Is there an equation for this?

[cris1102]

My son's math homework included the question ... "Three students came to the bake sale to shop. The sum of their ages is 33. The product of their ages is 1320. How old are they.

I know the answer is 10, 11, and 12 but I pretty much got lucky figuring it out. My question is, is there an equation he can use to figure out questions like this in the future?

Thanks

 

[stapel]

Since you would have only two equations in three unknowns, no, you cannot solve this strictly algebraically. You have to use the fact that your values will all be whole numbers, and also that students are generally between, say, five and twenty-five.

Note: For this sort of exercise, it is usually fairly helpful to factor first. In this case, you have:

. . . . .1320 = 2 × 2 × 2 × 3 × 5 × 11

Since 2 × 11 = 22 is probably too big, and since 3 × 11 = 33 is almost certainly too big, you can start with guessing one of the students' ages as being 11. This leaves you with 2, 2, 2, 3, and 5. Since these students are presumably friends, the ages probably aren't 2 × 3 = 6 and 2 × 2 × 5 = 20. How many other ways can you distribute the factors?

Then also check that the sum is 33. :wink:

Eliz.

 

[Denis]

Solving for 3 unknowns requires 3 equations; there's only 2 given:
a + b + c = 33
abc = 1320
Are you sure that's the complete problem?

 

[cris1102]

Solving for 3 unknowns requires 3 equations; there's only 2 given:

a + b + c = 33
abc = 1320
Are you sure that's the complete problem?

To be given 3 equations makes a lot more sense but yes, what I posted was the complete word problem straight from his math workbook (go figure). So we started with ages around 10 since that is how old he is and like i said, we got lucky lol. Math was never my favorite subject in school and it's been a few years since I've had to do any so, I'm learning this all over again right along with him lol

thank you Dennis and stapel for taking the time to respond

 

[Deleted member 4993]

Sometimes these questions "imply" a restriction - like all the ages are integers.

I would have also started "guessing" at 11 because 33/3 = 11 and \(\displaystyle 11^3\) = 1331