just a lil more help.. I am just not understanding the 2^19 part!! It's just not registering.. I know it is stupid but cud you please explain it in a bit more detailed fashion? I know that 20c0+ 20c1+.... 20c20= 2^20. but why is 20c1+20c3+20c5+... 20c19 =2^19??? please if you could elaborate on this a bit..!!
@koyal1990@koyal1990 : Is the answer 11 ?? I dont know whats wrong with this forum page , there is some technical glitch due to which i am not able to remove my previous post part , kindly ignore that ...
@koyal1990 said:@ashish0102 - yes, there is some prob with this page. I have faced a similar prob b4, yep, ans is 11. cud u kindly explain the method? thanks....
The funda is if r=s mod(b) and n=m mod(b) then , r.n= s.m mod(b)
Now in full detail :
30^3 =1 mod 19 , so our goal is to reduce the given expression in the form of 30^3
i.e. we have to check the expo 20^10^5 is a multiple of 3 or not ,
As 20^2= 1 mod 3 , the same way the whole expression will become :
30^ (3k+1)= 30^3^k * 30^1
so finally the question is : 30^3^(integer)*30^1 (mod 19)
@ipagal - eulers number of 11 is10. so if we can get 50^10 somehow, then the remainder wud become 1. so now we look at the power- it is 51^52. very obviously it ends n 1 hence the power is of the form 10k+1. so we can re write the um as (50^10k+1)/11? this means we can further simplify it as 50^10k * 50^1? now see, as I said before eulers number of 11 is 10, so the 1st part will give 1 as remainder, n 50/11 is 6. so 1*6 is our final remainder. hope it is clear now??
@redd said: what approach is to be used while solving questions asking number of digits in a square root,cube root or product??
bro in case of product i think u can simply calculate the range at first sight and make a guess work..
for square roots and cube roots the method is.. lemme tell u by example suppose u have 2 digit no say 81, we need to calculate the digits in it's square root.. write it as (81)^1/2.. take the 'log'if u get a fraction then the ans is the closest integer greater than the fraction.. solve for 729.. log[(729)^1/2]= 1/2(2.7)= 1.35 now take the integer larger than tis which is 2(27 has 2 digits)..
follow similar metjod for cube roots too. log[(729)^1/3]= 1/3(2.7)= .9 now take the integer larger than this which is 1(9 has 1 digit)..
try for some numbers by urself.. no need to take exact value of log, take a rough value..
@redd it's actually 0.5 * 2.7..it was my mistake,forgot to put the brackets.. and the log of any 3 digit number will lie between 2 and 3.. so it was closer to 1000 so i took the value closer to 3. u can take any value between 2 and 3,it will solve ur purpose..
Please help me out with the following problem : Rajesh found a decimal number which when represented in bases 2,3,4,5 and 6 ends in 1,2,3,4 and 5 respectively
A) What is the smallest positive integer satisfying this property ?
B) How many such three-digit numbers are possible ?
The 38th term and the 88th term of an arithmetic progression are 301/25 and 903/25 respectively.If the total number of terms in the progression is 175, what is the ratio of the sum of the first 75 terms to the sum of the last 100 terms of the progression?
