## silent 2 years ago Hi all - is anyone currently working on Assignment 1: McDiophantine: Selling McNuggets? I am working on it and would like to see if anyone has any input on how they would approach this (all parts). Thanks, please read down in the comments before posting.

1. ChapLeo

Problem 1 is the proving part. You can either write a program or use a pencil and paper. I did it in the latter method. Just list out the possibilities. For example; 6(0) + 9(6) + 20(0)= 54. Think of others like this one.

2. silent

do you know how to set up the code? the assignment my teacher gave me was to use "for loops" to prove if the combinations were possible but i keep getting stuck on just trying to set it up, i have written it down on paper and have tried to code it but i keep getting stuck.

3. ChapLeo

Don't know whether I can give you the direct answer since it's an assignment question, but here's a hint: you have to check each and every number combination possible. So you will need to use nested for..... loops. Does that help?

4. silent

what do you mean by nested? do you mean a "for loop" inside of a "for loop", because that is what i am currently doing but i am not sure what numbers i should be using in the for loops

5. ChapLeo

Yes! nested means a loop inside a loop. The number range is of your choice. Think wisely; the largest number is 55, right? So if you are creating a range for '6', it should be between 0 and 10, right? since 6 x 10 =60, which is above the range of our interest. think like that for 20 and 9 as well.

6. silent

I have this bit of code so far but i am unsure what i am doing wrong. int a, b, c, total; total = 55; for(a=1; a<total; a++) { for(b=1; b<total; b++) { for(c=1; c<total; c++) { System.out.println(a); System.out.println(b); System.out.println(c); can anyone help me?

7. TehmeerAli

You are using like Java. There is no a++ in Python. Python syntax is little bit different. I was doing same kind of mistake. However, I am stuck in first problem of problem set 1 that says get last module of 1000.