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 January 20th, 2016, 03:07 AM #1 Senior Member   Joined: Oct 2013 From: Far far away Posts: 429 Thanks: 18 counting problem If no repetitions are allowed, using the digits 0,1,2,3,4,5,6,7,8,9, how many A) two-digit numbers can be formed? B) of these are odd? C) of these are even? D) are divisible by 3? E) are less than 40? My attempt: A) since zero can't be first digit, the number of two-digit numbers that can be formed = 9*10 = 90. But repetitions are not allowed so 11, 22,..., 99 must be subtracted. So, the number of two-digit numbers without repetition = 90 - 9 = 81 B) number of odd numbers (with repetitions) = 9*5 = 45. Since repetitions are not allowed we have to minus 11, 33, 55, 77, 99. So, number of odd numbers (without repetitions) = 45 - 5 = 40 C) the number of even numbers = 81 - 40 = 41 D) I don't know how to solve this but here's an attempt: with repetitions allowed every third number in the list is divisible by 3. So the number of numbers divisible by 3 = 90/3 = 30. However, we must remove the multiples of 3 that are repetitions which are 33, 66, 99. That means 30 - 3 = 27 numbers in the list are divisible by 3 E) in the list with repetitions allowed 3*10 = 30 numbers are less than 40. But we have to remove the repetitions viz. 11, 22, 33 leaving us with the correct answer of 27 numbers in the list being less than 40. Am I correct? Is there a better way to solve this problem? Thanks. January 20th, 2016, 04:13 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra I think your answers are fine. I can't see any better ways to solve, only different. Thanks from shunya January 20th, 2016, 07:07 AM   #3
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 Originally Posted by v8archie I think your answers are fine. I can't see any better ways to solve, only different.
Next time I'll rephrase my question...i don't mind a different approach to the problem. Thanks. January 20th, 2016, 07:09 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra You can use modular arithmetic to find solutions where the digit sum is divisible by three. Thanks from shunya January 20th, 2016, 07:12 AM   #5
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 Originally Posted by v8archie You can use modular arithmetic to find solutions where the digit sum is divisible by three.
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