"Finding Smallest n for Congruence Modulo Homework

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Homework Statement



Find, with proof, the smallest positive integer n that satisfy all the congruences.

n = 1 (mod 2)
n = 2 (mod 3)
n = 3 (mod 4)
n = 4 (mod 5)
n = 5 (mod 6)
n = 6 (mod 7)
n = 7 (mod 8)
n = 8 (mod 9)
n = 9 (mod 10)

Homework Equations



Let a,b,m within Z with m > 0. Then, a = b mod m is m|a-b.
(not sure if relevant or not)

The Attempt at a Solution



I tried to figure out what some of the ristriction on n
(ie - has to be odd since n = 1 (mod 2))
but didn't get too far.
 
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The first congruence tells you that n must be odd. If you try odd numbers, starting from 3, on the other congruences, what do you get?
 
I sort of tried that approach using the other ristrictions but started to get into the 400s. (with only considering 3 of the 9 ristrictions).

plus, I need to provide a proof and this way is probably not going to work as a proof.
 
400 is hardly a large number :smile:.

Look, if you want strictly positive integers, the smallest that satisfy the first congruence is 3. Now, the smallest that satisfies the first two is 5. Now, what is the smallest that satisfies the first three?

When you get to the last one, you will have the smallest that satisfies all; the steps are a valid proof.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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