Math problem for a friend...

[Kyteland]

Originally posted by: Spencer278

Originally posted by: blahblah99
119 mod 2 = 1
119 mod 3 = 2
119 mod 4 = 3
119 mod 5 = 4
119 mod 6 = 5
119 mod 7 = 0

Maybe I'm reading the question wrong but when did subtract become mod?

Think of it like a game of telephone. Prof --> girl --> FallenHero --> ATOT. It got a little messed up along the way and we "corrected" it. 😉

 

[chuckywang]

Originally posted by: Kyteland

Originally posted by: aves2k

Originally posted by: chuckywang
Here's the proof: 119 satisfies all the conditions. QED.

LOL!

But it doesn't explain how to do the problem, so they learned nothing. The point is to teach them so that they can do it themselves next time.

Then you shouldn't post your proof as an intersection between two sets pertaining to two arbitrary rules (even though, in this case, it works). What if the question asks for all such numbers that satisfy the conditions, not just numbers between 100 and 150.

To get the answer, let X be the number(s) in question.

X = -1 (mod 2)
X = -1 (mod 3)
X = -1 (mod 4)
X = -1 (mod 5)
X = -1 (mod 6)

This implies X = -1 mod (LCD(2,3,4,5,6)) <-- Need to use something called the Chinese Remainder Theorem. It's pretty famous, so google it.

Therefore X = -1 mod 60. X = 60*C-1

Since X = 0 (mod 7), we get that X = 60(7a-5)-1 for a=1,2,3,4,.... (again by the Chinese Remainder Theorem)

X = 420a - 301 for a=1,2,3,4...

 

[Kyteland]

My "proof" was a perfectly arbitrary answer. The rule 100<X<150 simply defines the set {101,102...148,149}. The set I gave for rule 6) was the result you would get intersecting the countably infinite set {7*n: n= ...-2,-1,0,1,2...} with {101,102...148,149}. The result I gave for 5) ({119}) was that result intersected with the set {6*n-5: n = ...-2,-1,0,1,2...}. Nothing in my work prevents you from finding an answer using only rules 1) - 6).

 

[chuckywang]

Originally posted by: Kyteland
My "proof" was a perfectly arbitrary answer. The rule 100<X<150 simply defines the set {101,102...148,149}. The set I gave for rule 6) was the result you would get intersecting the countably infinite set {7*n: n= ...-2,-1,0,1,2...} with {101,102...148,149}. The result I gave for 5) ({119}) was that result intersected with the set {6*n-5: n = ...-2,-1,0,1,2...}. Nothing in my work prevents you from finding an answer using only rules 1) - 6).

I agree. I guess what I was trying to say was that your method is not the general method for solving Chinese Remainder Problems.

 

[silverpig]

Yeah, that was horribly worded.