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How many different three digit palindromes can you make using only the digits 1and 2?

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}.


How many three-digit palindromes can you make using only the digits 1 and 2?

111, 121, 222, 212


How many different three-digit palindromes can you make using only the digits 1 and 2?

111, 121, 212, 222


How many 5 digit palindromes are there that sum 6?

There are 6 solutions in all.The middle digit must be 0 , 2 or 4.If the middle digit is 0, then the first two digits must sum to 3. There are three possible solutions: 12, 21 and 30.If the middle digit is 2, then the first two digits must sum to 2. There are two possible solutions: 11 and 20.If the middle digit is 4, then the first two digits must sum to 1. There is one possible solutions: 10.


How many bit strings of length n are palindromes?

If n is even then number of palindromes is 2^(n/2). If n is odd then number of palindromes is 2*2^[(n-1)/2].


How many palindromes in 1 and 300?

There are 37 palindromes between from 2 to 299 inclusive.


How many palindromes are in the book HOLES?

2


What are the three digit palindromes that you can make only using the digits 1 and 2?

111, 121, 222, 212


Are there more 11 digit or 12 digit palindromic numbers?

There are more 12-digit palindromic numbers than 11-digit palindromic numbers. This is because the number of possible 12-digit palindromic numbers is greater than the number of possible 11-digit palindromic numbers. In general, the number of palindromic numbers of length n is 9 * 10^((n-1)/2), so for 11-digit palindromic numbers, there are 9 * 10^5 = 900,000 possibilities, while for 12-digit palindromic numbers, there are 9 * 10^6 = 9,000,000 possibilities.


How many palindromes between 1000 and 3000?

A palindrome reads the same forwards as backwards. Therefore the first and last digits must be the same, the second and penultimate the same, and so on. As 1000 & 3000 both have 4 digits, there are two pairs of digits that must be the same: the first and fourth, the second and third. The first digit (and hence last digit) can only be 1 or 2 - a choice of 2 The second digit (and hence third digit) can be any of 0, 1, 2, ..., 9 - a choice of 10 for each of the choices of the first/last digit. Thus there are 2 × 10 = 20 palindromes between 1000 and 3000.


How many different 2 digit numbers can be formed from the digits 2 6 7 and 8?

Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.


What is the greatest possible product of a 2 digit and a 1 digit number?

Highest 2-digit number = 99 Highest 1-digit number = 9 Highest possible sum from 2, 2-digit numbers = 198 Highest possible sum from 2, 1-digit numbers = 18