I am thinking of 3 different whole numbers which have a product of 36. 5 more than the sum of the 3 numbers is a perfect square. what is the sum of my 3 numbers?
Make a list of all possible combinations of three numbers with a product of 36. There are not very many.
The prime factors are 1*2*2*3*3. Make up the combinations from them, in groups of 3. Then compute the sums.
3 2 9 Sum = 15
18 1 2 Sum = 21
2 3 6 Sum = 11
3 3 4 Sum = 10
Add five to those sums. One number you get is a perfect square
To find the sum of the three numbers, we need to first determine what those numbers are. Let's break down the problem into steps:
Step 1: Find all the possible sets of three whole numbers whose product is 36.
To do this, we can start by listing the factors of 36:
1, 2, 3, 4, 6, 9, 12, 18, 36
Since we want three numbers, we can pair up these factors in different ways and calculate the third number by dividing 36 by the product of the first two. For example:
Pair 1: 1, 1, 36
Pair 2: 1, 2, 18
Pair 3: 1, 3, 12
Pair 4: 1, 4, 9
Pair 5: 1, 6, 6
Step 2: Calculate the sum of each set of three numbers.
Sum of Pair 1: 1 + 1 + 36 = 38
Sum of Pair 2: 1 + 2 + 18 = 21
Sum of Pair 3: 1 + 3 + 12 = 16
Sum of Pair 4: 1 + 4 + 9 = 14
Sum of Pair 5: 1 + 6 + 6 = 13
Step 3: Determine which sum is 5 more than a perfect square.
To do this, we can calculate the square root of each sum + 5 and check if the result is a whole number. For example:
Square root of (38 + 5) = √43 ≈ 6.56
Square root of (21 + 5) = √26 ≈ 5.10
Square root of (16 + 5) = √21 ≈ 4.58
Square root of (14 + 5) = √19 ≈ 4.36
Square root of (13 + 5) = √18 ≈ 4.24
We can see that only the sum of Pair 3 (16) has a square root that is a whole number (4).
Therefore, the sum of the three numbers you were thinking of is 16.
Note: In this case, the possible set of three whole numbers with a product of 36 and a sum of 16 is 1, 3, and 12.