First question is a beauty

First question is a beauty and the concept behind it z

1. The highest power of a in (m!/n!)= The highest power of a in m! - The highest power of a in n!. 2. The highest power of a in m! = [(m/a¹)] + [(m/a²)] + [(m/a³)] + . . . .

P1= 1 x 3 x 5 x 7. . . . x 55

     1 x 3 x 5 x 7. . . . x 55
  = ----------------------------
     2 x 4 x 6 x 8 . . . . x 54
  
            55!
  = ----------------------
     227 x 27!

The highest power of 3 in P1 = Highest power of 3 in 55! - Highest power of 3 in 27!
=[18 + 6 + 2] - [ 9 + 3 + 1]
= 13

i saw this method in Challenges and thrills for Olympiads

the correct answer is 15/6^5 i guess. the equation is x+y+z+u+v = 7 where x,y,z,u,v >= 1 so we can simplify the equation as x' + y' + z' + u' + v' = 2 where x' = x + 1 and so on. and x',y',z',u',v' >= 0 now let the solution be written as x'0y'0z'0u'0v'. Visualize the 0's here as separators. such that when a variable is 0 we leave it as blank and when it is a number we write it as 1's eg. 5 is written as 11111 . in this case lets take one of the solutions as 2,0,0,0,0. so we will write 2 as 11 and leave the other variables as blank. so one tuple looks like 110000. 1,0,1,0,0 could hence be written as 100100. so the number of ways of having arrangements of 6 where 2 are of one kind and 4 are of other is 6!/(2! * 4!). NOTE: This looks long and complicated now but believe me , once you get a hang of it can save you a lot of work. such as : find number of positive integral solutions of x + y = 20 , where x>=2 and y >= 4 . Imagine counting all the solutions.

Q1: even i am getting

Q1: even i am getting 35. Q2: i am getting 9/6^5 . the combinations i took is 1,1,1,1,3 and 1,1,1,2,2. in above two combinations 3 can be arranged in 5 ways and in secound combination 2,2 can be arranged in 4 ways so total is 9. please verify. Q3:aanswer i got is 30/6

I think you missed something

Q2. You are almost there There are two possible combinations 1,1,1,1,3 and 1,1,1,2,2. For 1,1,1,1,3 total possible arrangements are 5!/4! = 5 and for 1,1,1,2,2 total possible arrangements are 5!/(2!3!) =10 So the answer is 15/6⁵ am i correct? Q3. Correct 5: 1

rajorshi your method seems

rajorshi your method seems to be a nice one but i cldnt make much sense out of it even after reading it few times. How we ll proceed the other question you have mentioned x + y = 20 , where x>=2 and y >= 4 x'+ 2 + y' + 4 = 20 =>x' + y' = 14 let one of the solution be 14, 0 which we can write as 1 1 1 (14 1's) 0 and the possible arrangements are 15!/14! =15 is it ok? or i have messed it up

yep looks good ! in this

yep looks good ! in this case you could easily verify your answer by counting ! ( just for now so that you build confidence in this method.). i found ths in the book of Challenges and Thrills of Pre coleege Mathematics (for Olympiads). However this will work only when the coefficients of all the variables are 1 clearly. if you guys want to solve equations in positive integrs of the form ax+by = 1 or ax-by = 1 you could look in Hall and Knight Higher algebra. the method uses continued fractions but its really easy.

re:first question z beauty

..wot nishit suggestd dat sol 2 q no 1.....explanation is good bt wnt work in cat...so um puttin up ma way 2 solve the questn.. take 1*3*5............*55 these are all the odd nos in first series... jes calculate no of 3's 1,3*,5,7,9*,11,13,15*,17,19,21*,23,25,27*,29,31,33*,35,37,39*,41,43,45*,47,49,51*,53,55 these are the elements...in the first series.. * signifies da digits which contains 3 as a factor now calculte no of 3's in this bt mind it herez da catch...9 vil contain two(02) 3's and 27 vil contain three(03) threes...now calculate no of threes in dis it vil becum 1+2+1+1+3+1+1+2+1=13(threes) similarly calculate in second string also

Q.2

for questn no 2..i suggest ya to go by sekhar's method...it vil save tym in CAT