How many odd 1-, 2-, and 3-digit numbers are there?
A palindromic number is one that reads the same from left to right or from right to left. For example, 64746 is a palindromic number.
a/ How many odd six-digit palindromic numbers are there?
b/ How many odd seven digit palindromic numbers are there in which every digit appears at most twice?
i didn't understand if in both a and b there is each digit only twice so i will solve twiceOriginally Posted by mathsmad
a/ <each digit twice>
the sum of chosing including leading zero minus the sum of chosing with leading zero
5*9*8-9*8=288
<as many digits>
10*9*8-9*8=648
b/
5*9*8*7-9*8*7=2016
Consider the slots:
_ _ _ _ _ _
In the first slot you can only place 5 numbers (because since it is odd the last slot must be 1,3,5,7, or 9).
In the second slot you can place 10 numbers.
In the third slot you can place 10 numbers.
In the fourth slot you can only have one number (same as third slot because it is palindromic).
In the fifth slot you can only have one number.
In the sixth slot you can only have one number.
Thus by the Fundamental Counting Principle there are a total of 5*10*10 possibilities thus, 500 different numbers.
One interesting fact about palindromes is that if it is even digited then it is always divisible by 11. And if it is odd digited then is remainder when divided by 11 is the middle digit (thus if it is zero then it is divisible by 11 and this type of number can never be have remainder 10). Thus "there does not exist a palindromic number when divided by 11 leaves a remainder of 10!"
Consider an odd number <=999. Each of these gives a unique six-digitOriginally Posted by mathsmad
palindromic number as follows:
first add as many zeros to the front of the number as needed so that we
have exactly three digits. Reverse the digits and append reversed copy to
the front of the three digits.
It is self evident that any six-digit palindromic number can be generated in
this manner.
Therefore there are exactly as many six-digit palindromic numbers as there
are odd numbers <=999. There are exactly 500 odd numbers <=999, so
there are exactly 500 six-digit palindromic numbers.
For seven-digit odd palindromic numbers the three most and the three least
significant digits are the digits of a six-digit palindromic number. Thus for
each possible middle digit there are 500 seven-digit palindromic numbers. The
possible middle digits are 0, 1, 2, .., 9. Hence there are 5000 seven-digit
odd palindromic numbers.
RonL
Hello,Originally Posted by mathsmad
try this one about palindromic numbers:
You start with any number, which isn't a palindromic one. You write the ciphers of this number from end to start and add it to the original number. If the result is palindromic than you've got what you wanted. If the result is not a palindromic number you write the ciphers (of the result of course!) from end to start, add ... proof ... and so on, until you get a palindrom.
I've attached a simple example. If you start for example with 40793 than you need 22 steps to get a palindrom.
I don't know if this rule is proofed (I never found a proof for it). So maybe the proof is a nice job for a boring weekend ;-)
Bye