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Sum of 1s

Age 11 to 14 Short Challenge Level:
Doing part of the calculation
Imagine writing out a long sum with the digits in columns.
In the units column, there are 2016 1s, so the last digit in the answer will be a 6, and 201 will be 'carried' into the tens column.
 
There are only 2015 1s in the tens column, so the total for the tens column is 2015 + 201 = 2216. This means that the second last digit will be a 6, and 221 will be 'carried' into the hundreds column, as shown below.
 
There are only 2014 1s in the hundreds column, so the total for the hundreds column is 2014 + 221 = 2235. This means that the third last digit will be a 5, and 223 will be 'carried' into the thousands column, as shown below.
 
There are only 2013 1s in the thousands column, so the total for the thousands column is 2013 + 223 = 2236. This means that the fourth last digit will be a 6, and 221 will be 'carried' into the ten thousands column, as shown below.
 
There are only 2012 1s in the ten thousands column, so the total for the ten thousands column is 2012 + 223 = 2235. This means that the fifth last digit will be a 5, and we don't need to worry about what will be 'carried' into the ten thousands column.

So the last 5 digits will be 56566, as shown below.
 



Using place value
The first number is 1 digit long, the second number is 2 digits long, and so on, so since the last number is 2016 digits long, it must be with 2016th number - so there are 2016 numbers in the list.

The units (ones) digit of every number in the list is 1, so there are 2016 units (ones) in the sum.

The first number doesn't have a tens digit, but the tens digit of every other number is 1, so there are 2015 tens from the tens digits in the sum.

The first two numbers don't have a hundreds digit, but the hundreds digit of every other number is 1, so there are 2014 hundreds from the hundreds digits in the sum.

Similarly there are 2013 thousands from the thousands digits and 2012 ten-thousands from the ten-thousands digits in the sum.

This is filled in in the table below, which shows the place value of each digit. At the bottom, adding the digits in each of the last five columns gives the last five digits of the sum, which are 56566.

      TTh Th H T U
        2 0 1 6
      2 0 1 5  
    2 0 1 4    
  2 0 1 3      
2 0 1 2        
      5 6 5 6 6

You can find more short problems, arranged by curriculum topic, in our short problems collection.