10$ 5 is the only single digit; therefore, C=5. The second column now is evaluating 3B+1 because you have to "carry the one" you get by getting 15 after adding 5+5+5. By that logic: $3B+1=5$
$3B+1=15$
$3B+1=25$ Evaluating each, one gets: $B=4/3$
$B=14/3$ $B=8$ B is the only integer; therefore, B=8. The sum of the last column has to be exactly 5 because there aren't any more columns after that where you can "carry the one" to. By the same logic as for the second column, the last column is evaluating 2+3A. Therefore: $2+3A=5$ $A=1$ If C=5, B=8 and A=1, then ABC=185.

185 + 185 + 185 = 555; then ABC = 185