36
There are 36 permutations of two dice. Only one of them has a sum of two. Then probability, then, of rolling a sum of more than two is 35 in 36, or about 0.9722.
Because the sum of its proper divisors is greater than 36.
This is easier to solve by looking at the reverse problem, what is the probability of the sum being 11 or more. Out of the 6*6 = 36 outcomes, three (5,6), (6,5) and (6,6) satisfy this event. So the probability of getting a sum of 11 or more is 3/36. So the probability of less than 11 is 1-Pr(>=11) = 1 - 3/36 = 33/36 = 11/12 or 0.91667
There 36 possibilities, and out of the 36, there are 26 possibilities of getting a sum less than 9. 26/36 = 2/3
36
There are 36 permutations of two dice. Only one of them has a sum of two. Then probability, then, of rolling a sum of more than two is 35 in 36, or about 0.9722.
It is (36+28)-15 = 49
Because the sum of its proper divisors is greater than 36.
This is easier to solve by looking at the reverse problem, what is the probability of the sum being 11 or more. Out of the 6*6 = 36 outcomes, three (5,6), (6,5) and (6,6) satisfy this event. So the probability of getting a sum of 11 or more is 3/36. So the probability of less than 11 is 1-Pr(>=11) = 1 - 3/36 = 33/36 = 11/12 or 0.91667
There 36 possibilities, and out of the 36, there are 26 possibilities of getting a sum less than 9. 26/36 = 2/3
-36
The number is 4 (4+3) * (4+3) = 49 (4+2) * (4+2) = 36 49 - 36 = 13
Assuming you mean how to solve what three consecutive numbers sum to some value: Divide the sum by three to get the middle number. The other two numbers are one less than this and one more than it. example: Which three consecutive numbers sum to 108? 108 ÷ 3 = 36 → The other two numbers are 36 - 1 = 35 and 36 + 1 = 37 → The three numbers are 35, 36, 37. 35 + 36 + 37 = 108 as required.
The sum of 120 and 36 is 156.
13 and 7
-- Their sum is (36 + 34) = 70, not 72. -- Their difference is (36 - 34) = 2, not 4. -- Other than that, 36 and 34 are a terrific answer ! But 38 and 34 would be better.