Decipher the equality : \[(\overline{VER}-\overline{IA})=G^{R^E} (\overline {GRE}+\overline{ECE}) \] assuming that the number $\overline {GREECE}$ has a maximum value .Each letter corresponds to a unique digit from $0$ to $9$ and different letters correspond to different digits . It's also supposed that all the letters $G$ ,$E$ ,$V$ and $I$ are different from $0$.
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jeneva
12.02.2015 13:07
I feel like my solution is kinda brute force. First,I claim that $G=1$ since, $GRE+ECE\geq 100+200=300$ and $VER-IA<1000$ then $G^{R^E}\leq 3$ if $G=3$ then $R=1$ and $RHS>3(300+100)>1000$ which is not possible. if $G=2$ then $R=1$ and $RHS>2(200+300)=1000$ which is not possible. Then $G$ must equal to $1$ Notice that $869-27=1^{9^6}(196+646)$ I'll prove that it is not possible for the case$G=9,E>6$ If $E=7$ we get $GRE+ECE=197+7C7>900$ ,so $V=9$ which is not possible. if $E=8$ we get $GRE+ECE=198+8C8>1000$ which is not possible. so, $GRE=196$ I'll prove that it is not possible for the case$GRE=196,C=5,7,8$ if $C=5$ we get $GRE+ECE=852$ then $VER-IA=869-17\rightarrow I=1$ which is not possible. if $C=7,8$ we get $GRE+ECE\geq 872$ but $VER-IA<869$ which is not possible. So, $C=4$. It means that $GREECE=196646$ is the maximum possible value as desired.
dogukan
21.10.2015 21:35
jeneva wrote: I feel like my solution is kinda brute force. First,I claim that $G=1$ since, $GRE+ECE\geq 100+200=300$ and $VER-IA<1000$ then $G^{R^E}\leq 3$ if $G=3$ then $R=1$ and $RHS>3(300+100)>1000$ which is not possible. if $G=2$ then $R=1$ and $RHS>2(200+300)=1000$ which is not possible. Then $G$ must equal to $1$ Notice that $869-27=1^{9^6}(196+646)$ I'll prove that it is not possible for the case$G=9,E>6$ If $E=7$ we get $GRE+ECE=197+7C7>900$ ,so $V=9$ which is not possible. if $E=8$ we get $GRE+ECE=198+8C8>1000$ which is not possible. so, $GRE=196$ I'll prove that it is not possible for the case$GRE=196,C=5,7,8$ if $C=5$ we get $GRE+ECE=852$ then $VER-IA=869-17\rightarrow I=1$ which is not possible. if $C=7,8$ we get $GRE+ECE\geq 872$ but $VER-IA<869$ which is not possible. So, $C=4$. It means that $GREECE=196646$ is the maximum possible value as desired. R may be zero so I think answer is 701161.
sttsmet
15.05.2021 12:14
Exactly @above! @jeneva what if R=0???