2
49, 67, 85
45
81
50
The sum of your digits is the total number arrived at after adding two or more numbers.
Must end in 5 and be less than 2500, so only 1735 fits.
Eleven 1 + 1 = 2 1 x 1 = 1
22
13
The three-digit number you are describing is 730. It is a multiple of 10, greater than 600, and the sum of the first two digits (7 + 3) equals 10, while the difference (7 - 3) equals 4. However, the criteria provided do not seem to align correctly, as the difference cannot be 13 with the sum being 13. Please clarify the conditions for a correct answer.
To find the last two digits of the sum of the factorials of the first 100 positive integers, we can observe that for ( n \geq 10 ), ( n! ) ends with at least two zeros due to the factors of 10 in the factorial (from the pairs of 2 and 5). Therefore, we only need to calculate the sum of the factorials from 1! to 9!. The sum ( 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! ) equals 40320, and the last two digits are 20. Thus, the last two digits in the sum of factorials of the first 100 positive integers are 20.
It is: 59