IIT JEE , AIEEE Forum - Mathematics , Algebra

manasasambana

a palindrome number is always divisible by 11. possible no of digits n the number is?

sboosy

actually there is no restriction on the number of digits
reason:
divisibility by 11 :abcde(a,b,c,d,e are digits)
condition is (a+c+e)-(b+d)=11k where k is whole number
that is it can be equal to 0 or 11 or 22 etc
now consider palindromes
2 digit number (11 itself obviously is divisible)
3 digits 121
4 digits 1001
5 digits 11011
....
so it is so easy to generate a palindrome divisible by 11
technique is
even no of digits means put 1st and last digit as 1 rest as 0
odd no of digits means 1st 2 digits and last 2 digits as 1 and rest as 0
this ofcourse can include few of the possible palindromes
rest can be derived similarly
so there is no restriction on the no of digits

puneet

hii

well .. the answer is not hard to get ..

suppose n is even .. then wat we need to do is just find the number of possible ways in which n/2 digits can sum to equal other n/2 digits ..

so this can be found using the coefficient method .. right

the minimum sum should be 1 .. and the maximum shud be n/2.9

i think u will have to leave the answer exactly .. u will hav to leave it as series kinda ..

cheers

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