- #1

member 428835

- Homework Statement
- Given random integer ##x##, what's the probability ##x^3## ends in 111?

- Relevant Equations
- Nothing comes to mind

The last three digits of ##x^3## must be solely dependent on the last 3 digits of ##x##. So let ##x=a+10b+100c## for integers ##a,b,c##. Then ##x^3 = a^3 + 30 a^2 b + 300 a b^2 + 300 a^2 c +O(1000)## where of course ##O(1000)## don't affect the last 3 digits. Evidently ##a^3## is the only number that governs the 1's digit, and clearly must be ##a=1##. Then we have ##30b## being the only value that affects the 10's digit, which implies ##b=7## since ##30*7=210##. Thus, the value ##14900 + 300 c## must dictate the 100's digit, which implies ##c = 4## Since ##9+3*4 = 21## which ends in a 1. So this implies of all the numbers, the last 3 digits must be ##471##, which is 1/1000 odds, since there are 1000 possibilities from ##0-999## to choose from.

Have I missed something? Sure feels right, but you never know.

Have I missed something? Sure feels right, but you never know.