Question: A palindrome is a number that reads the same forward and backward. For example. 2442 and 111 are palindromes. If 5-digit palindromes are formed using one or more of the digits, 1, 2, 3, how many such palindromes are possible?
A) 12
B) 15
C) 18
D) 24
E) 27
“A palindrome is a number that reads the same forward and backward.”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.
Approach Solution 1
Since this is a five-digit palindrome, the middle number will act as a mirror.
Looking at Case 1
The Middle number is 1
The Number of options for the 1st place is 3
The Number of options for the 2nd place is 3
The Number of options for the 4th and the 5th place is 1
Total ways it could be done is 3x3 = 9
Looking at Case 2:
The Middle number is 2
Just Similar to the above reasoning as in case 1
Total ways it could be done is 3x3 = 9
Looking at Case 3:
The Middle number is 2
Just Similar to the above reasoning as in case 1 and case 2
Total ways it could be done is 3x3 = 9
so,
The Total number of ways is 9 + 9 + 9 = 27
The answer is E, 27 such palindromes are possible.
Correct Answer: E
Approach Solution 2
Lets try a different approach
A palindrome is a number which can be interpreted both forward and backward.
One or more of the digits 1, 2, 3 are used to create 5-digit palindromes.
Let's find the number of palindromes.
In this case, you only need to assign A, B, and C; the last two digits, and it will be assigned automatically.
The general form can be ABCBA.
The Total number of ways can be 3 × 3 × 3 = 27
The answer is E, 27 such palindromes are possible.
Correct Answer: E
Approach Solution 3
This palindrome has five digits, thus the middle number will serve as a mirror.
Considering Case 1,
1 is the middle number.
There are three alternatives for the first position.
There are three alternatives for the second position.
There is only one choice each for fourth and fifth place, making a total of 3x3 = 9 possible outcomes.
The middle number in Case 2 is 2.
Similar to the justification in instance 1 above, there are a total of 3x3 = 9 options to accomplish the task.
Case 3 reveals that the middle number is 2.
Similar to example 1 and case 2, the total number of possible solutions is 3x3 = 9.
Thus,
There are 9 + 9 + 9 ways in all, which is 27.
There are 27 potential palindromes, hence the solution is E.
Correct Answer: E
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