1 and 54 - added together.
55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
The sum of the numbers 1 through 10 is 55.
There are 11C2 = 11*10/(2*1) = 55 combinations.
There are 56C5 = 56*55*54*53*51/(5*4*3*2*1) = 3,819,816 combinations.
1 and 54 - added together.
55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
The sum of the numbers 1 through 10 is 55.
There are 11C2 = 11*10/(2*1) = 55 combinations.
11
1 2 5 11 22 55 110
It is 10*(10+1)/2 = 55
The sum_of_a_series = 1/2 x number_in_series x (first_number + last_number) The first odd number is 2 x 1 - 1 = 1, the 55th odd number is 2 x 55 - 1 = 109 Sum = 1/2 x 55 x (1 + 109) = 1/2 x 55 x 110 = 552 = 3025 It can be seen that the sum of the first n consecutive odd numbers is always n2: sum = n/2(1 + (2n - 1)) = n/2(2n) = n2
There are 56C5 = 56*55*54*53*51/(5*4*3*2*1) = 3,819,816 combinations.
Well, darling, that's an easy one. The two prime numbers that make 55 are 5 and 11. 5 is a prime number because it can only be divided by 1 and itself, and 11 is also a prime number for the same reason. So, 5 times 11 equals 55. Math made fun, you're welcome!
The two numbers are 125 and 55 You can make 2 equations 1) x + y = 180 2) x+ 40 =3y If you rearrange equation 2 you get x=3y-40 You can now put this in for x in equation 1 (3y-40) +y = 180 3y + y = 180 + 40 4y = 220 y=55 If you plug this in for y in equation 1 you get X + 55 = 180 x= 125 Check with equation 2 x+ 40 =3y 125+40=3x55 165=165 Therefore the two numbers are 125 and 55
1, 2, 5, 10, 11, 22, 55, 110