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# Help?

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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.

Feb 24, 2018

#2
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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.

Feb 24, 2018

#1
+107483
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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

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
#3
+107483
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Ah, yes...I see that, now.....thanks for the the correction!!!!

CPhill  Feb 24, 2018
edited by CPhill  Feb 24, 2018