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30 Aug 2018, 12:03
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:05) correct
63%
(02:03)
wrong
based on 355
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 0
B. 6
C. 9
D. 12
E. 18
30 Aug 2018, 17:07
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gracie
Given that the answer choices are relatively small, we might consider the strategy of listing and counting
To begin, if all 4 digits are the same, then the first two digits of a four digit palindrome form a multiple of the last two digits
For example, in the number 2222, the first two digits (22) is multiple of the last two digits (22)
So, 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999 all work.
At this point, we've already listed 9 possible palindromes.
So, we can ELIMINATE A and B
What else is there?
Well, numbers in the form n00n also work, since n0 must be a multiple of n
For example, in the number 2002, the first two digits (20) is multiple of the last two digits (02)
So, 1001, 2002, 3003, 4004, 5005, 6006, 7007, 8008, 9009 all work.
We now have a TOTAL of 18 possible palindromes.
Since 18 is the greatest answer choice, we can be certain that no more palindromes exist.
Answer: E
Cheers,
Brent
General Discussion
30 Aug 2018, 16:06
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gracie
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.
