The problem interests me, as I have always been fascinated by the old hand held calculator "check out" multiplication whereby you enter the amount 12345679, and then multiply it by any multiple of 9 and you will get an answer consisting of all digits that were the multiplier. For example, 12345679 * 63 = 777777777.

Is there any connection between these two interesting appearances of the serial sequence of digits, including all but the number 8?

Jack

we are after the sum of
1 + 11 + 111 + 1111 + 11111 + ... + 1111111....111111
in the last term, we have 2002 1s

i.e we are after

2002 + 10 * (2001) + 10 ^ 2 * (2000) + 10 ^ 3 (1999) + ... +
10 ^ 2000 * 2 + 10 ^ 2001

and with the calculator prob
I think it's because
12345679 x 9 is the magic number
111,111,111
if we see 12345679 as (1+11+111+1111+11111+...+111111111+1111111111) / 100

where (1+11+111+1111+11111+...+111111111+1111111111) = 1234567900

12345679 x 9 will = to
9 + 99 + 999 + 9999 + 99999 + ... + 9999999999
10 9s are 90, record 0, carry 9
9 9s + one 9 = 10 9s = 90, record 0, carry 9
8 9s + one 9 = 9 9s = 81
7 9s + one 8 = 8 9s - 1 = 71, carry 7
6 9s + one 7 = 7 9s - 2 = 61, carry 6
....
the resultant digits are all 1s, except the first two 00s

why, i donno haha... (

also,
it is interesting how
12 * 9 = 108
123 * 9 = 1107
1234 * 9 = 11106
12345 * 9 = 111105
123456 * 9 = 1111104
1234567 * 9 = 11111103
12345678 * 9 = 111111102
and to compensate the 2, we can plus one more 9
i.e time 9 by 12345679 instead of 12345679 to obtain 1111111

i know the sum of the product's digits has to be 9, since it's a product of 9, but, yeah,its just weird, how all the results begins with 111111s

first column
9 1s + = 9
second column
9 2s + = 18
third column
9 3s + = 27
the 2 carried from 27 making 18 into 20, then 20 carried making 9 + 2 = 11 becomes 1

there may be some connections to draw between the digit sum property of numbers and the fact that 1234xxxx * 9 behaves in this way, may be i am just too dumb to figure that out.

The original question asked for the sum of 2002 numbers.

As for your question... I haven't worked it out, but I think you have an excellent guess that they might be related.

yes, i started out with a very same method, then, once the pattern started to break (7901012) btw the 7 and 9, I though something must have gone wrong, and I wasn't at all convinced that the number will consists of 2002 digits only (for the very last column has a carried number of 200, then 220, then 222,

and when I started from the left most column, I thought the ending result may be 1234xxxx, then again, I am not too convinced, for the number of carries involved,

its not until I actually ran the problem through the computer did I begin to realize that there really is a pattern to it

I think this is the result because
1
1 1
1 1 1
if you add the series
you get
1 in the first digit
2
3 4 5 6 7 8 9
then when ur up to 10
the 10 would carry and 9 becomes 0, and 8 becomes 9, then you have two 0 0s, this will allow you to do plenty of carries? namely, the next 1?

or should I say that, when the carry is 222, and the number is 1998
we obtain the 0 (last digit in pattern) the next number 1997 222 results in a 9, but it intialiates a new carry number 221, which will decrease the next digit by 2 (because the carry number is now 1 less, and the number of 1s in that column is also 1 less)
then the next few numbers don't change the carry number, until, 89 came along,

now the same pattern initiates again, until the 9th number in the sequence comes along, and decreases the carry number by 1

the same pattern repeats, and the carry number eventually reaches 0, and in each sequence, there is a 1, thus we have 222 1s 2 1s we had before. Yeh, i guess this way, i can get around without thinking about the number of digits in the sum.

am I somewhat right?

.11111... + .01111... + .00111... + ...
= 1/9 + 1/90 + 1/900 + ...
= (1/9) (1 + .1 + .01 + ...)
= (1/9) (1.11...)
= (1/9) (10/9)
= 10/81
= .123456790123456790...

That is why I like this forum so much. Many interesting things turn up.

Jack