#Miscellaneous-1
Posts
#Numbers-2
A number N, when divided by a divisor D, gives a remainder of 52. The number 5N, when divided by D, gives a remainder of 4.How many values of D are possible? SOLUTION- We know that, Dividend = Divisor*Quotient + Remainder So, a number N will be of a type, N=D*k + 52 Now, from the above equation, we can say that D>52 Now, when 5N is divided by D, i.e, 5N = 5(D*k+52) = Dk 1 + 4 =>5Dk + 260 = Dk 1 + 4 Now, 5Dk is divisible by D and when 260 is divided by D remainder is 4, =>5Dk+256=Dk 1 =>5k + 256/D=k 1 So, D must be 256 or its factors as k 1 is always a naturak number, Factors of 256 = 2^8 Now, 52<D<=256 (where D is a natural number) Values of D satisfying are 64,128,256 Hence, 3 values of D are possible ANS
#SERIES-1
In a certain series , the nth term T n equals 4T n+1 + n-1. If T 1 = 4 ,then find the value of T 200. Solution- T 1 =4 T 2 = 4T 1 + (2-1) = 4^2 +1 T 3 = 4T 2 + (3-1) = 4^3 +4+2 T 4 = 4T 3 + (4-1) = 4^4 +4^2 +2*4 +3 T 5 = 4T 4 + (5-1) = 4^5 +4^3 +2*4^2 +3*4 +4 .... ...... ......... ........... T 200 = 4^200 + 4^198 +2*4^197 +3*4^196+......+198*4 +199 =4^200 + S n Now, S n =4^198 +2*4^197 +3*4^196 +.......+198*4 + 199------(i) S n /4 = 4^197 +2*4^196 +3*4^195+..........+198 + 199/4--------(ii) Eqn (i)-(ii), 3S n /4 = 4^198 +4^197 +4^196+.........+4+1-199/4 3S n /4 = (4^199 - 1)/(4-1) - 199/4 3S n /4 = (4^199 - 1)/3 - 199/4 S n =(4^200-4)/9 -3*199/9 S n =(4^200 - 4-597)/9 ...
