Recent content by Jonathanlikesmath

Well, I soon won't forget this method. Thank you all for the help!

I was working this and was thinking something along the lines of this for the modulo arithmetic. I got to a point where I need to use induction, so I might have taken a wrong turn... And have removed some things such as the first paragraph covering how the factors are spread across the first...

Its amazing you mentioned Bezout's Lemma, even though its not named as such in my text, its highlighted as a "fundamental proposition" and proved as an exercise in the current section of my text!

I did not recognize that a number is divisible by 120 iff it is divisible by all of 3,5, and this statement is going to require some investigation on my part. Originally went the route binomial route because I instantly recognized 120=5!. I think my way forward is to complete more problems and...

Done and done. Write it all out until somebody tells me that I am doing too much!

OK, I can see that being required to add some rigor. How do I know when I should add more details to formalize the proof or when I can proceed with a statement similar to "...it follows..."?

Proposition: The product of any five consecutive integers is divisible by 120. PF: Suppose for the sake of contradiction that there exist five consecutive integers whose product is not divisible by 120. Consider the sequence of integers starting from 0, every fifth integer is a multiple of 5...

Updated Proof. Proposition: The product of any five consecutive integers is divisible by 120. PF: Suppose for the sake of contradiction there exist five consecutive integers such that their product is not divisible by 120. Consider the five consecutive integers n, (n-1), (n-2), (n-3), and...

Starting from 0 every fifth integer is a multiple of 5, every forth integer is a multiple of 4, every third is a multiple of 3 and every other is a multiple of 2. From this it follows that every set of five consecutive integers must contain a multiple of 5, a multiple of 4, at least one...

Ok, so I didn't negate the statement properly, I only negate part of the statement, qualifier needs to be negated as well.

This is more a general question that this problem spurred and this is what I came up with. I do not feel it is acceptable but would like clarification moving forward. My text states the format for proof by contradiction is as follow; Proposition: P PF: Suppose ~P. ...a little math and...

Thank you for the input, I believe I have a better understanding now. Kinda feels like I am working the n choose k backwards by stating I have three 0s then I need to figure how many ways they can be uniquely arranged within six elements.

A = set of even 7-digit even numbers ## = 9 * 10 * 10 * 10 * 10 * 10 * 5 = 4500000 ## B = set of 7 -digit numbers with three 0s ## = 9 * { 6 \choose 3} * 9 * 9 * 9 = 131220 ## ## |A \cap B| ## Portion that does not end in ## 0 = 9 * { 5 \choose 3} * 9 * 9 * 4 = 9^3{ 5 \choose 3}4 = 29160...

Ok, so with those three facts, not only are you proving the inequality holds but also using the constraints of the initial problem. I would have never figured out what you meant, unless you showed me that or asked me to prove that the inequality holds. Thanks!

Yes, I found ##n_1 = 3## by brute force, thankfully it was small. I am not sure of a clever way to find it if the value was larger or the inequality more complex.