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Re: A palindrome is a number that reads the same forward and
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21 Apr 2018, 06:18
Top Contributor
enigma123 wrote:
A palindrome is a number that reads the same forward and backward, such as 121. How many odd, 4-digit numbers are palindromes?
A. 40 B. 45 C. 50 D. 90 E. 2500
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.
Re: A palindrome is a number that reads the same forward and
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13 Apr 2019, 17:51
enigma123 wrote:
A palindrome is a number that reads the same forward and backward, such as 121. How many odd, 4-digit numbers are palindromes?
A. 40 B. 45 C. 50 D. 90 E. 2500
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.
Re: A palindrome is a number that reads the same forward and
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01 Feb 2020, 18:15
enigma123 wrote:
A palindrome is a number that reads the same forward and backward, such as 121. How many odd, 4-digit numbers are palindromes?
A. 40 B. 45 C. 50 D. 90 E. 2500
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.
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?
Re: A palindrome is a number that reads the same forward and
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30 May 2020, 08:03
C
Odd Numbers: Units digit is odd. My approach: 4 digit number:: _ _ _ _
# of ways units digit can be filled (1,3,5,7,9): 5 ways # of ways tens digit can be filled (0,1, 2, 3,4, 5,6, 7,8, 9): 10 ways # of ways units digit can be filled (same as tens digit): 1 way # of ways units digit can be filled (same as units digit): 1 way
Tota # of ways: 1 * 1 * 10 * 5 = 50
_________________
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...
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?
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