A palindrome is a number that reads the same forward or backward. If t

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

Other questions on palindromes:
https://gmatclub.com/forum/a-palindrome- ... 29898.html
https://gmatclub.com/forum/a-palindrome- ... 81030.html
https://gmatclub.com/forum/a-palindrome- ... 59265.html
https://gmatclub.com/forum/a-palindrome- ... 61167.html
https://gmatclub.com/forum/a-palindrome- ... 19672.html
https://gmatclub.com/forum/the-country- ... 96679.html
https://gmatclub.com/forum/a-license-pl ... 68084.html
https://gmatclub.com/forum/a-man-while- ... 44093.html
https://gmatclub.com/forum/a-palindrome ... 14064.html
https://gmatclub.com/forum/the-country- ... 96679.html
https://gmatclub.com/forum/a-palindrome ... 71091.html
https://gmatclub.com/forum/a-palindromi ... 14196.html

Hope it helps.
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Those numbers wouldn't qualify, Karaan.
The question tells us that the first two digits form a multiple of the last two digits
Since 12 is not a multiple of 21, 1221 doesn't qualify.
Likewise, since 21 is not a multiple of 12, 2112 doesn't qualify.

I believe you're confusing product with multiple of

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.
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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



Hi BrentGMATPrepNow,

Would it be wrong to say that numbers such as 1221 and 2112 are palindromes ? Multiple of first two digits is equal to that of the last two digits and both the numbers read the same backwards too.

Looking forward to your response.

Thanks,
K

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