\section*{Problem 8}

Find four different three-digit numbers (in base 10) starting with the same digit, such that their sum is divisible by three of the numbers.


\section*{Solution}

Answer: \(108,117,135,180\). Sum \(540=108 \cdot 5=135 \cdot 4=180 \cdot 3\).

Try looking for a number of the form 3.4.5.n. We want \(12 n, 15 n\) and 20 n to have the same first digit. If the first digit is 1 , this requires \(n=9\). We must now check that the fourth number which must be \(60 n-12 n-15 n-20 n=13 n\) also has three digits starting with 1 . It does, so we are home. [In fact, in this case the first digit must be 1 , since \(20 n>3 / 212 n\).]

