Suppose the sum is SS = 100 + 101 + ... + 998 + 999
Also, writing it in reverse order gives
S = 999 + 998 + ... + 101 + 100
so, adding the two together,
2S = 1099 + 1099 + ... + 1099 + 1099 where there are 900 such terms.
So S = 1099*900/2 = 494550.
Positive integers are greater than negative integers. For positive integers: * The integer with more digits is larger. * If two integers have the same length, compare the first digit. If the first digit is the same, compare the second digit, then the third, etc., until you find a difference. In each case, the integer with the larger digit (at the first position where you find a difference) is the larger one.
The 3 consecutive odd positive integers are 7, 9 and 11.
The sum of two positive integers is positive
The sum of the first three even positive integers is 2 + 4 + 6 = 12.
To find the number of positive integers less than 900 with all odd digits, we consider the digits available: 1, 3, 5, 7, and 9. For a three-digit number, the first digit (hundreds place) can only be 1, 3, 5, 7, or 9 (5 options), while the tens and units places can also be any of the 5 odd digits (5 options each). Thus, there are (5 \times 5 \times 5 = 125) three-digit numbers. For two-digit numbers, the first digit can again be any of the 5 odd digits, and the second digit can also be any of the 5 odd digits, giving (5 \times 5 = 25). Finally, for one-digit numbers, there are 5 options (1, 3, 5, 7, 9). Adding these together gives (125 + 25 + 5 = 155) positive integers less than 900 with all odd digits.
Positive integers are greater than negative integers. For positive integers: * The integer with more digits is larger. * If two integers have the same length, compare the first digit. If the first digit is the same, compare the second digit, then the third, etc., until you find a difference. In each case, the integer with the larger digit (at the first position where you find a difference) is the larger one.
To find the number of three-digit positive integers with digits whose product is 24, we can break down 24 into its prime factors: 2 x 2 x 2 x 3. The possible combinations for the three digits are (2, 2, 6), (2, 3, 4), and (2, 4, 3). These can be arranged in 3! ways each, giving a total of 3 x 3! = 18 three-digit positive integers.
The 3 consecutive odd positive integers are 7, 9 and 11.
Find three consecutive positive even integers whose sum is 123 , Answer
Find three different positive integers that sum to seven?
The sum of two positive integers is positive
There are a total of 5 positive three-digit perfect cubes that are even. To find this, we first determine the range of three-digit perfect cubes, which is from 46 to 96. Then, we identify the even perfect cubes within this range, which are 64, 216, 512, 729, and 1000.
The sum of the squares of two consecutive positive even integers is 340. Find the integers.
The sum of the first three even positive integers is 2 + 4 + 6 = 12.
The numbers are 65 and 67.
To find the number of positive integers less than 900 with all odd digits, we consider the digits available: 1, 3, 5, 7, and 9. For a three-digit number, the first digit (hundreds place) can only be 1, 3, 5, 7, or 9 (5 options), while the tens and units places can also be any of the 5 odd digits (5 options each). Thus, there are (5 \times 5 \times 5 = 125) three-digit numbers. For two-digit numbers, the first digit can again be any of the 5 odd digits, and the second digit can also be any of the 5 odd digits, giving (5 \times 5 = 25). Finally, for one-digit numbers, there are 5 options (1, 3, 5, 7, 9). Adding these together gives (125 + 25 + 5 = 155) positive integers less than 900 with all odd digits.
There is a clever but tricky way to solve this. If we have to find the number of 4-digit integers that contains at least one 5, we can also find the number of all the 4-digit integers and the number of integers that do not contain any 5's and subtract it from the first number. This is called complementary counting.So, first of all, we must find the number of 4-digit numbers there are. There are 9000 of them.Now, we find the number of 4-digit integers without 5's. By thinking a little bit, we see the first digit must be from 0~9 excluding 5. That is a total of 9 numbers. This is the same for the next three digits. Therefore, there are 94 = 6561 4-digit numbers without a 5.Finally, we can subtract. 9000 - 6561 = 2439