Engineer's Puzzle

The Puzzle: The number of numbers

An engineer has found himself in a bit of a riddle. For reasons that will take far too long to explain here, he needs to quantify the digits between 1 and 1000. So in other words how many individual 0s,1s,2s… 9s are there in the sequential sequence 1 to 1000.

Of course you should try to apply logic here and find a pattern, rather than manually count the numbers, but how you do it is up to you.

The question is:

a) What is the most common digit, and why?

b) What is the least common digit, and why?

c) And, if you really want to impress, how many occurrences of each digit are there in the sequence 1 to 1000?

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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.

Justin Cunningham

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