y=9^x*(1+9+9^2+..+9^5) or 9^x*(9^6-1/8)

now 9^6/8 will leave reminder 1..hence 9^6-1 is divisible by 8
also last digit of 9^6 is 1..hence last digit of 9^6-1 is 0..i.e 9^6-1 is multiple of 10..hence 9^6-1 is divisible by both 8 and 5

so if we can conclude 9^x is integer y will be divisible by 5

only option b states that

hence ans option B

bownarrow wrote:q) x is a positive number. If 9^x+9^(x+1) +9^(x+2)+ 9^(x+3)+9^(x+4)+9^(x+5)=y, is y divisible by 5?

(1) 5 is a factor of x

(2) x is an integer

9^x+9^(x+1) +9^(x+2)+ 9^(x+3)+9^(x+4)+9^(x+5) = 9^x (1+9+9^2 + 9^3 + 9^4 + 9^5)
= 9^x(9^6 - 1)/(9-1) = 9^x . 66340

which is divisible by 5 anyway whatever be the value of x.
So D

bownarrow wrote:q) x is a positive number. If 9^x+9^(x+1) +9^(x+2)+ 9^(x+3)+9^(x+4)+9^(x+5)=y, is y divisible by 5?

(1) 5 is a factor of x

(2) x is an integer

all u need to know if a number is divisible by 5 is if the unitgs digit end with 0 or 5.
simplyfying the expression you get;
9^x(1+9+9^2+...+9^5)=y
working inside the bracket to get the units digit;
1+9+1+9+1+9 =30.
since it is given that x is a positive number, and we know from both statement 1 and 2 that
it is an integer, we know for both cases y will be divisible by 5.

analyst218 wrote:

bownarrow wrote:q) x is a positive number. If 9^x+9^(x+1) +9^(x+2)+ 9^(x+3)+9^(x+4)+9^(x+5)=y, is y divisible by 5?

(1) 5 is a factor of x

(2) x is an integer

all u need to know if a number is divisible by 5 is if the unitgs digit end with 0 or 5.
simplyfying the expression you get;
9^x(1+9+9^2+...+9^5)=y
working inside the bracket to get the units digit;
1+9+1+9+1+9 =30.
since it is given that x is a positive number, and we know from both statement 1 and 2 that
it is an integer, we know for both cases y will be divisible by 5.

THANKS FOR THE REPLY . BUT DO U THINK X SHLD BE AN INTEGER. SINCE THE LAST DIGIT IS 0 AND X IS POSITIVE ANYWAY Y IS DIVISIBLE 5????????????