Data Sufficiency

If X and Y are positive integers and n = 5^X and 7^(y+15), what is the units digit of n?
1.y = 2x - 15
2.y^2 - 6y + 5 = 0

The OA is B.

Can someone elaborate on how to solve this one? Seems like a simple problem of units digit using cyclicity?

Hope the question makes sense now.

Q is not clear, what do u mean by n = 5^X and 7^*(y+15)?

ANy power of 5 ends in 5 so no value of x required

y^2 - 6y + 5 = 0 =>(y-1)(y-5)=0=>y=1 o r5

7^16 ends in (7^4)^4 = 1^4 = 1 (7,9,3,1 is cyclicitty of power of 7) in 1.
7^20 ends in (7^4)^5 = 1^5 = 1

so B is fine even though it gives 2 values as both ends in 1

maihuna wrote:Q is not clear, what do u mean by n = 5^X and 7^*(y+15)?

ANy power of 5 ends in 5 so no value of x required

y^2 - 6y + 5 = 0 =>(y-1)(y-5)=0=>y=1 o r5

7^16 ends in (7^4)^4 = 1^4 = 1 (7,9,3,1 is cyclicitty of power of 7) in 1.
7^20 ends in (7^4)^5 = 1^5 = 1

so B is fine even though it gives 2 values as both ends in 1

Thanks...you understood the question correctly.. i figured out where i was making the mistake.