Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem9

Video Transcription

Question number 9, Damien wrote on the chalkboard all odd numbers from 1 to 2011 inclusive. Kosma erased all numbers divisible by 3. How many numbers were left on the chalkboard?  Let's enumerate some of the odd numbers that appear on the chalkboard, beginning with 1, then we have 3, 5, 7, all the way up to 2009, and then followed by 2011.  And we can see that the formula for these would look like taking an even number and then subtracting 1 from it, so 2k minus 1, where k ranges from 1, 2, 3, all the way up  to 1006, which would give us 2012 minus 1, so the last number on the chalkboard. And so we see that Damien wrote exactly 1006 numbers.  Now we have to find how many were erased by Kosma, and he erased all the odd ones divisible  by 3, so 3, then 9 is next, then 15, and remember that the test for divisibility by 3 is add up all the digits of a number and see if they are divisible by 3.  So 2010 works, but that's not odd, so 2009 is not divisible by 3, 2008 is not, 2007 is the last number on the list.  And these are odd numbers, so they have to look like 2n minus 1, but they also have to be divisible by 3.  So they're generated by 6n minus 3, where n here ranges from 1, 2, all the way up to 335. So we see that Kosma erased exactly 335 numbers.  So then it's a matter of subtracting, there are 1006 minus 335, or 671 numbers remaining.  And so we see that our answer is C.

Video Summary

Damien wrote all odd numbers from 1 to 2011 on the chalkboard, totaling 1006 numbers. Kosma erased those divisible by 3. The divisible sequence follows the form \(6n - 3\), resulting in 335 numbers being erased. Subtracting these gives 671 numbers remaining on the chalkboard. The final answer is 671 numbers.