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palindrome

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j_shreyans GMAT Destroyer! Default Avatar
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palindrome Post Sun Oct 12, 2014 3:21 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    A palindrome is a number that reads the same forward and backward, such as 121. How many odd, 4-digit numbers are palindromes?

    A)40
    B)45
    C)50
    D)90
    E)2500

    OAC

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    Post Sun Oct 12, 2014 3:36 am
    Shouldn't the answer be 90?
    The first place and similarly the 4th place can be filled in 9 ways i.e digits from 1 to 9
    The second and the third place can be filled in 10 ways i.e digits from 0 to 9
    hence total number of palindromes = 9 X 10 = 90.

    Maybe I am missing out some duplication W.r.t the OA.

    Post Sun Oct 12, 2014 4:43 am
    j_shreyans wrote:
    A palindrome is a number that reads the same forward and backward, such as 121. How many odd, 4-digit numbers are palindromes?

    A)40
    B)45
    C)50
    D)90
    E)2500

    OAC
    For the 4-digit integer to be a palindrome:
    The THOUSANDS digit must be the SAME as the UNITS digit.
    The TENS digit must be the SAME as the HUNDREDS digit.

    Since the integer must be ODD, the number of options for the units digit = 5. (1, 3, 5, 7, or 9.)
    Number of options for the thousands digit = 1. (Must be the SAME as the units digit.)
    Number of options for the hundreds digit = 10. (Any digit 0-9.)
    Number of options for the tens digit = 1. (Must be the SAME as the hundreds digit.)
    To combine the options above, we multiply:
    5*1*10*1 = 50.

    The correct answer is C.

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    Post Sun Oct 12, 2014 1:05 pm
    j_shreyans wrote:
    A palindrome is a number that reads the same forward and backward, such as 121. How many odd, 4-digit numbers are palindromes?

    A)40
    B)45
    C)50
    D)90
    E)2500

    OAC
    Take the task of building palindromes and break it into stages.
    Begin with the most restrictive stage.

    Stage 1: Select the units digit
    We can choose 1, 3, 5, 7 or 9
    So, we can complete stage 1 in 5 ways

    Stage 2: Select the tens digit
    We can choose 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9
    So, we can complete stage 2 in 10 ways

    IMPORTANT: At this point, the remaining digits are already locked in.

    Stage 4: Select the hundred digit
    This digit must be the SAME as the tens digit (which we already chose in stage 2)
    So, we can complete this stage in 1 way.

    Stage 5: Select the thousands digit
    This digit must be the SAME as the units digit (which we already chose in stage 1)
    So, we can complete this stage in 1 way.

    By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus build a 4-digit palindrome) in (5)(10)(1)(1) ways (= 50 ways)

    Answer: C
    --------------------------

    Note: the FCP can be used to solve the majority of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

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    Brent

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    Post Sun Oct 12, 2014 1:06 pm
    Here's a palindrome question I created for BTG a longggg time ago (for their Math Challenge Question contest):

    Quote:
    A palindrome is a word that is read the same backwards as forwards. For example, the words “BADAB,” “IAGAI,” and “HHHHH” are all palindromes.

    How many 5-letter palindromes can be created using the letters A, B, C, D, E, F, G, H, I and J?
    For a full solution, watch the following YouTube video: http://youtu.be/qfiPnXIBx7g

    Cheers,
    Brent

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    Post Sun Oct 12, 2014 8:45 pm
    Here's my approach.

    Our palindrome can have four identical digits (such as 1111) or two distinct pairs of identical digits (such as 1331). Let's find each pair.

    FOUR IDENTICAL DIGITS:
    We could have 1111, 3333, 5555, 7777, or 9999, for a total of five numbers.

    TWO PAIRS OF IDENTICAL DIGITS:
    The digit that begins and ends our number must be ODD, so we have five choices there. The other number must be DISTINCT (i.e. not the same as the last digit), so we have NINE other digits. (0 through 9, excepting the odd digit we already used. For instance, if the digit that begins and ends our number is 3, the other digit can be one of 0, 1, 2, 4, 5, 6, 7, 8, or 9, giving us numbers such as 3003, 3113, etc.)

    Since we have FIVE choices for the first/last digit pair and NINE for the second/third digit pair, this gives us 5 * 9 = 45 options.

    Summing the cases, we have 45 + 5, or 50.

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    Post Sun Jun 26, 2016 11:40 pm
    let the no be XYYX. eg X=1 and y 2 so the no is 1221

    odd Int 1,3,5,7,9

    X at unit's digit has to be odd for the 4 digit no to be odd

    X at unit's digit=No of ways it can be odd =5
    No of options for X at 1000th digit will be= 1


    Y at tens digit the no of options= any no from 0,1,2,3,4,5,6,7,8,9 digits=10
    Y at 100th digit the no options will be 1

    1*1*10*5=50.

    Post Mon Jun 27, 2016 9:19 am
    Hi All,

    Since the 4th digit has to match the 1st digit and the 3rd digit has to match the 2nd digit, you don't have as many options as you might think.

    Since the palindrome has to be ODD, both the 1st and 4th digits have to be ODD and the SAME...

    The 1st digit could be 1, 3, 5, 7 or 9
    The 4th digit must MATCH the first digit, so once you choose the 1st digit, there is ONLY ONE possible number for the 4th digit.

    The 2nd digit could be any of the 10 options (0 - 9, inclusive).
    The 3rd digit must MATCH the 2nd digit, so once you choose the 2nd digit, there is ONLY ONE possible number for the 3rd digit.

    Thus, there are...

    (5)(10)(1)(1) = 50 options

    Final Answer: C

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