Q1 3 digit number in which middle one is perfect square is formed using 1-9.Find sum
Q2 Number of 6 digit numbers in which sum of digits is divisble by 5.
Aslo find sum of such numbers
29Ans 1...the condition is middle term can have only 1,4,9..
so total 3 digit numbers possible satisfying the abv condition ( 9*3*9) = 243
then the required sum will be like = 3*9*(1+2+3+4+5+6+7+8+9)*100 + 81(1+4+9)*10 + 3*9*(1+2+3+4+5+6+7+8+9)
= 121500 + 11340 + 1215 = 134055
I hope i have not made any calculation error
1answer for 2 .is \sum_{k=3}^{7}{\text{coefficient of } x^{5k} \ \text{in the expansion of}(1+x+x^2+...+x^9)^6}
???
24ya govind u r rite.....but my real dbt is how did u coem to knwo that repition was allowed ????
i was trying with no repition..and htere were so many cases...........
1@ govind....i think 9*3*10=270 possibilities are there....u missed zero in the one's position.....
29@Ronak.....Eureka mentioned that the didgits have to be 1-9
Q1 3 digit number in which middle one is perfect square is formed using 1-9.Find sum
This one's for eureka...Q1 modified ..without repetitions...
Without repetition there will be only 168 cases that means we need to subtract the values of 75 cases...
134055 - [[ { 2(2+3+4+5+6+7+8+9) + 9(1)}*100 + 25(1)*10 +{ 2(2+3+4+5+6+7+8+9) + 9(1)}] +
[{2(1+2+3+5+6+7+8+9) + 9(4)}*100 + 25(4)*10 + {2(1+2+3+5+6+7+8+9) + 9(4)} +
{2(1+2+3+4+5+6+7+8)] + 9(9)}*100 + 25(9)*10 + {2(1+2+3+4+5+6+7+8) + 9(9)}]]
on simplifying it u will get
134055 - [ 36800+ 3500 +368]
134055 - 40668
= 93387
24ya thats what i was saying..without repition it gets tooo long....anyways thx..
plz see Q2 eveyone
