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Nine. The sum of the digits must be a multiple of 9; because of the repeated digits, this is only possible if the first two digits add up to 9.

Lvl 1

Nope there is also 9999 so there is 10

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90 four-digit palindromes are divisible by 9.

4 four-digit palindromes are divisible by 18

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There are 90 4 dight ones

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great answer tyyy

Q: How many 4 digit palindromes are multiples of 9?

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There are 90 four-digit palindromes

There are 90 such numbers.

None. 1221 and 3443 are both 4-digit palindromes but no digit has remained the same between the two. First and fourth, second and third.

There are 90 palindromes with 4 digits.The first digit can be any digit from the set {1,2,3,4,5,6,7,8,9}.With each choice of the first digit, the second can be any digit from the set {0,1,2,3,4,5,6,7,8,9}.That makes 9*10 = 90 permutations for the first two digits. These determine the palindrome since the third and fourth digits are the same as the second and first, respectively.

There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.

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-4

There are 90 four-digit palindromes

There are 90 such numbers.

999

None. 1221 and 3443 are both 4-digit palindromes but no digit has remained the same between the two. First and fourth, second and third.

There are 90 palindromes with 4 digits.The first digit can be any digit from the set {1,2,3,4,5,6,7,8,9}.With each choice of the first digit, the second can be any digit from the set {0,1,2,3,4,5,6,7,8,9}.That makes 9*10 = 90 permutations for the first two digits. These determine the palindrome since the third and fourth digits are the same as the second and first, respectively.

There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.

-1000

900 This explains it. A positive integer is a palindrome if it reads the same forward and backwards such as 1287821 and 4554. Determine the number of 5-digit positive integers which are NOT palindromes. We start by counting the total number of 5 digit positive integers. The first digit is between 1 and 9, so we have 9 choices. Each of the other 4 digits can be anything at all (10 choices for each). This gives us 9(10)4 = 90000 five-digit positive integers. Now we need to count the number of 5 digit palindromes. Again, we have 9 choices for the first digit and 10 choices for each of the next two. The tens and units digits however are fixed by our choices so far. Therefore, there are only 900 five-digit palindromes. Therefore, the total number of five-digit positive integers which are not palindromes is 90000-900 = 89100.

There are 75 multiples of 12 between 100 and 1000.

There are 400. Assuming the number must be at least 10,000, then: In a 5 digit palindrome, the first and last digits must be the same, and the second and fourth digits must be the same; and: For the first and last digit there is a choice of 4 digits {2, 4, 6, 8}; For each of these there is a choice of 10 digits {0, 1, ..., 9} for the second and fourth digits; For each of the above choices these is a choice of 10 digits {0, 1, ..., 9} for the third digit; Making 4 x 10 x 10 = 400 possible even 5 digit palindromes.

There are 2000 4-digit numbers that are multiples of 5, so, instead of listing them all, it is equally valid to say: Any4-digit number whose final digit is either a 5 or a 0 is a multiple of 5. Get Right? :P