For n in N (Natural Numbers), use modular arithemetic to establish each of the following divisibility statements:(adsbygoogle = window.adsbygoogle || []).push({});

a) 7 | 5^{2n}+ 3 ( 2^{5n-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 | 5^{2n}+ 3 ( 2^{5n-2}), so wouldn't I have to show that:

5^{2n}+ 3 ( 2^{5n-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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Mod question I thought I got but I don't.

**Physics Forums | Science Articles, Homework Help, Discussion**