For n in N (Natural Numbers), use modular arithemetic to establish each of the following divisibility statements: a) 7 | 52n + 3 ( 25n-2) Attempt: Well I've set up a mod 7 table and was figuring out the values from n = 0,....6 for the two individual expressions, then I was going to combine them. But I of course hit a rough patch when I got to 25n - 2. Since I'm in the Natural numbers, fractions do not exist, so how do I treat 25n - 2 when n = 0 ? (Note: This is what I attempted prior to getting help originally, then I was told that 0 would not be included in the Natural numbers for this question.)......but I was working through this question but it occurred to me that I'm trying to show that 7 | 52n + 3 ( 25n-2), so wouldn't I have to show that: 52n + 3 ( 25n-2) "congruent" 0 (mod 7)?.......but how can I show this if 0 is not suppose to be included in my work? A follow up to that question is that the question asked me to establish that the above expression is true. Is the only way I can establish its truth is by using a mod table or is there another method? and does this apply to most questions of this nature?