# Birthdays

• Aug 28th 2008, 03:15 PM
Obsidantion
Birthdays
Cool puzzle taken from a magazine:
"Three children of different ages share the same birthday. On one of their birthdays, one of their ages was the sum of the other two ages. On another birthday a few years later, the youngest observed that one of their ages was half the sum of the other two ages. When the number of years since the first occasion was half the sum of the ages on the first occasion, one celebrated her 18th birthday. What birthdays were the other two celebrating at this time?"
• Aug 29th 2008, 06:06 PM
Soroban
Hello, Obsidantion!

An interesting problem ... a bit tricky for a magazine, though.

Quote:

Three children of different ages share the same birthday.
On one of their birthdays, one of their ages was the sum of the other two ages.
On another birthday a few years later, one of their ages was half the sum of the other two ages.
When the number of years since the first occasion was half the sum of the ages
on the first occasion, one celebrated her 18th birthday.
What birthdays were the other two celebrating at this time?

We assume that their ages are positive integers.

Let the ages be: .$\displaystyle a \:<\: b \:<\: c$

On the first occasion: .$\displaystyle c \:=\:a+b\;\;{\color{blue}[1]}$

Their ages were: .$\displaystyle \begin{array}{|c|c|c|} a & b & a +b \end{array}$

A few years later: .$\displaystyle b \:=\:\frac{a+c}{2}$ . . . $\displaystyle b$ is the average of $\displaystyle a$ and $\displaystyle c$

Substitute [1]: .$\displaystyle b \:=\:\frac{a + (a+b)}{2} \quad\Rightarrow\quad b \:=\:2a$

Their ages were: .$\displaystyle \begin{array}{|c|c|c|} a & 2a & 3a \end{array}$

$\displaystyle \frac{a+b+c}{2}$ years after the first occasion .$\displaystyle = \;\frac{a+2a+3a}{2} \:=\:3a$ years later

They are all $\displaystyle 3a$ years older. .Their ages are: .$\displaystyle \begin{array}{|c|c|c|} 4a & 5a & 6a \end{array}$

At this time, one of is 18 years old . . . It must have been the oldest, $\displaystyle 6a$
. . $\displaystyle 6a = 18 \quad\Rightarrow\quad a = 3 \quad\Longrightarrow\quad 4a = 12,\;5a = 15$

Therefore, the other two were 12 and 15 years old.

• Aug 29th 2008, 11:00 PM
Obsidantion
Quote:

Originally Posted by Soroban
Hello, Obsidantion!

An interesting problem ... a bit tricky for a magazine, though.

We assume that their ages are positive integers.

Let the ages be: .$\displaystyle a \:<\: b \:<\: c$

On the first occasion: .$\displaystyle c \:=\:a+b\;\;{\color{blue}[1]}$

Their ages were: .$\displaystyle \begin{array}{|c|c|c|} a & b & a +b \end{array}$

A few years later: .$\displaystyle b \:=\:\frac{a+c}{2}$ . . . $\displaystyle b$ is the average of $\displaystyle a$ and $\displaystyle c$

Substitute [1]: .$\displaystyle b \:=\:\frac{a + (a+b)}{2} \quad\Rightarrow\quad b \:=\:2a$

Their ages were: .$\displaystyle \begin{array}{|c|c|c|} a & 2a & 3a \end{array}$

$\displaystyle \frac{a+b+c}{2}$ years after the first occasion .$\displaystyle = \;\frac{a+2a+3a}{2} \:=\:3a$ years later

They are all $\displaystyle 3a$ years older. .Their ages are: .$\displaystyle \begin{array}{|c|c|c|} 4a & 5a & 6a \end{array}$

At this time, one of is 18 years old . . . It must have been the oldest, $\displaystyle 6a$
. . $\displaystyle 6a = 18 \quad\Rightarrow\quad a = 3 \quad\Longrightarrow\quad 4a = 12,\;5a = 15$

Therefore, the other two were 12 and 15 years old.

It was a Mensa magazine. Good work, you are right.