
4. If we divide the number 13 by the three numbers 5,7 , and 9 , then these divisions leave remainders: when dividing by 5 the remainder is 3 , when dividing by 7 the remainder is 6 , and when dividing by 9 the remainder is 4 . If we add these remainders, we obtain $3+6+4=13$, the original number.

(a) Let $n$ be a positive integer and let $a$ and $b$ be two positive integers smaller than $n$. Prove: if you divide $n$ by $a$ and $b$, then the sum of the two remainders never equals $n$.

(b) Determine all integers $n>229$ having the property that if you divide $n$ by 99,132 , and 229 , the sum of the three remainders is $n$.

## 4. Version for klas 6

(a) Let $r$ be the remainder upon dividing $n$ by $a$. We will first prove that $rn$, we have $r=n-a$ (since we already knew that $a229>t$. We know that $r+s+t=n$, and hence $n-t=r+s$. We can conclude that $r+s$ is a positive multiple of 229 . Since $99+132<2 \cdot 229$, we have $r+s<2 \cdot 229$, which implies that we must have $r+s=229$.

Since $r \leqslant 98$ and $s \leqslant 131$, the fact that $r+s=229$ implies that $r=98$ and $s=131$. Therefore, the number $n+1$ is divisible by both 99 and 132 , and hence by their least common multiple $\operatorname{lcm}(99,132)=\operatorname{lcm}(9 \cdot 11,3 \cdot 4 \cdot 11)=4 \cdot 9 \cdot 11=396$. Also, from $n=229+t$ and $t<229$ we deduce that $n+1 \leqslant 458$. It follows that the only possibility is $n+1=396$, hence $n=395$.

When $n=395$ the three remainders are $r=98, s=131$, and $t=166$, and indeed satisfy the equation $n=r+s+t$.
