Mind your C's and D's

$C+ \overline {CC}=\overline {D4}$

Now, For this to be true, C can be Either $2$ or $7$ (Because the units digit is 4)

Case 1: C is 2

The crytarithm Becomes:- $2+22=24$

Here, D becomes 2, And since C and D are distinct, The above crytarithm becomes false.

Case 2; When C is 7

The crytarithm Becomes:- $7+77=84$

The crytarithm above satisfies all the given conditions

Hence, The only value of C can be 7

$CC +C = 11C + C = 12C$

So $D4$ must be divisible by $12$ thus $D4 = 24$ or $84$ .

If $D = 2$ then $12C = 24$ so that $C = \frac{24}{12} = 2$ but $C$ and $D$ need to be distinct.

So $D=8$ such that $12C = 84$ where $C = \frac{84}{12} = \boxed{7}$ .

Simple, you have to get 4 at units place by adding same digits. This can be done by either using 2 or 7 i.e.C=2 or 7. When adding 2 and 2 , although result will be 4 but at ten's place of answer(i.e. D) we have to get a different no. Means a no which will yield carry at unit's digit i.e. 7. Add 7 and 7, you will get 4 and carry 1. Add this carry with 7 and you get 8. Therefore, C=7 and D=8.

if we try the c 2=4 we find 2 but then c have to equal d, and that's unacceptable, so we need c 2=a 4 , where a is a digit, and then we get 7*2=14.

The lowest place value digit of the sum $D4$ is the clue. $C+C$ in the units column causes a carry-over of $1$ to the next place value, incrementing the $C$ on the left to a $D$ on the number line implied by the problem statement.

Only $2$ and $7$ are possible values of $C$ that result in a 4 in the units place of the sum. Only $7$ as a value for $C$ would cause a carry-over of $1$ when evaluating $C+CC$ .

let X be the carry, C + C = X4, then X + C = D, but C and D are consecutive whole numbers, so we can say D = C + 1, from X + C = D = C + 1, therefore X = 1... from C +C = X4 , 2C = 14, then C = 7

This is easy. So since 2C =4 we can try C = 2 but it will not work so it should be 2C = 14 and C is 7

My solution: $x + ( 10x + x ) = ( x + 1 ) 10 + 4$

$12x = 10x + 14$

$2x = 14$

$x = 7$

Edit: $D = C + 1$ because the largest single digit is 9, for example the last digits are both 9, then we have 18, so you can carry the '1'. It is also $C + 1$ because in the problem it was shown that from C it turned to D, so there is a change of digit.

When C = 2 The answer is 24 It is false, Because D = 2, So C = 7. Then D= 8: 7 + 77 = 84.

cc+c=d4; so cc+c should result in a number that ends in 4 so ether 7 or 2 then since the result is d4 that means by adding the two c's it should result in a number larger then ten so 7 fits the bill because 7+7=14 while 2+2=4 // 22+2=24 would not work since the d would be c but 77+7=84 does because d remains ect.. I don't really care though....

44+4 and 99+9 are not equal to 10D+4. If C=2, then 22+2=24. However, since C and D are distinct, D is not equal to 2. If C=7, then 77+7=84, giving you D=8, which is distinct from 7.

simply! there're 4 possible answers, test one by one, then find 7+77=84