1thanku :)
Find teh sum of all four digit palindromes.............
u must be knowing wat a palindrome is
if not
it is a word or number that reads the same backwards as forwards
i want a soln :)
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15 Answers
62This is kind of simple...
FInd the number of palindromes with the first digit 1...
There will be 10 such palindromes... (Each will have last digit 1 too)
Same with 2, 3, .... 9
The sum of all these digits will be (1001+2002+...9009) x (10)
Now lets find the number of numbers where 1 is the 2nd digit.. there will be 9 such palindromes...
The sum of all these digits will be (110+220+....+990) x 9
SO the required sum is
(1001+2002+...9009) x (10) + (110+220+....+990) x 9
1nishant bhaiya, with the method u hav posted, the sum is coming out to be 495000. but i'm getting 495050.
dunno where is the discrepancy.
i did by actually summing all palindromes starting with 1...and then all palindromes with 2 and then 3..(which was a very easy task)...and then a pattern was evident....
1btw i was jus wondering that it is more of a lenghty soln....i mean summing up whole lot of 20 numbers...doenst seem good :P...i guess some elegent soln do exist...
btw shreyan 495000 is the rit ans
24good thread for refrence
http://targetiit.com/iit-jee-forum/posts/23-09-09-what-is-the-10000th-palindrome-11401.html
62It is not such a lengthy method..
It is infact very simple.. bcos most of the terms there form an AP...
:P
1001 x (1+2+...9) x (10) + 110 x (1+2+....+9) x 9 =10010 x (1+2+...9) + 990 x (1+2+....+9) = 11000(45) = 495000
So there is no discrepancy either [1]
1palindrome::it is a word or number that reads the same backwards as forwards..........e.g..NITIN,NAMAN,12321,1234321,
amstrong no.::a no. whose sum of cubes of each digits is equal to tht number....e.g..153=13+53+33=1+125+27=153..........hope u understand