Best Answer

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

Study guides

☆☆

More answers

Lvl 2

90 four-digit palindromes are divisible by 9.

4 four-digit palindromes are divisible by 18

Lvl 2

There are 90 4 dight ones

Lvl 1

great answer tyyy

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

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

The first digit can be any one of nine (all except zero). For each of those . . .The second digit can be any one of ten.Total possibilities for the first two digits = 9 x 10 = 90.Since the 4-digit number is a palindrome, the 3rd and 4th digits are determinedby the 1st and 2nd ones.So the total number of 4-digit palindromes is the same as the number of possibilitiesfor the first 2 digits = 90 .

even numbers, specifically the last digit is 0, 2, 4, 6, or 8.

1000, 1004, 1008 and just keep adding four until you get to 9996.

12 is the only number which satisfies this. Multiples of 4 are: 4, 8, 12, 16, . . . .

17

Related questions

-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 6,750 such numbers.

8 of them.

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.

People also asked

Featured Questions