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Guest K Sengupta
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N is a duodecimal (base 12) positive integer having precisely fifty digits which is divisible by the duodecimal number 157.

Determine the units digit of N given that each digit of N is the same with the exception of the units digit.

Note: N cannot contain any leading zero.

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^ Oh, yeah silly question, I should've googled it in the first place but then I couldn't edit my post...

Anyways, I have a solution plan but I have a problem, there's something wrong either with my theory or the question itself:

First, we have the duodecimal integer K which is 50 digits of all 1's (K=1111.....111), now I wrote a program that'll do the following:

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Create an array of integers, 50 long, each has the number 1 in it (these will represent the digits of K)

Find the integer digNum which is the position of the first non-zero digit counting from left to right.

Subtract 1 from the integer in the array in the position digNum+1, subtract 5 from digNum+2, subtract 7 from digNum+3

Go over the digits of the array from right to left, if a digit is negative, add 12 to it and subtract 1 from the digit left to it.

Stop when digNum is more than 47 (as in the array is less than 3 digits)

Display the array.

////

This is not brute force, this is more like (or even exactly like) second grade subtraction method, it's not brute force because it has done all that in half a second and when I use brute force on numbers with more than 6 digits usually I get warning messages saying that the script is taking too much memory...

What the program basically does is if you have K:

11111111111111111111111111111111111111

-1570000000000000000000000000000000000

^ Is what you subtract from it and continue like that until you have a 3 or less digit number.

I have found that K12%15712=1612 (% means remainder of division), from that you can conclude that (K*2)12%15712=3012 and (K*3)12%15712=4612...(K*11)12%15712=14612

To make a number divisible by 157 you must either subtract from it to make the remainder 0 or add to it to make the remainder 157, and that by only changing the first digit, I can't see how it can be done in any of these results, anyone have an idea why?

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I've looked over the script and tested it over and over, the closest number to what the question is looking for is:

111111111111111111111111111111111111111111111110A7

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I think there's a typo in the problem. The 15712 should be 13712.

8. Where N is:

5555555555555555555555555555555555555555555555555812

Oh, and 14712 also works with N of:

9999999999999999999999999999999999999999999999999512

in which case the answer is 5.

Ty SP.

Indeed, it was a typographical lacuna and, it should have been "duodecimal number 137" in the first line inclusive of the OP. Given the maximum time limit of 10 minutes to edit my post, it was not possible to rectify the OP

Edited by K Sengupta
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^ You could've corrected in a new post...

Yeah I wrote a program in Python a couple of days ago ... trying to figure it out and thinking I must be missing something ... and then memorial day came and I had to do some family stuff ... Anyways I got the answer(s) now ... I was mad for two days thinking I'm an Idiot ... I'm glad to know I'm not

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