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Four-digit integer palindromes with positive integer initial digits are:

1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, 1991, 2002, 2112, 2222, 2332, 2442, 2552, 2662, 2772, 2882, 2992, 3003, 3113, 3223, 3333, 3443, 3553, 3663, 3773, 3883, 3993, 4004, 4114, 4224, 4334, 4444, 4554, 4664, 4774, 4884, 4994, 5005, 5115, 5225, 5335, 5445, 5555, 5665, 5775, 5885, 5995, 6006, 6116, 6226, 6336, 6446, 6556, 6666, 6776, 6886, 6996, 7007, 7117, 7227, 7337, 7447, 7557, 7667, 7777, 7887, 7997, 8008, 8118, 8228, 8338, 8448, 8558, 8668, 8778, 8888, 8998, 9009, 9119, 9229, 9339, 9449, 9559, 9669, 9779, 9889, 9999

This is a total of 90 numbers.

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Lvl 2

there is 10 million

Lvl 1

4000

Q: How many 4 digit palindromes are there?

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There are 900 6-digit palindromes.

35.

-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.

90000. With 10 digit palindromes, the last 5 digits are the same as the first 5 digits in reverse, eg 12345 54321. So it comes down to how many 5 digit numbers are there? They are the numbers "10000" to "99999", a total of 99999 - 10000 + 1 = 90000.

Related questions

There are 90 four-digit palindromes

-4

There is 90 four digit palindromes.

There are 90 such numbers.

There are 90 four-digit palindromes

There are no four-digit perfect squares that are palindromes.

There are 900 6-digit palindromes.

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.

There are 10 3-digit odd palindromes that are divisible by five.

90

90 of them.

9 of them.