A positive integer N is a Palindrome if the number obtained by reversing the sequence of the digits of N is equal to N. The year 1991 was the only year of the last century which was a palindromic year.
Read this document to solve this interesting problem.
Write your answers in the comment section.
Read this document to solve this interesting problem.
Write your answers in the comment section.
1661/11 = 151
ReplyDelete1441/11= 131
2002/11= 182
1111/11= 101
this series can be understood
eg: 1551/11= 141
now,
555/11= 50.45.......aroung 14 times
767/11= 69.72...14 times and 73
10101/11=918.27...i4 times and 3
sir can u explain this one ?????
Well sir.
ReplyDeleteA Palindrome year with
FOUR DIGITS is ALWAYS divisible by 11
THREE DIGITS is SOMETIMES divisible by 11
TWO DIGITS is ALWAYS divisible by 11
This is a direct conclusion of divisibilty rules of 11...
All paindrome years with 4 digits will have the difference between the sum of dgits at odd places and at even places zero as it is a symmetrical number...
same applies to a 2-digit palindrome number..
But with a three digit palindrome year, it is only possible if difference between the sum of digits at odd places and at even places is 0...
OR
In simple words if the middle digit is equal to the sum of the digits at the ends, it will definitely get divided by 11, for eg: 363( where 3+3=6, 6 which is the middle term)
That's all I could conclude...Please explain other things regarding this, if I missed out ...Looking forward to your formal explaination....
1991/11 =181
ReplyDelete1661/11 =151
1441/11 =131
similarly 1881/11 =171
2002/11 =182
and the next palindromic year will be 2112
nd 2112/11 =192
similarly 2222/11 =202
All palindrome years with an even number of digits are divisible by 11.
The first four digit palindrome is 1001 which is divisible by 11. The next four digit palindrome is 1111, which is also divisible by 11 and is 110 bigger than 1001. 110 is important, because not only is it divisible by 11, but every four digit palindrome is x(1001) + y(110), where x is any integer from 1 to 9, and y is any integer from 0 to 9. It is not that difficult to expand this argument to 6 digits, 8 digits and so on.
Well of the 90 triple digit palindrome numbers, 15 of them are prime.so certainly they cant be divisible by 11.
so
A palindrome year that has 4 digits can always be divided by 11.
A palindrome year that has 3 digits can sometimes be divided by 11.
A palindrome year that has 2 digits can always be divided by 11
yes... you all have explained this very well.
ReplyDeleteThere is no special explanation as such required.
here, FOUR DIGITS is ALWAYS divisible by 11
THREE DIGITS is SOMETIMES divisible by 11
TWO DIGITS is ALWAYS divisible by 11
We may think some questions like how many 2 digit, 3-digit and 4-digit palindromic numbers are possible?
now i got it !!! thankzx sir , gunjan and dude ( neo -pentane )
ReplyDelete