35.
21.
2,4,6,8,10,12,14,16,18
89
There are 5460 five digit numbers with a digit sum of 22.
Since the first integer of the 3-digit numbers must be 2, the sum of the second and the third digits would be 13.Since the largest digit is 9, the 3-digit numbers could be 249, 294, 258, 285, 267, and 276.Thus, there are 6 integers between 200 and 300 whose sum of their digits is 15.
81
The smallest digit palindrome that is the sum of two 3-digit palindromes is 121. This is achieved by adding the two 3-digit palindromes 101 and 20, both of which are palindromic. Therefore, 101 + 101 = 202, but if we consider a valid case with two different palindromes, we can use 111 and 110, which gives us 221, the next smallest palindrome. However, the smallest individual palindrome formed by the sum of any two 3-digit palindromes remains 121.
The smallest 4-digit palindrome is 1001. To find if it can be expressed as the sum of two 3-digit palindromes, consider the smallest 3-digit palindromes, which are 101, 111, 121, etc. The combination of 101 and 900 (another 3-digit palindrome) gives 1001, making 1001 the sum of two 3-digit palindromes. Thus, the answer is 1001.
Nine. The sum of the digits must be a multiple of 9; because of the repeated digits, this is only possible if the first two digits add up to 9.
1221, 2112, 3003, 10401, 11211, 12021
There are five such numbers.
Six. 1551 2442 3333 4224 5115 6006
54
Three of them.
21.
30,25,15,20, or 10.
2,4,6,8,10,12,14,16,18