The product of the ages, in years of three (3) teenagers is 4590. None of them have the same age. What are the ages of the teenager? UNDERSTANDING THE PROBLEM:

DEVISE A PLAN:

CARRY OUT THE PLAN:

REVISE THE SOLUTION:

Check: 15*17*18 = 4590 and all ages are distinct.

Explanation:

Let's say the three ages are x,y and z.

Since they are teenagers, this means x,y,z are values drawn from this set {13,14,15,16,17,18,19}

The number 4590 is even, so 2 is a factor. This means 2 is a factor of either x, y or z.

We see that 2 is a factor of 14,16 and 18. So it's possible that these are the x,y,z values we're after. But note how 5 is not a factor of any of these values, but it is a factor of 4590 (since the value ends in 0).

Furthermore, 3 is a factor of 4590 because the digits of 4590 add to a multiple of 3. We have 4+5+9+0 = 18 which is a multiple of 3.

The teenage years that are multiples of 3 are {15, 18}. Those two values either are even or have 3 as a factor or have 5 as a factor.

Let's see what we get when we multiply 15 and 18. So 15*18 = 270.

Divide 4590 over 270 to get 4590/270 = 17

Through luck we found the answer. Though usually with trial and error problems like this, it takes a bit more work.

So therefore the three ages are 15, 17, 18.

Here's another approach:

Start with the set {13,14,15,16,17,18,19} which again is the set of all possible teen ages.

Divide 4590 over each item in that set

We see that the denominators 15, 17, and 18 produce whole value results. Everything else leads to a non-whole decimal value. So we can eliminate 13, 14, 16, and 19.

The only thing left is 15,17,18 which are the three different ages.

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)

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