Brent Hanneson wrote:x and y are positive integers such that xy = 10^n. If x>y, the units digit of x-y CANNOT be
A) 1
B) 3
C) 5
D) 7
E) 9

parallel_chase wrote:

Brent Hanneson wrote:x and y are positive integers such that xy = 10^n. If x>y, the units digit of x-y CANNOT be
A) 1
B) 3
C) 5
D) 7
E) 9

xy=10^n

x>y

x =5 , y= 2 , x-y = 3

x = 25, y = 4, x-y = 1

x = 10, y = 1, x-y = 9

x = 25, y = 8, x-y = 7

Hence 5.

aroon7 wrote:

parallel_chase wrote:

Brent Hanneson wrote:x and y are positive integers such that xy = 10^n. If x>y, the units digit of x-y CANNOT be
A) 1
B) 3
C) 5
D) 7
E) 9

xy=10^n

x>y

x =5 , y= 2 , x-y = 3

x = 25, y = 4, x-y = 1

x = 10, y = 1, x-y = 9

x = 25, y = 8, x-y = 7

Hence 5.

x=200, y=5, x-y = 195. so not 5
Ans is 7
when x= 25, y=8, we get xy = 200. this is not nth power of 10...

what is the OA?

thanks!

Brent Hanneson wrote:Man, what a bad day for me.
I think this might be my last post, since I have now posted 2 questions with missing/incorrect information.

The question needs the proviso, "x and y" are not multiples of 10.
If we add that proviso, the correct answer is C (5).

Yikes - pretty sloppy stuff.
Sorry to all those who spent a great deal of time on this poorly-constructed question.