Question

How many positive integers n satisfy 127 \equiv 7 \pmod{n}? n=1 is allowed.

Answer 1

$$127\equiv7\pmod{n}\\127=kn+7\\120=kn$$so essentially it boils down to how many divisors \(120\) has.

Answer 2

So, how many positive integers n satisfy 127=*7 (mod n)? n=1 is allowed.

Where =* is the cungruent symble

Answer 3

oops congruent not cungruent

Answer 4

why kn?

Answer 5

to count the number of divisors observe we can factorize \(120=2^33^15^1\) hence we have \(4\times2\times2=16\) divisors (since each divisor must be of the form \(2^i3^j5^k\) for \(0\le i\le3,0\le j\le 1,0\le k\le 1\))

$$1\times120=120\\2\times60=120\\3\times40=120\\4\times30=120\\5\times24=120\\6\times20=120\\8\times15=120\\10\times12=120$$

Answer 6

@orple8 that is what congruence means; \(a\equiv b\pmod{n}\) means that \(a=kn+b\) for some integer \(k\)

Answer 7

oh right *slaps forehead* I'm stupid

Answer 8

so the answer would be 12?

Answer 9

no wait, 16. the answer would be 16. ooohhh! I see...thanks!

Answer 10

i made a few silly mistakes at first :-p like incorrect prime factorization

Answer 11

thanks so much! :)

Answer 12

closing? the question i mean XD

Answer 13

oh shoot actually maybe we don't need to ignore the first couple

Answer 14

I keep confusing myself :-p yeah we don't need to ignore the small ones, see:$$127\equiv7\equiv0\mod1\\127\equiv7\equiv1\mod2\\127\equiv7\equiv1\mod 3\\127\equiv7\equiv3\mod 4\\127\equiv7\equiv2\mod 5\\127\equiv7\equiv1\mod 6$$

Answer 15

\(16\) is correct though yep :-p

Answer 16

heehee :) that's kind of funny. no offense.

Answer 17

no offense taken! The only modular arithmetic I've done in school was in 6th grade so about 6 years ago now :-p

Answer 18

you're lucky! I didn't do modular arithmetic until 7th grade :D