orl wrote: A magician asked a spectator to think of a three-digit number $ \overline{abc}$ and then to tell him the sum of numbers $ \overline{acb}$, $ \overline{bac}$, $ \overline{bca}$, $ \overline{cab}$, and $ \overline{cba}$. He claims that when he knows this sum he can determine the original number. Is that so? The sum is $ 122a+212b+221c$ and the question is to know if $ 122a+212b+221c=122a'+212b'+221c'$ with $ a,b,c,a',b',c'\in\{0,1,2,3,4,5,6,7,8,9\}$ implies $ a=a'$, $ b=b'$ and $ c=c'$ So the question is to know if $ 122u+212v+221w=0$ with $ u,v,w\in\{-9,-8,..., 8, 9\}$ has a solution different from $ u=v=w=0$ This equation implies $ w=2t$ even and so $ 61u+106v+221t=0$ with $ u,v\in\{-9,-8,..., 8, 9\}$ and $ t\in\{-4,-3,-2,...,2,3,4\}$ Taking this equality mod $ 61$, we get that : $ v\in\{-9,-8,..., 8, 9\}$ $ \implies$ $ 106v\in\{0,3,6,10,13,16,19,22,26,29,32,35,39,42,45,48,51,55,58\}\pmod {61}$ $ t\in\{-4,-3,-2,...,2,3,4\}$ $ \implies$ $ -221t\in\{0,8,15,23,30,31,38,46,53\}\pmod{61}$ And so, since $ 106v=-221t\pmod{61}$, $ v=t=0\pmod{61}$ and so $ u=v=w=0$ And so the result : the magician is right.

Denote $ a+b+c = \sigma$. By adding all the numbers the magician gets, he obtains $ 222\sigma - \overline{abc}$. Assume the same value is obtained as $ 222\sigma' - \overline{a'b'c'}$ with, let's say, $ \sigma \geq \sigma'$. Then $ \overline{abc} - \overline{a'b'c'} = 222(\sigma - \sigma') = 222k$. Now, from $ \sigma - \sigma' = k$ one gets $ (c-c') = k - (a-a') - (b-b')$, while $ 221k = \overline{abc} - \overline{a'b'c'} - k = 100(a-a') + 10(b-b') + (c-c') - k = 99(a-a') + 9(b-b')$. This means that $ 9$ divides $ k$, only possible when $ k = 0$, since $ 222k = \overline{abc} - \overline{a'b'c'} < 1000$. Therefore $ \overline{abc} = \overline{a'b'c'}$, and so the number is uniquely determined.