[Guest]

All the terms of an arithmetic progression having the first term as 2 with common difference 4 and the last term as 2010, are written successively without commas or spaces, resulting in the following pattern:

2610141822263034384246........20062010

Determine the sum of all the digits in the above pattern.

Edited January 29, 2010 by K Sengupta

[Guest]

Sum of digits = 2

[Guest]

The one's place digit is always one of the even digits (0,2,4,6,8)... and the first few numbers are 2 6 10 14 18 22... the pattern repeats after 20 numbers.

Total number of times 2 comes in the end = 2002 - 2 /20 = 100

Similarly number of times that 8 comes in the end is 1998 - 18 / 20 = 99

For tens place, we have all digits from 0 to 9 while even digits have 2 numbers each, odd digits have 3 numbers each

So, number of times that 0 occurs in 10s place is 40, 1 occurs 61 times, 2 ocurs 40 times, 3 occurs 60 times and so on

For hundreds place, from 2 to 102, there are 25 numbersand this pattern repeats every 100 numbers. So, each digit occurs 50 times in 100s place

In thousands place, 1 occurs 250 times and 2 occurs 3 times.

Adding all these possibilities we get the sum of digits

So the sum of all digits is 6793

[plainglazed]

6815 similarly to DeeGee but with 100 each of the even numbers in the ones place plus the extra 2 and 6 in 2002 and 2006 (2008 total in the ones column). 60 of each odd number and 40 of each even number plus one extra 1 in 2010 (2301 total in the tens column). 50 each 1-9 and 53 zeros (2250 total in the hundreds column). 250 ones plus three 2s in 2002, 2006, 2010 (256 total in the thousands column)

[Guest]

6815 similarly to DeeGee but with 100 each of the even numbers in the ones place plus the extra 2 and 6 in 2002 and 2006 (2008 total in the ones column). 60 of each odd number and 40 of each even number plus one extra 1 in 2010 (2301 total in the tens column). 50 each 1-9 and 53 zeros (2250 total in the hundreds column). 250 ones plus three 2s in 2002, 2006, 2010 (256 total in the thousands column)

I think yours is correct... I made a mistake by counting 1 less for each of the even digits in ones place

[Guest]

All the terms of an arithmetic progression having the first term as 2 with common difference 4 and the last term as 2010, are written successively without commas or spaces, resulting in the following pattern:

2610141822263034384246........20062010

Determine the sum of all the digits in the above pattern.

Sum of all digits is 6813.