Young's Double Slit experiment question 
Q-A beam of light consisting of two wavelenghts 600 nm and 450 nm is used to obtain interference in Young's Double Slit experiment (YDSE). Find the least distance from the central maximum where the bright fringes due to both wavelengths coincide. The distance between the two slits is 0.4 mm and screen is placed at a distance of 1.0 m from the slits.

Now my problem is not with the question but the solution given in my book.
It says 

Let at distance $x$ from central bright maxima, the bright fringes due to both the wavelengths coincide for the first time. It is possible only if within the distance $x$ there are $n$ fringes for light of 600nm and $(n+1)$ fringes for light of wavelength 450 nm.

Okay so my problem is that I am unable to figure out how have they made this argument about $n$ and $n+1$ fringes? It's isn't mention anywhere in the question. Can't it be ($n$+any number)?
If there's any better solution than this I would be really thankful for that
 A: In this case whoever set the question has chosen wavelengths with a nice ratio and not too far apart, so the first coincidence is at $n_{600}$ = 3 and $n_{450} = 4$. However you're quite correct that this isn't a general rule. For example if the two wavelengths, $\lambda_1$ and $\lambda_2$ were both primes they would coincide when $n_1 = \lambda_2$ and $n_2$ = $\lambda_1$.
A: Yes, you are right it can be any number but in this case it's 1. 
Let's see :
Consider longer wavelength making n fringes then shorter wavelength will make n+x fringes( where X is any integer equal to or greater than 1). Now we see that in our case how it comes out to be 1.

600(n)=450(n+x)
600n-450n=450x
15n=45x
n=3x

Since n has to be the least value and an integer(>0) and is proprtional  to x ,we can guess the value of x,  It must be such that it should give least value for n and must be an integer therefore in this case it comes out to be 1. 
One may say why x must be integer . This is so because fringes are integers.(or else other wavelength fringe will be n+x and become a non integer which wrong)
If wavelengths were prime then also we can aplly same logic to get the value of x. In those cases fringes will be equal to others wavelength, as pointed out by John Rennie.
Consider the example where wavelengths are 11nm and 17nm,

17(n)=11(n+x)
6n=11x
n=11x/6
To make n least integer such that x is also an integer we get
  x=6
So n_11=(n+x) = 11+6= 17 and n_17=11

Now you can see why we don't take x any fraction.
A: Yes, the ratio is for example $4/5$ and they are asking for minimum distance 
$n_1=4k$; $n_2=5k$ (for some $k$), the distance will be minimum when $n$ is minimum, so $k=1$ and $n_1=4$ and $n_2=5$.
