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palindrome

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by krishnasty » Sat Aug 06, 2011 12:41 pm
A palindrome is a number that reads the same forward and backward, such as 242. How many even five-digit numbers are palindromes?
40
400
500
5,000
100,000

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by GmatKiss » Sat Aug 06, 2011 12:43 pm
should be 400/500 !
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by maihuna » Sat Aug 06, 2011 12:45 pm
Even it means ending in 2/4/6/8 excluding 0, as leading zero can not be counted for 5 digits.
2nd and 2nd last element can be any of 0-9
middle element any of 0-9

so it should be : 4*10*10 i.e. 400
krishnasty wrote:A palindrome is a number that reads the same forward and backward, such as 242. How many even five-digit numbers are palindromes?
40
400
500
5,000
100,000

No OA provided
Charged up again to beat the beast :)
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by krishnasty » Sat Aug 06, 2011 8:52 pm
i am still not able to understand. Can any1 elaborate on the solution?
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by Frankenstein » Sat Aug 06, 2011 9:11 pm
krishnasty wrote:i am still not able to understand. Can any1 elaborate on the solution?
Hi,
Palindrome is of the form abcba.
a is even -> it can be 2,4,6,8 -> 4 values
b is any number from 0 to 9 -> 10 values
c is any number from 0 to 9 -> 10 values

So, total number of even 5-digit palindromes = 4*10*10 = 400
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by Anurag@Gurome » Sat Aug 06, 2011 9:13 pm
krishnasty wrote:i am still not able to understand. Can any1 elaborate on the solution?
As the number has to be even, the last digit must be either 0 or 2 or 4 or 6 or 8. But as the number also has to be a palindrome, if the last digit is zero then the first digit will be also zero. That will make the number a four digit number. Hence, last digit cannot be equal to zero.

As the number has to be a palindrome, the number will be of the form ABCBA.

Now A can be either 2 or 4 or 6 or 8. Thus, we have 4 choices for A.
And B and C can be any of the 10 digits (0 to 9). So 10 choices for both B and C.

Hence, the number of such five digit numbers = 4*10*10 = 400

The correct answer is B.
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