Find a number $\mathrm{N}$ with five digits, all different and none zero, which equals the sum of all distinct three digit numbers whose digits are all different and are all digits of $\mathrm{N}$.
Answer: 35964

There are $4.3=12$ numbers with a given digit of $n$ in the units place. Similarly, there are 12 with it in the tens place and 12 with it in the hundreds place. So the sum of the 3 digit numbers is $12.111(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e})$, where $\mathrm{n}=\mathrm{abcde}$. So $8668 \mathrm{a}=332 \mathrm{~b}+1232 \mathrm{c}+1322 \mathrm{~d}+1331 \mathrm{e}$. We can easily see that $\mathrm{a}=1$ is too small and $\mathrm{a}=4$ is too big, so $\mathrm{a}=2$ or 3. Obviously e must be even. 0 is too small, so $\mathrm{e}=2,4,6$ or 8. Working mod 11, we see that $0=2 \mathrm{~b}+2 \mathrm{~d}$, so $\mathrm{b}+\mathrm{d}=11$. Working $\bmod 7$, we see that $2 \mathrm{a}=3 \mathrm{~b}+6 \mathrm{~d}+\mathrm{e}$. Using the $\bmod 11$ result, $\mathrm{b}=2, \mathrm{~d}=9$ or $b=3, d=8$ or $b=4, d=7$ or $b=5, d=6$ or $b=6, d=5$ or $b=7, d=4$ or $b=8, d=3$ or $\mathrm{b}=9, \mathrm{~d}=2$. Putting each of these into the $\bmod 7$ result gives $2 \mathrm{a}-\mathrm{e}=4,1,5,2,6,3,0,4$ mod 7. So putting $\mathrm{a}=2$ and remembering that $\mathrm{e}$ must be $2,4,6,8$ and that all digits must be different gives a, b, d, e = 2,4, 7, 6 or 2, 7, 4, 8 or 2, 8, 3, 4 as the only possibilities. It is then straightforward but tiresome to check that none of these give a solution for $\mathrm{c}$. Similarly putting $\mathrm{a}=4$, gives $\mathrm{a}, \mathrm{b}, \mathrm{d}, \mathrm{e}=3,4,7,8$ or $3,5,6,4$ as the only possibilities. Checking, we find the solution above and no others.