First we find the sum of the integers from 2 to 100 including 2 and 100, then we divide by 99 since there are 99 numbers. There are 100 numbers between 1 and 100 and we are excluding only the number 1. If we want to exclude the number 2, and count only the numbers 3,4,5....100, this can be done with the same procedure and a slight modification. So the sum of the numbers 1 to 100 can be found by writing the numbers 1, 2,3,...100 Now write them backwards, starting at 100,99,98....1 Each column has a sum of 101 and there are 100 columns. So the total is 100x101, but we wrote the list twice so we must divide by 2 The sum is 100x101/2=50x101=5050 Now remember we want only 2 to 100 so the sum we seek is 5049. Since there are 99 number, the mean is 5049/99=51 In general the sum of the first n positive integers is n(n+1)/2 This can be proved the way we did or by induction.
(7*100*101)/2 = 35,350 jpacs * * * * * What? How can there be 35,350 integers in the first 100 integers? There are 14 of them.
They are 2n+2
It is 100*(100+1)/2 = 50500.
there are 999 - 100 + 1 = 900 positive triple digit positive integers, between 100 and 999.(e.g. there are 102 - 100 + 1 = 3 triple digit integers between 100 and 102,namely 100, 101 and 102.)multiply that by 2 to take in consideration of the negative integers,you have 1800 triple digit integers.
33/100 !
(7*100*101)/2 = 35,350 jpacs * * * * * What? How can there be 35,350 integers in the first 100 integers? There are 14 of them.
48
They are 2n+2
It is 100*(100+1)/2 = 50500.
2550
203
you can say that 2/10 of every 10 integers is divisible by 5, so multiplying 2/10 by 100, giving you 200/1000 total integers are divisible by 5. half of all integers are odd, so divide 200/1000 by 2 is 100/1000, so you can correctly state that 100 odd integers under 1000 are divisible by 5.
If you mean "between" 100 and 1000 to mean excluding 100 and 1000 then 449. If you want to include 100 and 1000 then 451.
there are 999 - 100 + 1 = 900 positive triple digit positive integers, between 100 and 999.(e.g. there are 102 - 100 + 1 = 3 triple digit integers between 100 and 102,namely 100, 101 and 102.)multiply that by 2 to take in consideration of the negative integers,you have 1800 triple digit integers.
33/100 !
The sum of the integers from 1 to 100 inclusive is 5,050.
ℤ is the symbol for the set of all integers, that is {..., -2, -1, 0, 1, 2, ...}.