A palindrome is a number that reads the same forward and

@enigma123: Here is my explanation:

A palindrome is a number that reads the same forward and backward. Examples of four digit palindromes are 1221, 4334, 2222 etc
You basically get to choose the first two digits and you repeat them in opposite order. Say, you choose 45 as your first two digits. The next two digits are 54 and the number is 4554.
Also, you need only odd palindromes. This means that you need an odd digit at the end i.e. 1/3/5/7/9. This means that you need to start the number with an odd digit. Only then will it end with an odd digit.

In how many ways can you pick two digits such that the first one is an odd digit?
The first digit can be selected in 5 ways.(1/3/5/7/9)
The second digit can be selected in 10 ways.(0/1/2/3...8/9)
Total = 5*10 = 50 ways
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We want to determine how many odd 4-digit numbers like XYYX are there. Notice here that in a palindrome X and Y can be the same, for example 1111 or 3333 are also palindromes.

Since, the number must be odd then the first and the last digits (X's) can take 5 values (1, 3, 5, 7, or 9) and the two middle digits (Y's) can take 10 values (0, 1, ..., 9), so total such numbers possible is 5*10=50.

Answer: C.

Hope it's clear.
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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. So, be sure to learn it.

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First recognize you only need to consider the first two digits (because the second two are just the first two flipped)

There are 90 possibilities for the first two digits of a 4 digit number, 10-99 inclusive. Everything starting with a 1,3,5,7,9 will be odd, which is 5/9ths of the combinations.

5/9*90 = 50 C

Similar questions to practice:
a-palindrome-is-a-number-that-reads-the-same-forward-and-129898.html
a-palindrome-is-a-number-that-reads-the-same-forward-and-backward-181030.html
a-palindrome-number-reads-the-same-backward-and-forward-159265.html
a-palindrome-is-a-number-that-reads-the-same-forward-and-bac-161167.html

Hope it helps.
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HI Buneul,

A quick question with this problem. We can also make a palindrom out of event number also right.. such as 8228. Why does it have to be odd numbers.

Sorry a very old question.

Because the question specifically asks about the number of odd palindromes.
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Bunuel

Why is it not 2500? 5*10*10*5?

Hi dubyap,

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

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

Hi Bunuel..

Can u tell me why we r doing 5*10 ways..

Thanks in advance

Because of Principle of Multiplication.

Principle of Multiplication
If an operation can be performed in ‘m’ ways and when it has been performed in any of these ways, a second operation that can be performed in ‘n’ ways then these two operations can be performed one after the other in ‘m*n’ ways.
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So we have the 4-digit numbers in the form of ABBA where A is an odd number and B can be any digit including B = A.

Therefore, we have 5 choices for the first A and 10 choices for the first B. However, since the second A and B must be the same as the first A and B, respectively, there is only 1 choice for each of the second A and B. So we have 5 x 10 x 1 x 1 = 50 such numbers.

Answer: C
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We can use the “slot method” to count all the 4-digit, odd palindromes. __ __ __ __

Since the last digit must be odd, our only choices are 1, 3, 5, 7, or 9 for the first/last digit. There are no restrictions on the inner digits, so we have 10 choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Notice that the outer two numbers must match and the inner two numbers must match, creating numbers such as 1221 or 5665. We have 5 choices for the outer two digits and 10 choices for the inner two digits. Our “slot method” diagram looks like this: 5 10 1 1. Once a digit is selected for the left outer digit, there is only one possible choice for the right outer digit, which must match it. Similarly for the two inner digits, the left choice determines the right. Using the counting principle, we have 5 × 10 × 1 × 1 = 50 choices for our 4-digit number.

Notice that we do not set the problem up as 5 10 10 5 and multiply, giving 2500. There are really only two choices to be made – number of possibilities for inner digits and number of possibilities for outer digits.

The correct answer is C.

Bunuel @experts I tried to do as follows

all 4 digit nos palindrome can be formed in 9*10*1*1 ways = 90 (first digit cant be 0)
we only need odd hence divide by 2 therefore 45.
could you help with where i am going wrong?

Hi Kritisood,

Since we're creating a palindrome, both the first digit AND the last digit must be the SAME. The question asks us for all of the ODD-numbered palindromes, which means that the last digit can only be 1, 3, 5, 7 or 9... and the same holds true for the first digit (since it has to match the last digit). Thus, the number of ODD integer palindromes is...

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

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

thanks for the response EMPOWERgmatRichC
I wanted to understand what specifically am I doing wrong in my approach. There are a few cases im overcounting evidently. Could you pls help with this?

Hi Kritisood,

A palindrome is a number that reads the same forwards AND backwards. For example, the numbers 121 and 8558 are both palindromes. From your calculation, you correctly understand that with a 4-digit number, the "3rd" digit MUST match the "2nd" digit and the "4th" digit" MUST match the "1st" digit.

The question asks us for all of the ODD-numbered palindromes; an ODD number is an integer that ends in 1, 3, 5, 7 or 9. This means that the "4th" digit can only be one of those five options - and since we're dealing with a palindrome, the FIRST digit can only be a 1, 3, 5, 7 or 9. There are no restrictions on the "2nd" and "3rd" digits though.

Options for the 1st digit: 5
Options for the 2nd digit: 10
Options for the 3rd digit: (must match the 2nd digit): 1
Options for the 4th digit: (must match the 1st digit): 1

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

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

Personally I would break it down to slots.

___ ____ ____ ____

Last slot has to be odd, either 1,3,5,7,9 --> thats 5 options
Last and First slot have to be the same, so first slot has 1 option

2nd slot has 10 options. any of 0,1,2,3,4,5,6,7,8,9
3rd slot has to be same as 2nd slot so 2nd slot only has 1 option

therefor 1 * 10* 1 * 5 ways = 50

C