The three-digit positive integer has a ones digit of 3. What is the probability that is divisible by 3? Express your answer as a common fraction.

Maplesnowy
Feb 24, 2018

#2**+1 **

CPhill: The question says "The three-digit positive integer ** has a ones digit of 3**". Doesn't that mean that the 3rd digit is fixed at 3? E.G. 103, 113, 123, 133, 143, 153 163, 173, 183, 193......and so on for a total of 9 x 10 = 90 3-digit numbers ALL ending in 3. And there would be: 90/3 =30 numbers out the 90 divisible by 3.** Or 30/90 =1/3.**

Guest Feb 24, 2018

#1**+1 **

For this to be divisible by 3, the sum of the first two digits must be divisible by 3

There will be (99-12)/ 3 + 1 = 30 such numbers

And there are (999-100) + 1 = 900 three digit numbers....so

The probability that one ends in "3" and is divisible by 3 is just

30 / 900 = 1/30

CPhill
Feb 24, 2018

#2**+1 **

Best Answer

CPhill: The question says "The three-digit positive integer ** has a ones digit of 3**". Doesn't that mean that the 3rd digit is fixed at 3? E.G. 103, 113, 123, 133, 143, 153 163, 173, 183, 193......and so on for a total of 9 x 10 = 90 3-digit numbers ALL ending in 3. And there would be: 90/3 =30 numbers out the 90 divisible by 3.** Or 30/90 =1/3.**

Guest Feb 24, 2018