Good numbers

Relevant wiki: Arithmetic and Geometric Progressions Problem Solving

Just use $AP$

$\text{Let, }a_1, a_2, a_3, a_4, \ldots, a_n \text{ be the terms of an AP such that, } 4a_n + 1 \text{ is a multiple of 5} \\ a_1 = 501 ~~ \boxed{\text{Since } 4\times 501 + 1 = 2005, 5|2005} \\ \text{Since }4a_n + 1 \text{ is a multiple of 5, }a_n \text{ should differ by 5(=d). If you want we can check below} \\ a_2 = 506 \\ d = a_2 - a_1 = 506 - 501 = 5 \\ \text{Therefore, last term between 500 - 1000 that is a good number is} \\ a_n = 996 \\ a + (n-1)d = 996 \\ 501 + (n-1)5 = 996 \\ (n-1)5 = 495\\ n-1= 99 \\ \boxed{n = 100}$

Therefore there are 100 terms.

The integers which give a multiple of 5 need to give a 4 or 9 as last number after multiplying by 4. A 9 will never happen (even times odd is even and even times even too) The only numbers wicht give a 4 are 1 and 6. So 501, 506,511, 516 and so on satisfy 4n+1 as a multiple of 5. So per 100 there.are 10 x 2 = 20 options. Since there are it is from 500 to 1000 there are 5x20=100 such integers.

Just modify it

$4n+1\\ =5n-n+1 \\ =5n-(n-1)$

So, $5|n-1$ then $n=5m+1$