Constant Limit = 1000; % Sufficient digits. Constant Base = 10; % The base of the simulated arithmetic. Constant FactorialLimit = 365; % Target number to solve, 365! Array digit[1:Limit] of integer; % The big number. Integer carry,d; % Assistants during multiplication. Integer last,i; % Indices to the big number's digits. Array text[1:Limit] of character; % Scratchpad for the output. Constant tdigit[0:9] of character = ["0","1","2","3","4","5","6","7","8","9"]; BEGIN digit:=0; % Clear the whole array. digit[1]:=1; % The big number starts with 1, last:=1; %
2 equivalent fractions for 15 and 23 = 30/46, 45/69
The two fractions are -181/2 and 191/2.
The sum of those fractions is 9 over 10 or 9/10.
7272 * * * * * WRONG! Sum from 23 to 123 = (sum from 1 to 123) minus (sum from 1 to 22) = 123*124/2 - 22*23/2 = 7626 - 253 = 7373. NOT 7272!
You must convert the fractions to equivalent fractions with a common denominator, which in this case is 20.
2 equivalent fractions for 15 and 23 = 30/46, 45/69
The sum of two fractions will be equal to one when the numerator and the denominator of their sum are the same. Example: 1/3 + 2/3 = 3/3 or 1
The two fractions are -181/2 and 191/2.
The sum of those fractions is 9 over 10 or 9/10.
1/2 + 1 1/2=2
1/2 + 1/2 = 1
7272 * * * * * WRONG! Sum from 23 to 123 = (sum from 1 to 123) minus (sum from 1 to 22) = 123*124/2 - 22*23/2 = 7626 - 253 = 7373. NOT 7272!
the sum the sum
Consider a denominator of r; It has proper fractions: 1/r, 2/r, ...., (r-1)/r Their sum is: (1 + 2 + ... + (r-1))/r The numerator of this sum is 1 + 2 + ... + (r-1) Which is an Arithmetic Progression (AP) with r-1 terms, and sum: sum = number_of_term(first + last)/2 = (r-1)(1 + r-1)/2 = (r-1)r/2 So the sum of the proper fractions with a denominator or r is: sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2 Now consider the sum of the proper fractions with a denominator r+1: sum{r+1} = (((r+1)-1)/2 = ((r-1)+1)/2 = (r-1)/2 + 1/2 = sum{r) + 1/2 So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2 The first denominator possible is r = 2 with sum (2-1)/2 = ½; The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½; And there are 100 - 2 + 1 = 99 terms to sum So the required sum is: sum = ½ + 1 + 1½ + ... + 49½ = 99(½ + 49½)/2 = 99 × 50/2 = 2475
if you are adding two fractions that are both greater than 1/2, what must be true about the sum?
11/15.
1/3 + 1/2