abhirup1711 wrote:A Palindrome is a number that the same forward and backward eg 242. How many even 5 digit numbers are palindromes?
40
400
500
5000
100000

So, our 5-digit number will look like this XYZYX (where X, Y and Z represent digits)

We'll take the task of building 5-digit numbers and break it into stages.

Stage 1: choose the value of X
Since we need to have a 5-digit number, the first digit cannot be zero (e.g, 07470 is not a 5-digit value). Also, since the 5-digit number must be even, the units digit must be even.
So, the possible value of X are 2, 4, 6 and 8 (4 options)
So, we can complete stage 1 in 4 different ways

Stage 2: choose the value of Y
Y can be any digit from 0 to 9
So, we can complete stage 2 in 10 different ways

Stage 3: choose the value of Z
Z can be any digit from 0 to 9
So, we can complete stage 3 in 10 different ways

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus build a 5-digit number) in (4)(10)(10) ways ([spoiler]= 400 ways = B[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

Hi Sana.noor,

The reason why there are 10 options is because the question allows for it. IF the question limited the number of times that a digit could occur, then the math would change. Since the question asks for ALL of the 5 digit palindromes, then we're supposed to consider all of the possibilities.

For example, if we make the first and last digit = 2, we'd have...

2 - - - 2

Those middle 3 digits could be anything, BUT the second and fourth digits have to be the same, so THAT rule affects the math.

So we could have

23432

But NOT:

23452

Here, we'd only have 100 numbers that fit the above format (not 1,000).

GMAT assassins aren't born, they're made,
Rich