When we take a closer look at the problem, F in FOUR must equal 1, since no two 3-digit numbers add up to equal or exceed 2000. Also, T in TWO must be higher than 4. No 3-digit number lower than 500 would be multiplied by 2 and equal 1000 or higher.
But trial and improvement shows that T must be higher than 6, because no combination where T equals 5 or 6 satisfies the conditions of TWO + TWO = FOUR. Therefore, the 3-digit numbers in TWO + TWO = FOUR combinations must be higher than 699.
I’ve found 5 possible combinations: 734+734=1468, 765+765=1530, 836+36=1672, 867+867=1734, and 938+938=1876. 969+969=1938 does not work, so those are all the solutions.
Meanwhile, ONE+ONE=TWO is possible.
ONE+TWO=THREE is impossible. It is impossible for a three-digit number to be multiplied by two and equal to a product higher than 9999.
ONE+THREE=FOUR is also impossible. Five digit numbers are not lower than four digit numbers.
FOUR+FIVE=NINE is possible.
Cryptarithm subtractions can be composed and said to be possible if they comply with the rules of arithmetic. Here are some possible cryptarithm subtractions:
TWO-ONE=ONE
FOUR-TWO=TWO
NINE-FIVE=FOUR
NINE-FOUR=FIVE
SEVEN-FOUR=THREE
Solution
196082
Problem / game
First name
Youyou
School
Sha Tin Junior School
Country
Age
9