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Julie ask her teacher, "how old were you in 2008?" "My age in 2008 was the sum of all the digits of my year of birth,' replied the teacher. How old was the teacher in 2008?

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We can safely assume that the teacher is more than eight years old, so the teacher was born in $19ab$ for some digits $a$ and $b$. The sum of the digits of the teacher’s birth year is therefore $10+a+b$. The year written $19ab$ is numerically $1900+10a+b$, so in $2008$ the teacher was


years old. Thus, $108-10a-b=10+a+b$, or $11a+2b=98$.

Now $2b$ and $98$ are both even, so $11a$ must be even, and therefore $a$ must be $0,2,4,6$, or $8$. On the other hand, $2b$ is at most $18$, so $11a=98-2b$ is at least $98-18=80$. And from here it takes just one small step to finish the solution.

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Assuming the teacher was born in the XX century, she cannot have more than 28 years (1+9+9+9). In this case she was born in 1980. From 1980 to 1989 the sum of the digits is strictly increasing. It is easy to find 1985 as a solution (23 years old).

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$$19AB + 10 + A + B = 2008\Longrightarrow 1000+ 900 + 11 A + 2B = 1998\Longrightarrow$$


Can you take it from here...and, of course, explain what was done above ?

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