Bunuel
Tough and Tricky questions: Remainders.
If x and y are positive integers and \(n = 5^x + 7^{(y + 15)}\), what is the units digit of n?
(1) \(y = 2x – 15\)
(2) \(y^2 – 6y + 5 = 0\)
given that x and y are positive integers they are >0
\(5^1\) has a unit digit of 5
\(5^2\) has a unit digit of 5. This means that regardless of the exponential power of 5 it will always have a digit of 5, as long as the power is a positive integer.
So we are interested in knowing the y value.
Statement 1) does not provide us any information.
Insufficient
Statement 2) \(y^2 – 6y + 5 = 0\)
(y-5)(y-1) = 0
y = 1 or 5
The unit digit of powers of 7 are below.
\(7^1\)=7
\(7^2 = 49\)
\(7^3 = 343\)
\(7^4 = 2401\)
This cycle repeats in multiples of 4, so 8,12,16,20 would have a unit digit of 1.
if y =1 then the power is 16, meaning it will have a units digit of 1
if y = 5 then the power is 20, meaning again it will have a digit of 1.
sufficient.
B is the answer.