1. ## need help!

ok i dont know how to solve this problem but i got this far and idk what to do next ok here is the question:
For which integers are 4x-6 divisible by 14?

how far i got
4x-6== 0 (mod 14)
4x==6 (mod 14)
and now im clueless what to do next

2. ## Re: need help!

Hello, Petrus!

Here's a rather primitive solution.

$\displaystyle \text{For which integers is }4x-6\text{ divisible by 14?}$

$\displaystyle \text{We want: }\,4x-6 \:=\:14a\,\text{ for some integer }a.$

$\displaystyle \text{Solve for }x\!:\;\;x \:=\:\frac{7a + 3}{2} \quad\Rightarrow\quad x \:=\:3a + 1 + \frac{a+1}{2}\;\;[1]$

$\displaystyle \text{Since }x\text{ is an integer, }(a+1)\text{ is even . . . }\:a\text{ is odd.}$
. . $\displaystyle \text{That is, }\,a \:=\:2k-1,\,\text{ for some integer }k.$

$\displaystyle \text{Substitute into [1]; }\:x \;=\;3(2k-1) + 1 + \frac{(2k-1)+1}{2}$

$\displaystyle \text{Therefore: }\:x \:=\:7k-2\,\text{ for any integer }k.$

3. ## Re: need help!

is it any other way to solve it with using (mod 14) thats how we should solve it

4. ## Re: need help!

Well, there is always "brute strength"- 4(1)= 4, 4(2)= 8, 4(3)= 12, 4(4)= 16= 2 (mod 14), 4(5)= 20= 6 (mod 14)- Aha! But we don't want to stop there:
4(6)= 24= 10 (mod 14), 4(7)= 28= 0 (mod 14), 4(8)= 32= 4 (mod 14), 4(9)= 36= 8 (mod 14), 4(10)= 40= 12 (mod 14), 4(11)= 44= 2 (mod 14), 4(12)= 48= 6 (mod 14)!!, 4(13)= 52= 10 (mod 14). So we have two solutions, mod 14: 5 and 12. Of course, because those are modulo 14, the equation is satisfied by those plus any multiple of 14: 5+ 14i, and 12+ 14j.

The first, 5+ 14i is is equivalent to Soroban's 7k- 2 for k odd and 12+ 14j to 7k- 2 for k even.

A bit more formal method is to say the 4x= 6 (mod 14) means that there exist an integer n such that 4x= 6+ 14n which is the same as 2x= 3+ 7m or 2x- 7n= 3. Now, because 2 and 7 are "relatively prime" we can use Euclid's algorithm: 2 divides into 7 three times with remainder 1 so (1)7- (3)2= 1. Multiplying both sides by 3, (3)7- (9)2= 2(-9)- 7(-3)= 3. That is, x= -9 and n= -3 is a solution to that equation. Of course, x= -9+ 7i, n= -3+ 2i also satisfies that equation: 2(-9+ 7i)- 7(-3+ 2i)= -18+ 14i+ 21- 14i= 3 for all i. Replacing "i" with "k+ 1" gives x= -9+ 7(k+ 1)= -9+ 7k+7= -2+ 7k, Soroban's answer.