5050, according to the program I quickly whipped up.
Yes, 100 is divisible by 3. To determine if a number is divisible by 3, you can sum its digits. The sum of the digits in 100 is 1 + 0 + 0 = 1, which is not divisible by 3. Therefore, 100 is not divisible by 3.
The sum of the digits of the number 10 is calculated by adding its individual digits together. The digits in 10 are 1 and 0. Therefore, the sum is 1 + 0 = 1.
The factors of 3 are 1 and 3. The sum of the digits of these factors is calculated as follows: for 1, the sum of its digits is 1, and for 3, the sum of its digits is 3. Therefore, the total sum of the digits of the factors of 3 is 1 + 3 = 4.
5050 ie 101 x 50
5,050
100
Yes, 100 is divisible by 3. To determine if a number is divisible by 3, you can sum its digits. The sum of the digits in 100 is 1 + 0 + 0 = 1, which is not divisible by 3. Therefore, 100 is not divisible by 3.
The sum of the digits of the number 10 is calculated by adding its individual digits together. The digits in 10 are 1 and 0. Therefore, the sum is 1 + 0 = 1.
The factors of 3 are 1 and 3. The sum of the digits of these factors is calculated as follows: for 1, the sum of its digits is 1, and for 3, the sum of its digits is 3. Therefore, the total sum of the digits of the factors of 3 is 1 + 3 = 4.
5050 ie 101 x 50
1 + 1 = 2 The sum of the digits is therefore 2.
The sum of all the digits of all the positive integers that are less than 100 is 4,950.
Digits are from 1 to 9the sum of the 9 greatest digits is = 9+9+9+9+9+9+9+9+9 = 81the sum of the 9 lowest digits is = 1+1+1+1+1+1+1+1+1+1= 9and so :81 + 9 = 90
Since a dozen is 12, then the digits are 1 and 2. To get the sum, you simply add 1 + 2, which = 3.
5,050
1 + 23 + 4 + 5 + 67 = 100 is one way. 1 + 2 + 34 + 56 + 7 = 100 is another. 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 so you need to move two non-adjacent digits that sum to 8 into the tens' places. This is because: Start with a sum of 28 Lose two digits that sum to 8 from the units so that you have 28 - 8 = 20 Gain two digits that in the tens place that sum to 8 so that 20 + 80 = 100.
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.