Take the task of building palindromes and break it into stages.

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

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

Stage 3: Select the thousands 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 hundreds digit
This digit must be the SAME as the thousands digit (which we already chose in stage 3)
So, we can complete this stage in 1 way.

Stage 5: Select the tens digit
This digit must be the SAME as the ten-thousands digit (which we already chose in stage 2)
So, we can complete this stage in 1 way.

Stage 6: Select the units digit
This digit must be the SAME as the hundred-thousands 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 6 stages (and thus build a 6-digit palindrome) in (9)(10)(10)(1)(1)(1) ways (= 900 ways)

Answer: D
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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

Cheers,
Brent
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The hundred-thousand's digit won't be zero or else it's no longer a 6 digit integer (you could form a palindromic number with five digits though). The subesquent two digits, ten-thousand's and thousand's digit, can be 10 different numbers each, zero - nine. Thus,
\(9 * 10 * 10 * 1 * 1 * 1 = 900\) Answer D

Plus, it won't be 1,000 since the first three digits can't be more than 999 and the first three must be the same as the last three..

Kr,
Ron

trying to grab this..
Stem needs to get clearer to me.
Palindromic numbers read the same way forward and backward.
as in
100001
131313
555555
770077


They all read the same forward or backward.
It didn't state it must be the alternating example given in the stem.

100001
200002
.
.
900009
nine

101101
201102
301102
.
.
.

My!!
There shud be a formulae for this.
Gmatprepnow solution above isn't quite clear to me either.
Thanks

Posted from my mobile device

Why were the last two digits tied to one possible value? why the sudden change of rule?
I know you guys are right. you are pro. but I need to get it. I'm the learner. pls elucidate.

The basic idea is that, once we have selected the first 3 digits of the number, the last 3 digits automatically follow.
For example, if the first 3 digits are 356---, then (to be a palindrome), the last 3 digits must be ---653, so we get the number 356653.
Likewise, if the first 3 digits are 197---, then (to be a palindrome), the last 3 digits must be ---791, so we get the number 197791.

I hope that helps.

Cheers,
Brent
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Hi Brent,

Should the number in the highlighted portion not be 356653?

Also, may be the example used in the question stem does not provide complete clarity into the concept of Palindrome numbers. I was trying hard to replicate a 6-digit number that has first and second digits repeated in the remaining 4 positions. But now I understand from Brent's solution that Palindromes are of the form:

In case of 6-digit integers -> A B C C B A e.g. 123321
In case of 5-digit integers -> A B C B A e.g. 18981
In case of 4-digit integers -> A B B A e.g. 9889
In case of 3-digit integers -> A B A e.g. 989

Warm Regards,
Pritish
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\(Asked:\) How many 6-digits number are Palindromic numbers? A Palindromic number reads the same forward and backward, example 12121.

Let the 6-digit Palindromic number be xyzzyx

Options for x = 9
Options for y = 10
Options for z = 10

Number of 6-digit Palindromic numbers = 9*10*10 = 900

IMO D
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You're absolutely right. Lazy error on my part.
I've edited my response.

Cheers and thanks,
Brent
_________________

1st digit can have 9 options. 2nd digit 10. 3rd digit also 10.
6th digit = 1st digit = 1 option
5th Digit = 2nd digit = 1 option
4th digit = 3rd digit = 1 option.

Multiply all cases = 9*10*10*1*1*1 = 900

this pallindrome would exists as ; xyzzyx ;
so x= 9 ; y=10 & z= 10
total pairs ; 9*10*10 ; 900
IMO D

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