brainbuster no. 22 - Out for the count

This competition closed on 2 April 2007.

the brainbuster

The sum of all the digits used in the series of numbers from 1 to 12 inclusive is 51:

i.e. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + (1 + 0) + (1 + 1) + (1 + 2) = 51

The sum of all the digits used in the series of numbers from 1,020 to 1,022 inclusive is 12:

i.e. (1 + 0 + 2 + 0) + (1 + 0 + 2+ 1) + (1 + 0 + 2 + 2) = 12

What is the sum of all the digits used in the numbers from 1 to 1,000,000 inclusive?

(In addition to providing the correct answer, you must also show the method that you used to calculate the sum.)

the solution

Answer: 27,000,001

The solution as provided by our winner:

Leave 1,000,000 on its own, then pair up the numbers as follows:

(999,999) with (0)
(999,998) with (1)

And so on until:

(500,000)+(499,999)

so all the numbers from 1 to 999,999 appear in a pair. The digits in each pair add up to 54, and there are 500,000 pairs, giving a total of 27,000,000. Then add in the (1) from (1,000,000) to give 27,000,001.

and the lucky winner is...

Congratulations to Peter Wilkins - currently studying for a PhD in Aerodynamics at Cranfield University - who wins £200 of Firebox vouchers.

Many thanks to all entrants - do try brainbuster no. 23!

thinkingpeoplefirst