Please help! Find all possible digits P and N satisfying the equation P+P+P+N=NP (bar over NP), where NP (bar over NP) is the two-digit number with N in the tens place and P in the ones place. The solution must include a convincing argument that you have found all possible answers!
\(\large \begin{align} \color{black}{\normalsize \text{NP (bar over NP) is the two-digit number } \hspace{.33em}\\~\\ \normalsize \text{with N in the tens place and P in the ones place} \hspace{.33em}\\~\\ \normalsize \text{this can be expressed as} \hspace{.33em}\\~\\ \overline{NP} =10N+P\hspace{.33em}\\~\\ \normalsize \text{so} \hspace{.33em}\\~\\ P+P+P+N=\overline{NP} \hspace{.33em}\\~\\ 3P+N=10N+P \hspace{.33em}\\~\\ 2P=9N \hspace{.33em}\\~\\ P=\dfrac{9}{2}N \hspace{.33em}\\~\\ \normalsize \text{here only even numbers will work for N} \hspace{.33em}\\~\\ \normalsize \text{as odd number cannot be divisible by 2} \hspace{.33em}\\~\\ so ~~N=\{2,4,6,8\cdot \cdot \cdot \cdot \infty \} \hspace{.33em}\\~\\ \normalsize \text{example, when N=2} \hspace{.33em}\\~\\ P=\dfrac{9}{2}N \hspace{.33em}\\~\\ P=\dfrac{9}{2}\times 2 \hspace{.33em}\\~\\ P=9 \hspace{.33em}\\~\\ \normalsize \text{check} \hspace{.33em}\\~\\ 9+9+9+2=29 \hspace{.33em}\\~\\ }\end{align}\)
@mathmath333 thank you so much!! but what did it mean when you put "check 9+9+9+2=29"? what does that show/prove?
the assumption was given in the problem that the where \(\large P\) is the unit's digit and \(\large N\) is the ten's digit of \(\large \overline{NP}\) \(\large \begin{align} \color{black}{P+P+P+N=\overline{NP} \hspace{.33em}\\~\\ \color{blue}{9+9+9}+\color{red}{2}=\color{red}{2}\color{blue}{9} }\end{align}\) just gave that to confirm that our approach and the answer is \(\bf correct\)
oh okay, so with that being said, is this a convincing enough argument that we have found all possible answers?
not their is restriction that NP is a two digit number so , u have to find even numbers for N upto which the NP(bar) is a two digit number only
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