My Friend, 4 kids and gold and silver coins

Question

The other day I met an old college friend in a coffee shop. We both were students in the Math department of that college. Here was our conversation (after the small talk)

We are meeting first time after college! She said

Yes. What is it, 25-30 years?

Yup. Just celebrated my __th Birthday.

Wow. That is a milestone.

Everything downhill from here!

So how many kids do you have?

Four

How old are they?

You will have to guess that, Mr. Math from my college.

OK

So on my Birthday, I gave them some gold and silver coins. I had same number of gold and silver coins each. Same number (each) as my age.

So number of gold coins equals number of silver coins equals your age?

Right. This is how I distributed the Gold coins. Each kid got same number of gold coins as their age number. I used up all the gold coins that way.

How about the silver coins?

I did it differently. Each kid got silver coins equal to the next higher age number of their sibling. Used up all silver coins also. Can you guess their ages?

Higher age number of just one sibling?

Right

No I need more info.

Two kids got same total number of coins each. The other two kids also got same total number of coins each but this number was different than what the other 2 got. Can you guess?

No. I need more.

No. I think you just need a paper and pencil. Give me a call when you solve it. I gotta go.

You in a rush?

Yup. Kids are home. The teenager needs a ride.

Bye then

After a few minutes it hit me. I knew their ages and how many coins each kid received. Do you? Can you do it with just paper and pencil?

Answer 1

This answer uses the following assumptions about statements in the riddle:

Assumption 1: All children must be younger than the maximum 30 year period that the friends had not met.

Assumption 2: The fact that one of them was referred to as "the teenager" implies that exactly one of them is from 13-19 years old.

Assumption 3: If two children were twins, neither would count as the others next oldest child for the purpose of the silver coin amount. However, in the case of two children born in the same year without being twins, the older would be used for the younger's silver coin amount.

Assumption 4: The parent must be at least age 43 (18 to enter college + 25 period they have not met).

Assumption 5: The parent is an age that is a multiple of 10 to qualify as "a milestone".

First conclusion:

The second and third oldest must be twins.

This is because:

Assuming all 4 children have distinct ages, and that their ages are A, B, C, and D from highest to lowest. The number of gold coin is equal to A+B+C+D. The number of silver coins is equal to A+B+C.
For this to work, D must be 0. However, if D=0, then the youngest obtains C coins. The others would obtain A coins, A+B coins, and B+C coins. The only way for these to form two matching pairs is for A=B=C. This contradicts Assumption 2.

Assuming the youngest 2 are twins, and the ages of siblings are A, B, C, and C. The number of gold coins would be A+B+C+C and the number of silver coins would be A+B+B. In this case, the children get totals of A, A+B, B+C, and B+C coins. These can not form two pairs without B=C=0.

Assuming the oldest 2 are twins, and the ages of siblings are A, A, B, and C. The number of gold coins is A+A+B+C. The number of silver coins is A+B. This could only be the case if A=B=C=0.

With distinct ages, the two eldest being twins, and the two youngest being twins disproven, the second and third children must be twins.

Moving on to the coins themselves:

Let the siblings' ages be A, B, B, and C from oldest to youngest. Let the parent's age, the total number of silver coins, and the total number of gold coins be D.
D = A + B + B + C
D = 0 + A + A + B
Given B+A = B+A, the other pair would have to be 0+A = B+C. Coincidentally, A=B+C can be obtained from the combining and simplifying the two above formulae.

Since only one child is a teenager, it must be the eldest or the youngest. Assuming it is the youngest, they are at least 13 and the twins are at least 20. A=B+C, the oldest would be at least 33 in this case. This contradicts Assumption 1.

Given that the eldest is the teenager, 12<A<20 and C<B<13.
The minimum parent's age is at A=13, B=7, C=6. The maximum parent's age is at A=19, B=12, C=7. This makes the parent's possible age range 38 - 50.

The only value in this range that fits Assumptions 4 & 5 is 50.

This leads to the conclusion that:

The children's ages are 19, 12, 12, and 7.
They received 19G+0S, 12G+19S, 12G+19S, and 7G+12S.

Answer 2

It seems like an answer is

the youngest is 0 (i.e., less than 1), and the other three are the same age as each other. The only factor that affects the triplets' ages is that they are a teenager and have to each be $\frac13$ of the age of the parent. If it’s been ~25-30 years since the parent was in college, that would make them ~ 50, so the teenagers could reasonably be 15-18 years old.

Explanation (sorry for terrible formatting):

Number the children $1$ (youngest) through $4$ (oldest).  Child $i$ gets $\text{age}_i$ gold coins and $\text{age}_{i+1}$ silver coins (with child $4$ getting $0$ silver coins). \begin{align}\text{sum(gold)} &= \text{age}_1 + \text{age}_2 + \text{age}_3 + \text{age}_4\\\text{sum(silver)} &= \text{age}_2 + \text{age}_3 + \text{age}_4 + 0\end{align} $$\text{sum(gold)} = \text{sum(silver)} \implies \text{age}_1=0$$
From there, the four childrens' coin sums form two pairs, i.e.

    a_2 = a_2 + a_3 AND a_3 + a_4 = a_4 

 OR a_2 = a_3 + a_4 AND a_2 + a_3 = a_4 

 OR a_2 = a_4       AND a_2 + a_3 = a_3 + a_4 

The first pair implies

a3 = 0 => a2 = 0 => a4 = age of parent 

Not great.

The second pair also implies

a3 = 0 => a2 = 0 => a4 = 0

Also not great.

The last pair implies a2, a3, and a4 are all equal. That works.

So, the ages and coins are

(Using 18 for the triplet age)
Youngest: Age: 0, Gold: 0, Silver: 18
Youngest triplet: Age: 18, Gold: 18, Silver: 18
Middle triple: Age: 18, Gold: 18, Silver: 18
Oldest triplet: Age: 18, Gold: 18, Silver: 0