|By Co0 (Co0) on Tuesday, April 06, 2004 - 03:58 pm: Edit|
Will someone repeat the question from the March SAT regarding the palindromes if they remember it? I forgot it and want to know why the answer is C
|By Homer2568 (Homer2568) on Tuesday, April 06, 2004 - 04:10 pm: Edit|
A palindrome is defined as a number that is read the same both forwards and backgrounds (e.g. 202, 303, etc.).
Column A: The number of 3 digit palindromes.
Columb B: The number of 4 digit palindromes.
The reason its C is because the 3 digit palindromes are (202, 232, 252, 242, 343, 545, etc. etc. etc. ). The 4 digit palindromes are (2002, 3003, 2112, 2222, 2332, 2442, 2552, 3553) so they end up being the same. Just having hte first and last numbers being the same doenst make it a palindrome (e.g. 3233...becuase read backwards its 3323...not the same).
|By Gmf05 (Gmf05) on Tuesday, April 06, 2004 - 04:57 pm: Edit|
Mathematically it's proven using combinations/permutations.
For 3 digit palindromes...you have 9 choices for the first digit (1-9 since you can't use zero.) Then you have 10 choices for the second, 0-9. However, the third digit must be the same as the first in order for the number to be a palindrome...so you have 1 choice for it.
Thus you have 9*10*1 = 90 choices for 3 digit palindromes
For 4 digits you again have 9 choices for the first digit and 10 for the second. But you have both the third and the fourth down to 1 choice, because they are pre-determined by the earlier digits. (Again, it wouldn't be a palindrome if these two digits were different.)
So, you have 9*10*1*1 = 90 for 4 digit palindromes as well...
At least that's how I did it.
|By Qwert271 (Qwert271) on Wednesday, April 07, 2004 - 03:00 pm: Edit|
I'm glad you did it that way instead of BS prep book method. Way to be!
|By Gmf05 (Gmf05) on Wednesday, April 07, 2004 - 06:10 pm: Edit|
Why thank you.
I agree, the prep books/classes teach some very crappy methods of problem solving. For instance, when I took a free SAT from Princeton Review, they were pitching their classes to us beforehand and told us to basically "pick an answer you don't think is right" on hard math questions...
|By Conker (Conker) on Thursday, April 08, 2004 - 05:17 am: Edit|
Yep, I did it the same way as Gmf. Was this considered to be one of the harder problems on the SAT?
|By Supernova (Supernova) on Thursday, April 08, 2004 - 12:08 pm: Edit|
...i just did it like homer... it's common sense that in a 4 digit palindrome the 2 middle numbers would have to be the same, thus there are the same number of 3 and 4 digit palindromes
comb/perm confuse me, personally, i find it much easier to just reason it out logically.
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