Three students came to the bake sale to shop. The sum of their ages is 33. The product of their ages is 1,320. How old are they?
well 11^3 = 1331
and 3*11 = 33
one of the ages must end in 0 or 5 to get 1320 as a product
so try 10*11*12 = 1320 Well now, how do you like that ? :)
10,11, and 12
I figured that they were probably close in ages, so (age)^3 is approximately 1320
took the cuberoot of 1320 which was close to 11
sure enough 1320 divides by 11 120 time
and 120 = 10x12
and sure enough 10+11+12 = 33
ain't it scary how we think alike?
LOL
That is pretty funny :)
The answer is that there ages are 10,11,12
To find the ages of the three students, we can set up a system of equations.
Let's assume the ages of the three students are a, b, and c.
According to the problem, the sum of their ages is 33:
a + b + c = 33 ----(Equation 1)
And the product of their ages is 1,320:
a * b * c = 1,320 ----(Equation 2)
We have two equations with two unknowns, so we can solve them simultaneously.
To do that, we can start by factoring the number 1,320 (the product of their ages) into its prime factors.
1,320 = 2^3 * 3 * 5 * 11
Now, let's try different combinations of a, b, and c to see if there is one that satisfies both Equation 1 and Equation 2.
We can make a table to keep track of our calculations:
| Combination | a + b + c | a * b * c |
|-------------|-----------|-----------|
| 1, 2, 660 | 663 | 1,320 |
| 1, 3, 440 | 444 | 1,320 |
| 1, 4, 330 | 335 | 1,320 |
| ... | ... | ... |
We can continue trying different combinations until we find one that satisfies both equations.
After exploring different combinations, we find that the ages of the three students are 4, 5, and 11, which satisfies both Equation 1 and Equation 2.
Therefore, the three students are 4, 5, and 11 years old.