what is the smallest four=dogot palindrome that is the sum of two different three-digit palidromic nubmers?
202+909 = 1111
maybe someone else can improve on that. It has been shown that every number can be written as the sum of not more than three palindromes...
505 + 606 = 1111
ur welcome
1111
232 + 323 = ?
yo sup kids
To find the smallest four-digit palindrome that is the sum of two different three-digit palindromic numbers, we can follow these steps:
1. Start by listing all the three-digit palindromic numbers. A palindromic number is one that reads the same backwards as forwards. In this case, we are looking for three-digit palindromes, so the numbers we are interested in will have the form "ABA", where A and B can be any digit from 0 to 9.
The three-digit palindromes are: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, ..., 919, 929, 939, 949, 959, 969, 979, 989, 999.
2. Once we have the list of three-digit palindromes, we can try adding them together in pairs to find the sum that gives us a four-digit palindrome. Start by adding the first palindrome (101) with each subsequent palindrome:
101 + 111 = 212 (not a palindrome)
101 + 121 = 222 (not a palindrome)
101 + 131 = 232 (not a palindrome)
101 + 141 = 242 (not a palindrome)
...
Continue this process until you find a sum that results in a four-digit palindrome.
3. The smallest four-digit palindrome that is the sum of two different three-digit palindromic numbers is the answer to your question.
Note: It is important to skip the cases where we are adding the same three-digit palindrome to itself, as the question asks for two different three-digit palindromes.