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Q: Find the sum of the odd numbers from 1 to 99?

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answer:2500

The sum of all odd numbers 1 through 99 is 9,801.

100

99^2 == 9,801

2500

99

100

9801 * The sum of the first two odd numbers (1+3) is 4, or 22 * The sum of the first three odd numbers (1+3+5) is 9, or 32 * The sum of the first four odd numbers (1+3+5+7) is 16, or 42 * ...and so on So the sum of the first 99 odd numbers, using the pattern above, would be 992 or 9801.

printf ("%d\n", (1+99)/2*50).

97 and 99

99 is an odd number. Even numbers are evenly divisible by two, while odd numbers are not.

The sum of 1+ 3 + 5 + ... + 99 = 50 x (1 + 99) ÷ 2 = 2500.

let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34

The sum of the even numbers is (26 + 28 + ... + 100); The sum of the odd numbers is (25 + 27 + ... + 99) Their difference is: (26 + 28 + ... + 100) - (25 + 27 + ... + 99) = (26 - 25) + (28 - 27) + ... + (100 - 99) = 1 + 1 + ... + 1 There are (100 - 26) ÷ 2 + 1 = 38 terms above which are all 1; their sum is 38 x 1 = 38. So the difference of the sum of all even numbers and all odd numbers 25-100 is 38.

97 + 99 = 196

The composite odd numbers from 9 to 99.

99

They are: 99+67 = 166 and 99-67 = 32

The sum of all the integers between 1 and 99 inclusive is 4950.

Use the formula for the sum of an arithmetic sequence. Start with 11, end with 99; the interval is 2.

The numbers are 31, 33 and 35.

The numbers must average 33, so the answer is 32, 33 and 34.

4950

Since we are talking for the sum of digits, then we are talking about 2-digit numbers between 1 and 100, which are 90 numbers (10 to 99). From these numbers only half of them satisfy the above condition, that is 45 numbers. So 50% of 2-digit numbers between 1 and 100 have an odd sum of their digits.

Infinity

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