MathRevolution wrote:

If m and n are positive integers, is \(10^n+m\) divisible by 3?

1) n=1

2) m=2

\(m,n\,\,\, \geqslant 1\,\,\,{\text{ints}}\)

\(\frac{{{\text{1}}{{\text{0}}^n} + m}}{3}\,\,\,\mathop = \limits^? \,\,\operatorname{int}\)

To prove (1) is insufficient, let us present what our method calls a BIFURCATION (ALGEBRAIC, in this case):

\(\left( 1 \right)\,\,\,n = 1\,\,\,\left\{ \begin{gathered}

\,{\text{Take}}\,\,m = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\, \hfill \\

{\text{Take}}\,\,m = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\

\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,\left\{ \begin{gathered}

m = 2 \hfill \\

{10^{n\,\,\,}} = \left( {{{10}^n} - 1} \right) + 1 \hfill \\

\end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\frac{{{\text{1}}{{\text{0}}^{n\,}} + m}}{3}\,\,\, = \,\,\frac{{\left( {{{10}^n} - 1} \right) + 1 + 2}}{3}\,\,\,\,\mathop { = \,}\limits^{n\,\, \geqslant \,\,1} \,\,\,\,\frac{{\overbrace {\left( {{\text{1}}{{\text{0}}^n} - 1} \right)}^{\left\langle {9 \ldots 9} \right\rangle }}}{3} + 1\,\,\,\, = \,\,\,\,\left\langle {3 \ldots 3} \right\rangle + 1\,\,\,\, = \,\,\,\,\operatorname{int} \,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)

The above follows the notations and rationale taught in the GMATH method.

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Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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