a) Following a logic argument, 1 is going to be the most common digits as the sequence ends at 1000, which will make it appear once more than the other digits…
b) …that is except for 0, which will be the least common as it doesn’t have to sit in front of any number between 1 to 99 in the same way as the others do, i.e. 11,12… 22,23 etc.
c) So we know 0 and 1 will be different, but how many times do the other numbers show up. I found it easiest to add up the total of times a number appears in each column, i.e. single, tens, hundreds, thousands. We can make a generalisation of 2-9, and add one on the 1s.
1-9 1 occurrence of each number (except 0)
10-99(each number will appear 10 times in the tens column, and 10 times in the single column)
100-999 Each number shows up 99 times in the hundred column + 10 times in the ten column + 10 in singles
Then 20 times for each other hundred, not with that number i.e. 20 x 8 = 160.
So, 1 + 20 + 119 + 160 = 300 appearances for numbers 2 to 9.
And of course, 301 for the number 1…
And for zeros.
10-999 times
100-999(10 times in each hundred x 9) 90 in hundred columns, 90 times in each single column.
1000+3
9 +90 +90 +3= 192‘0’s in the sequence.
Too easy? Why not email your puzzle to jcunningham@findlay.co.uk and put your wits against the intellect of Engineering Materials’ readership.