36 is a perfect square number.
How about: 36+61 = 97
They are: 36 and 64
It is 9+36 = 45
x2 + 9 = 36
The number is 4 (4+3) * (4+3) = 49 (4+2) * (4+2) = 36 49 - 36 = 13
sum of 14th square number and 10th square number
"the sum of 2" and "twice" "a number" "is" "36"2+2x=36 â–
There are two solution: -3 and +2.
64 = 28 + 36
Part1: Finding probability of getting sum as a perfect square. Maximum sum of both the dice is (6+6) equal to 12. Up to 12, the perfect squares are: 1, 4 and 9. Getting a sum of 1 from two dice is not possible. So, we are left with 4 and 9. To get 4, the combination can be: (2,2) or (1,3) or (3,1). This means, to get the sum as 4, the probability is [3/36]. To get 9, the combination can be: (3,6) or (6,3) or (5,4) or (4,5). This means, to get the sum as 9, the probability is [4/36]. Therefore,the total probability of getting the sum as a perfect square is: [(3/36)+(4/36)]=[7/36]. Part2: Finding the probability of getting sum as an even number. The possible even numbers can be 2, 4, 6, 8, 10 and 12. But, as 4 is already considered in part1, it should be ignored in this case. The probability of getting sum as 2 is: [1/36] The probability of getting sum as 6 is: [5/36] The probability of getting sum as 8 is: [5/36] The probability of getting sum as 10 is: [3/36] By adding all the above, the probability of getting sum as an even number (ignoring 4) is: [(1/36)+(5/36)+(5/36)+(3/36)]=[14/36]. From part 1 and part 2, we get the total probability as [(7/36)+(14/36)]=[7/12]=0.583333.
10^2 = 6^2 + 8^2 ie 100 = 36 + 64
25 is a square number, and the sum of 9 and 16 is 25. 9 and 16 are also square numbers.