We need to use the divisibility rule of 11 :

Form an alternating sum of the digits. The result must be a multiple of 11 or it must be 0.



We start the first digit as a positive (+) and subsequently alternate the signs.


Examples :

46827 ( = 11 x 4257)

+4 - 6 + 8 - 2 + 7 = 11 ✓


8684874 ( = 11 x 789534)

+8 - 6 + 8 - 4 + 8 - 7 + 4 = 11 ✓


3666663 ( = 11 x 333333)

+3 - 6 + 6 - 6 + 6 - 6 + 3 = 0 ✓

So now we need to solve for n '2008's and 623 as the last 3 digits.


Now 2008 has 4 digits.


For every 2008, the alternating sum is :
+2 - 0 + 0 - 8 = -6

So for n '2008's , their alternating sum is -6n (n times of -6)


We then continue for the last three digits 623. Their alternating sum is +6 - 2 + 3 = 7



In total their alternating sum is : -6n + 7 or 7 - 6n


So we can try a few values of n.


For n = 1,

7 - 6n = 7 - 6(1) = 7 - 6 = 1 ✖


For n = 2,

7 - 6n = 7 - 6(2) = 7 - 12 = -5 ✖


For n = 3,


7 - 6(3) = 7 - 18 = -11 ✔

(Since -11 = 11 x (-1) )


So smallest n = 3

Do u have any method. For this, I do not understand. Thanks

Which part is not understood?

Anyway, the official answer given by the SMOPS also uses the divisibility rule of 11 and a similar working.

You might want to read up on the divisibility rules online

Why is there negative in 2008 ? Actually I very confused . Maybe I don't know how to do at all. Thank anyway for help. So kind of you.

Because the series goes like this :

200820082008....623

Since we have to apply the alternating signs,

In every 2008 , the alternating sum of digits
= +2 - 0 + 0 - 8
= -6


There are 4 digits in 2008. That's an even number of digits so we always end with a - for the last sign.

No matter how many 2008s are there, it is always ending with a - for the last sign.

So for n 2008s, the sum is -6 × n = -6n.

Then when we finish up the at 623, we continue with + 6 - 2 + 3 = + 7

So the alternating sum over all is -6n + 7 = 7 - 6n

Note that for this alternating signs, we can start with - first also.


For every 2008,

-2+0-0+8 = 6

For n 2008s, that would equal 6 × n = 6n
It always ends with a + as the last sign.

When we reach the last three digits 623,
-6 + 2 - 3 = -7


Total alternating sum = 6n - 7

Thank you so much. I kind of understood. I know what do u means. But is it only apply on 11x ? Or for what this type of question ? Are u a tutor?

If we list it out it would be :


2 - 0 + 0 - 8 + 2 - 0 + 0 - 8 + 2 - 0 + 0 - 8 + ... + 2 - 0 + 0 - 8 + 6 - 2 + 3
(n times)


= -6 - 6 - 6 - ... - 6+ 7
(n times)

= -6n + 7

= 7 - 6n


Or


-2 + 0 - 0 + 8 - 2 + 0 - 0 + 8 - 2 + 0 - 0 + 8 - ... - 2 + 0 - 0 + 8 - 6 + 2 - 3
(n times)


= 6 + 6 + 6 - ... + 6 - 7
(n times)

= 6n - 7

Yes, this only applies for division by 11. Yup I'm a tutor.

Here's a resource for it :

https://www.mathsisfun.com/divisibility-rules.html

I believe for Olympiad competitions, students have to be familiar with all the divisibility rules in general.

Your students must be very lucky to have you as their teacher for been so patience and nice .