Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem5

Video Transcription

Five, all four-digit positive numbers which have the sum of their digits equal to four were arranged in decreasing order. At which position is the number 2011 found?  So let's make a list of all such possible numbers, noting that a four-digit number cannot begin with a zero, and so we begin with the number four.  The largest such number we can make is the number 4,000. Then if we use a three, 3,100 is possible, 3,010 and 3,001 are all the possibilities here with three as the first digit.  And with two as the first digit we have 2,200, 2,022, and 2,002 as the possibilities, noting that zero cannot be the first number.  And then we can have 2,110, 2,101, 2,111 if we use a two first and then some ones, and then we can stop because in between these numbers we don't have any more possibilities.  So now let's order them from least to greatest. 4,000 would be the biggest number so that comes first, then 3,100, 3,010, 3,001, then  2,200, 2,110, 2,101, 2,022, then 2,011, and in 10 position is 2,002. So 2,011 would appear in position 9. The answer is D.

Video Summary

The number 2011 is positioned 9th when all four-digit positive numbers, whose digit sum equals four, are arranged in decreasing order. The sequence is: 4,000; 3,100; 3,010; 3,001; 2,200; 2,110; 2,101; 2,022; and 2,011.