# A puzzle

• Dec 9th 2006, 09:13 AM
getmath
A puzzle
John was born in 19ab. The two digit number ab when divided by 2 gives his age on his birthday in 1999.

Can you tell me how old was he?
• Dec 9th 2006, 09:26 AM
TriKri
Quote:

Originally Posted by getmath
John was born in 19ab. The two digit number ab when divided by 2 gives his age on his birthday in 1999.

Can you tell me how old was he?

1999 - age = 1900 + ab
age = ab/2

1999 - ab/2 = 1900 + ab
99 = 1.5 ab
ab = 66
age = 33
• Dec 9th 2006, 11:33 AM
Soroban
Hello, getmath!

Another approach . . .

Quote:

John was born in $\displaystyle 19ab$.
The two-digit number $\displaystyle ab$ when divided by 2 gives his age on his birthday in 1999.

Can you tell me how old was he?

Let $\displaystyle x\,=\,ab$

He was born in the year $\displaystyle 1900 + x$

$\displaystyle \frac{x}{2}$ years later, it is 1999.

Hence: .$\displaystyle (1900 + x) + \frac{x}{2} \:=\:1999\quad\Rightarrow\quad x\:=\:66$

Therefore, he was born in $\displaystyle 1966$; he was $\displaystyle 33$ in 1999.

• Dec 9th 2006, 12:45 PM
OReilly
Quote:

Originally Posted by TriKri
1999 - age = 1900 + ab
age = ab/2

1999 - ab/2 = 1900 + ab
99 = 1.5 ab
ab = 66
age = 33

Use of symbol "ab" here is not good because it can represent a*b.

You could've use x=ab as Soroban did.

Beside that, everything is ok.

I used formula
$\displaystyle 1999 - \frac{{10a + b}}{2} = 1900 + 10a + b$
where $\displaystyle 10a + b$ is that ab two digit number.
• Dec 9th 2006, 07:40 PM
getmath
Quote:

Originally Posted by OReilly
Use of symbol "ab" here is not good because it can represent a*b.

You could've use x=ab as Soroban did.

Beside that, everything is ok.

I used formula
$\displaystyle 1999 - \frac{{10a + b}}{2} = 1900 + 10a + b$
where $\displaystyle 10a + b$ is that ab two digit number.

Mathematically you are correct.