gracie wrote:
A palindrome is a number that reads the same forward or backward. If the first two digits of a four digit palindrome form a multiple of the last two digits, how many such four digit palindromes are there?
A. 0
B. 6
C. 9
D. 12
E. 18
Beautiful problem!
\(?\,\,\,\,:\,\,\,\,\# \,\,{\text{special}}\,\,{\text{palindromes}}\)
\(\underline {a \ne 0} \,\,\,\underline b \,\,\underline b \,\,\underline a\)
\(1{\text{st}}\,\,{\text{case:}}\,\,\,{\text{a}}\,\,{\text{ = }}\,{\text{b}}\,\,\, \Rightarrow \,\,\,9\,\,{\text{possibilities}}\,\,\,\,\,\left( {1111\,\,,\,\, \ldots \,\,,\,\,9999} \right)\)
\(2{\text{nd}}\,\,{\text{case:}}\,\,\,{\text{a}}\,\, \ne \,{\text{b}}\,\,\)
\({\text{1}} \leqslant \,\,\,{\text{a}}\,\,{\text{ < }}\,\,{\text{b}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {ab} \right\rangle < \left\langle {ba} \right\rangle \,\,\,\, \Rightarrow \,\,\,0 < \frac{{\left\langle {ab} \right\rangle }}{{\left\langle {ba} \right\rangle }} < 1\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{\left\langle {ab} \right\rangle }}{{\left\langle {ba} \right\rangle }} \ne \operatorname{int}\)
\(b = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{10a}}{a} = \operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,9\,\,{\text{possibilities}}\,\,\,\,\,\left( {1001\,\,,\,\, \ldots \,\,,\,\,9009} \right)\,\,\,\,\)
The problem ends here, in terms of GMAT environment: we found 18 possibilities and there are not choices greater than that. We are safe!
Now let´s proof the problem is really with the correct answer, LoL ... (The inspections below are not pretty, but at least there are only a few of them!)
\(1 \leqslant b < \,{\text{a}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{\,\left\langle {ab} \right\rangle }}{{\left\langle {ba} \right\rangle }} = \operatorname{int} \geqslant 1\,\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\,a\left( {10 - \operatorname{int} } \right) = b\left( {10\operatorname{int} - 1} \right)\)
\(\operatorname{int} = \left\{ \begin{gathered}
1\,\,\,\, \Rightarrow \,\,\,9a = 9b\,\,\,\,{\text{impossible }}\,\,\left( {{\text{in}}\,\,{\text{this}}\,\,{\text{case}}} \right) \hfill \\
{\text{2}}\,\,\, \Rightarrow \,\,\,72 \geqslant 8a = 19b\,\,\, \Rightarrow \,\,b \leqslant 3\,\,\,\,\,\, \Rightarrow \,\,\,{\text{no}}\,\,{\text{solutions}}\,\,\,\left( {{\text{by}}\,\,{\text{inspection}}} \right)\, \hfill \\
3\,\,\, \Rightarrow \,\,\,63 \geqslant 7a = 29b\,\,\, \Rightarrow \,\,b \leqslant 2\,\,\,\,\,\, \Rightarrow \,\,\,{\text{no}}\,\,{\text{solutions}}\,\,\,\left( {{\text{by}}\,\,{\text{inspection}}} \right) \hfill \\
4\,\,\, \Rightarrow \,\,\,54 \geqslant 6a = 39b\,\,\, \Rightarrow \,\,\,b \leqslant 2\,\,\,\, \Rightarrow {\text{no}}\,\,{\text{solutions}}\,\,\,\left( {{\text{by}}\,\,{\text{inspection}}} \right) \hfill \\
\end{gathered} \right.\)
The above follows the notations and rationale taught in the GMATH method.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here:
https://gmath.net