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.

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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