Smart McDonalds Employee

Perhaps this is not a good solution but I will try my best to explain it.

First of all, this question is the variant form of the famous math puzzle and the puzzle appears at movie Fermat's Room . By using the same logic, first thing that I should do is finding prime factorization of $16170$ . We obtain: $2, 3, 5, 7, 7, 11$ . Let $x$ , $y$ , and $z$ be the ages of three customers and $t$ be the age of the fry cook. Therefore, we have $$ $xyz=16170$ and $x+y+z=3t.$ $$ Do not use any sophisticated method, use the easiest one: trial and error , because the combinations are not so many. Just make sure that $x$ , $y$ , $z$ , and $t$ are reasonable numbers, which means all of them are integers and each number is less than $100$ . After using trial and error, we will find that there are two combinations that yield $x+y+z=90$ , which are: $11, 30, 49$ and $14, 21, 55$ . These two combinations that made the fry cook could not determine the each age of the customers at the first time he got a clue from the cashier. After he got the second clue: "None of them are the same age as you!", he could determine the answer with 100% certainty, which is: $14, 21, 55$ since $$ $t=\frac{1}{3}(11+30+49)=30\;\Rightarrow\;\text{the same age as one of the three customers}.$ $$ Thus, the positive difference between the age of the oldest customer and the youngest customers is $55-14=\boxed{41}$ .

Anyway, if you are interested to try 'similar' problem, you may want to try this one: Two Math Professors .

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

First, let's try to figure out what is going on:

Since the fry cook knows how old he is, he knows both the sum and the product of the ages. Additionally, his age is the average of the three ages (make sure you see why). So he knows the sum and product, yet he still doesn't know their ages! That means there must be more than one case!

How can we find these cases?

The problem implies that there exists a case with the average being the fry cook's age. Since one is his age, one must be k larger than his age, and the other must be k less. If the fry cook's age is f, then we have $(f-k)(f)(f+k)=f^3-fk^2=16170$ . Here it gets a bit guess and checky, but I tried to be nice when I wrote the problem, so the first thing most people try (f=30) works so we have $27000-30k^2=16170\implies{k=19}$ so we obtain the triple (11,30,49), and this sums to 90!

How to find the other case?

Not sure of a good way actually. All we have to do is find integers (a,b,c) such that abc=16170 and a+b+c=90 which after a bit of guess and check, we find is (14,21,55), so the difference is $\boxed{41}$

I had made this question by easiest method :-

as I had firstly taken out the positive divisors which are 2,3,5,7,11 after that the as we know the sum of ages is three times the cook therefore the sum of ages of three customers should be a multiple of 3 and when the sum is divided by 3 it should not be equal to the age of any customers

searching that qualification you will get only 4 sets (10,21,77) , (55,21,14) , (65,35,7) , (70,21,11)

In these you will find the most appropriate the 2nd set which is (55,21,14) .

I have taken this set because of following reasons:-

1) this can fulfill all the qualifications mentioned above. 2)this the set in which the age of younger customer is highest. 3)this is also the set which can be taken as a example.

OK ..i did this question by trial and error, but till a trick left...u might come across the combination 35,22,21 which satisfies all the conditions ..but that cannot the ans ..Because the sum of these 3 is 78 ..and no other combination has a sum 78 ..so the person would not need the 3rd hint ..but because the 3 rd hint was given we had to reject this odered triplet