Play with N.

Let $N$ = abcd, where a,b,c,d are single digit whole numbers, and a>0.

So, we may write, N= $10^{3} a+\ 10^{2}b+10c+d$ .

Therefore, $\frac{N}{29}= 10^{2}b+10c+d$ .

Now, from the two equations, we get, $10^{3}a=28(10^{2}b+10c+d)$ So, $28=7*2^{2}$ is a factor of $10^{3}a$ , which means that $10^{3}a$ must contain a power of 7 in its prime factorisation.

So, $a$ has to be one among $7,14,21,...$ . But, $a$ is a one-digit number, so only possible value of $a\text{ is 7}$ . After, putting $a=7$ in the third equation, we get

$100b+10c+d=250.$

Each of b,c and d is a one-digit number, so, their values are:

$b=2,$ $c=5,$ $d=0.$

Hence, $N$ is $7250$ .