Divide 3903 between amar and akbar such that amars share at the end of 7 yrs is equal to akbars share at the end of 9 yrs at 4 percent pa rate of compound interestAns is amar is 2028 and akbar is 1875
let amar's be x and akbar's will b 3903-x akbar's 9 yr's is equal to amar's 7 yrs share. it's given (3903-x)[1+1/25]^9=x[1+1/25]^7
solving .. (3903-x)[26*26/25*25] = x 3903*26*26=x+ x676/625 or x = 3*676 x= 2028
6??n^5 - 17n^3 + 16n = n^5 - 16n^3 -n^3 + 16n = n^3 ( n^2 - 16) - n (n^2 -16 )(n^3 -n) (n^2 - 16 ) = n (n+1)(n-1) (n+4)(n-4) = [(n-1)*n * (n+1) ] is always divisible by 6. ;No..perfect rule for (n-4) (n+4 ) so it has to be 6?/
Ravi, i think The highest number that always divides n(n^2-1) is 24. (if n is odd.) and it is 6 if n is even. Since n>=4 which will include odd values too, it should be 24.
your answer is coincidentally right but your logic is wrongaccording to your logic, the answer of this question -Sum of all positive integers less than 6 and relatively prime to 3 should be 3 but it actually is 12.This is because n*E(2n) is not equal to 2n*E(n)
sir prime factors of 3 = 3 and prime factors of 6 = 2 and 3 we cannot apply it here
but there prime factors of 150 = 2 , 3 and 5 and prime factors of 300 = 2 , 3 and 5
prime factors of both the numbers are same so we can apply this
Ravi, i think The highest number that always divides n(n^2-1) is 24. (if n is odd.) and it is 6 if n is even. Since n>=4 which will include odd values too, it should be 24.
sir prime factors of 3 = 3 and prime factors of 6 = 2 and 3 we cannot apply it here but there prime factors of 150 = 2 , 3 and 5 and prime factors of 300 = 2 , 3 and 5 prime factors of both the numbers are same so we can apply this
your answer will never match if n is odd.
try this question - Sum of all positive integers less than 2250 and relatively prime to 1125.
The 150 contestants of Miss India 2010 are given individual numbers from 1 to 150. Several rounds happen before the final winner is selected. The elimination in each round follows an interesting pattern. In the 1st round starting from first contestant, every 3rd contestant is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) contestant in the next round (leaving 3, 5, 8, 9, ... ). This process is carried out repeatedly until there is only the winner left. What is the number of Miss India 2010?
The 150 contestants of Miss India 2010 are given individual numbers from 1 to 150. Several rounds happen before the final winner is selected. The elimination in each round follows an interesting pattern. In the 1st round starting from first contestant, every 3rd contestant is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) contestant in the next round (leaving 3, 5, 8, 9, ... ). This process is carried out repeatedly until there is only the winner left. What is the number of Miss India 2010?(a) 93 (b) 48 (c) 119 (d) 38 (e) 140
i have been trying this for the past 20 mins....got 140 by brute force....cld not find a pattern... :(
The 150 contestants of Miss India 2010 are given individual numbers from 1 to 150. Several rounds happen before the final winner is selected. The elimination in each round follows an interesting pattern. In the 1st round starting from first contestant, every 3rd contestant is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) contestant in the next round (leaving 3, 5, 8, 9, ... ). This process is carried out repeatedly until there is only the winner left. What is the number of Miss India 2010?(a) 93 (b) 48 (c) 119 (d) 38 (e) 140
150 in base 3 - 12120 left shift operator two times - 12012 12012 in base 10 = 2 + 3 + 54 + 81 ==140
you are not getting my point i said prime factors of 150 = prime factors of 300 and are same i.e. 2 , 3 , 5 but here prime factors of 2250 = x , 5 and 2 and prime factors of 1125 = a , b , 5 (2 is not here) so , that won't work here
so why are you advocating a method that won't work in half of the cases?
