Count multiples of 7 between range using positioning I am interested in finding the number of multiples of $7$ between $[10, 500]$
In other cases I have seen that a neat trick is to check if there is a pattern between the sequence of numbers and their corresponding position (e.g. for multiples of $3$ the position is $1/3$ of the number and hence it is easy to figure out the last position hence total numbers).
For this case I tried to follow this same process:




14
21
28
35
42
49
........
490
497




1
2
3
4
5
6
........

?




but I could not find any relation between the numbers in the upper row and the numbers in the lower row so as to deduce the last number in the lower row and figure out the number of multiples.
Is it possible to apply this in this case? Am I missing a pattern?

HINT: Separate the interval into subintervals of length $7$. How many multiples of $7$ are there in each of them?

On the other side, the pattern in the sequence is $7(n+1)$ for $n \le 70$. We use $n+1$ instead of $n$, as we need to shift since $7 < 10 < 2 \times 7$. Also we need $n \le 71$, since $71 \times 7 < 500 < 72 \times 7$.
