With this submition there is a picture of all the answers for the cryptarithms task that I have completed.....
Today, I will be concentrating on question 23
Equation:
A B A
A B A
A B A
A B A
+A B A
= C D B A
This looks quite hard but you just have too think. There is one good way to experiment which is trial and error. This may be helpful in some equations but not all. For this one quite a bit of thinking is involved.
-All the letters have to be nine or below.
-A+A+A+A+A has to make a two digit number as when you add them all the answer is two digits. This means that A cannot be one.
-A+A+A+A+A = A on the left side. So this means A has to be 5 because 5 is the only number that can have the answer A.
- Now we move on to B. It is the same concept for B. B+B+B+B+B has to equal B. This means B has to equal 7 as it's the only remaining number. A could have also been seven but it didn't work in the end.
This means that C=2. D=8. B=7. A=5 This works if you add them all up together.