Q: What is the answer of 1 plus 2 plus 3 plus plus 99 plus 100 plus 99 plus 98 plus 97 plus plus 3 plus 2 plus 1?

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99 + 98 + 97 = 294

294

100 + 99 + 98 + 2 + 1 + 0.3 = 300.3

The median for 85, 93, 95, 96, 98, 99, 100, 100 is 97. The number in the middle is 96, and 98. Add - 96 + 98 = 194 Divide - 194 ÷ 2 = 97

97 is a prime number. 98 and 99 are not.

Since 1+99=100. And 2+98=100 and 3+97=100 etc to 49+51=100 it's 49 of those plus 50 or 4,950. The sum of the first 99 whole numbers is 4,950

97, 98, 99 and 100.

303

5,050

294

The median is (98+99)/2 = 98.5

100! / 97! = 100 * 99 * 98 = 970200

92 93 94 95 96 97 98 99

100-1-1-1= 99-1-1= 98-1= 97 ..... ..... ..... ..... ..... ..... ..... ..... 100-1-1-1= 100-3= 97

The three consecutive numbers whose sum is 294 are 96, 98 and 100. Also, 97, 98 and 99.

The answer is 409. Simply add the numbers together like you would for two plus two. To check your work, a calculator is the best option.

First add all the addition together and add all the subtraction together. Then there will be the total addition and the total subtraction. Subtract the total subtraction from the total addition.99+98-97+96-95+94-93+92-91+90=(99+98+96+94+92+90)-(97+95+93+91)=569-376=193

95

98+1 100-1 97+2 96+3 93+6

1 + 98 = 99

printf ("%s\n", "1, 2, 3, 4, 5, 6, ..., 98, 99, 100"); printf ("%s\n", "100, 99, 98, 97, ..., 3, 2, 1");

100 + 71 = 171, as does 99 + 72, 98 + 73, and so on.

There are infinitely many sets: For example, -99, -98, -97, ... -2, -1, 0, 1, 2, ... 97, 98, 99, 100 or 1, 1, 1, 1 ... [100 times] or 0.0001, 0.0001, 0.0001, ... [ a million times].

100 + 99 = 199

101

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