6545454545454.....(54 comes 100 times) / 5454545454545...(45 comes 100 times) == ?
Please help me in solving this question
hey... since numerator and denominator both are 201 digit number and numerator > denominator so the remainder, simply, will be the difference of num and deno. i.e (6545454545454........545454) -(5454545454545......454545)= 1090090909.........(09 comes 100 times)
The questionh is to divide both of them yaar not the difference
hey... since numerator and denominator both are 201 digit number and numerator > denominator so the remainder, simply, will be the difference of num and deno. i.e (6545454545454........545454) -(5454545454545......454545)= 1090090909.........(09 comes 100 times)
sry i thought u were asking for remainder. for division here it goes.
after dividing first time 6545454....5454 by 54554545....4545 we are getting remainder 1090909....0909(09 100 times) now multiply 1090909...0909 by 10 we get 109090909...09090 now the question turns to 10909090......09090/545454545.....4545 it is clear that numerator is 2 times of denominator.
The questionh is to divide both of them yaar not the difference
SINCE the question is of the form x/y where the number of digits in both are same so since numerator has 6 as frst digit and denominator has 5 so we can see the quotient will be 1..from this we can represent the number x=y*1+remainder which implies remainder =x-y
1) 240. lets do in reverse way. let us find out the number of factors that are divisible by 2 3 5 and den subtract.. n(2)= 900/2= 450 n(3)= 900/3= 300 n(5)=900/5= 180 now in this we have included numbers which are multiles of both 2 and 3 like 6.. and 3 and 5 like 15 and 2 and 5 like 10 we have a formula n(AUBUC) = n(A) +n(B) +n(C) -n(AIntersection B) -n(AIntersection C)- n(C Intersection B) +n(AIntersection B Intersection C) n(2*3)=n(6) = 150 n(10)= 90 n(15) = 60 n(2*3*5) = n(30) = 30 sub the values we get 660 these many numbers are divisible by 2 or 3 or 5 so 900- 660 = 240 gives the answer..
2) the least number is 1000000000000 and the highest number is 999999999999 so to have sum as 2, 2 cases 1) when 1st digit is 1, then the other 1 can be in 12 other places so 12 values possible 2) when digit is 2.. it has to be the first digit so only 1 value so 12+1 = 13 3) pls follow http://www.pagalguy.com/discussions/official-quant-thread-for-cat-2011-part-4-25070678
I didn't actually find it, it just looked too small to be the sum so I thought maybe he's made a mistake reading the question and found the 20th term, because he sounded confident.
