Question

If (HE)^H=SHE, where the alphabets take the values from (0-9) & all the alphabets are single digit then find the value of (S+H+E)?

Answer 1

$\underline{\underline{\Huge{\textbf{Solution:}}}}$

Given that,

The alphabets E, S, H can take any value from 0-9

$\LARGE{\mathsf (HE)^H = SHE}$

$\underline{\Huge{\textbf{Step-1:}}}$

$\textsf{Note the Range of values that RHS can take}$

From this it can be concluded that,

$\begin{itemize}\begin{enumerate}\item \textsf{RHS is a 3-digit No. }\item \textsf{None of the digits in the RHS are same}\end{enumerate}\end{itemize}$

$\implies \sf SHE \in (111, 999)$———[1]

$\underline{\Huge{\textbf{Step-2:}}}$

$\textsf{Find the Range of values that HE(in LHS) can take}$

$\quad\maltese\;$Consider the least possible value for E and using $\textsf{trial and Error Method}$ to find the Range of values within which LHS falls so that [1] gets satisfied.

Here, Least possible value for an alphabet = 0

(Given that an alphabet can take any value from 0-9)

$\therefore Take\: E = 0$

$\begin{itemize}\item Say H = 1\begin{itemize}\item \rm HE^H \implies 10^1 =10\begin{itemize}\item 10 \notin (111, 999) \item \therefore\rm H \neq 1 \end{itemize}\end{itemize}\end{itemize}$

$\begin{itemize}\item Say H = 2\begin{itemize}\item \rm HE^H \implies 20^2 = 400\begin{itemize}\item 400 \in (111, 999) \item \therefore\rm H = 2 \end{itemize}\end{itemize}\end{itemize}$

$\begin{itemize}\item For H > 2\begin{itemize}\item \sf HE \notin (111, 999) \item \because\texttt{RHS has to be a 3-digit No. } \end{itemize}\end{itemize}$

$Clearly\: HE \in (20, 29]$

$\implies H= 2$ ———[2]

$\quad$In the interval, 20 was excluded due to the fact that the Result obtained for $(20)^2$ has two similar digits which doesn't satisfy the RHS (which has different values.)

$\underline{\Huge{\textbf{Step-3:}}}$

$\textsf{Find the values that E(in LHS) can take}$

$\mathsf{HE^H =\underbrace{HE\timesHE \times HE\times HE \cdots\cdots HE}_{H\: times}}$

Consider the terms contributing in LHS to Result the Unit digit E.

$(H\underline{E})^H = SH\underline{E}$

$\underbrace{E\timesE \times E\times E \cdots\cdots E}_{H\: times}\longrightarrow\textsf{Resulted Unit digit E}$

$\quad\maltese\;$It is observed that,

$\textbf{E is number which when multiplied} \\\textbf{ by itself a certain number of times results} \\ \textbf{the same number at it's unit's place. }$

If the No. has Unit digit 1, 5, 6, then any power of that No. will have the same unit digit.

In other words,

$\boxed{\begin{minipage}{5cm}\maltese \; \; \textsf{Let xy^z = k ,\\\\ If \; y } =\begin{cases}1 &\textsf{Unit digit in k is 1}\\5 &\textsf{Unit digit in k is 5}\\6 &\textsf{Unit digit in k is 6}\end{cases}\; \; \end{minipage}}$

$\therefore E= 1\: or\: 5\: or\ 6$ ———[3]

$\underline{\Huge{\textbf{Step-4:}}}$

$\textsf{Check the possibilities by combining}\\\textsf{the results of [2] and [3] and note the result}$

Combining [2] & [3],

We have

$\bf HE = \begin{cases}21\\25\\26\end{cases}$

$\underline{(HE)}^H= S\underline{HE}$

$\quad\maltese\;$It is also observed that,

$\textbf{the tens Digit and Unit digit in}\\ \textbf{base of LHS is same as the tens Digit}\\ \textbf{and unit's digits in result of product.}$

21 doesn't satisfy this condition $\sf(21^2= 441)$

26 doesn't satisfy this condition $\sf(22^2 = 484)$

$\textsf{The only Number that satisfies this}\\\textsf{condition is \Large{\textbf{25.}}}$

$\boxed{\huge{\boxed{\LARGE{\mathsf ( \underline{25})^2 = 6 \underline{25}}}}}$

$\therefore \begin{Large}\begin{array}{ccc}\bf S = 6;&\bf H = 2 ;& \bf E= 5\end{array}\end{Large}$

$\underline{\Huge{\textbf{Step-5:}}}$

$\textsf{Find S+H+E}$

$\underline{\textsf{S+H+E =}}$

$\underline{\textsf{6+2+5 = 13}}$

$\blacksquare\; \; \textsf{The value of S+H+E = \Huge{\underline{\LARGE{\textbf{13}}}}}$