Formula for number of multiples of $n$ in a given interval [a,b]
$begingroup$
Given an interval $[a,b]$ and $n$,
find the number of multiples of $n$ in the interval.
($a,b,n$ are natural numbers)
Examples :
1) There are $2$ multiples of $3$ in the interval $[0,3]$
2) There are $1$ multiples of $3$ in the interval $[1,3]$
3) There are $0$ multiples of $5$ in the interval $[1,4]$
By trial and error I manged to cookup below formula
$$leftlfloor dfrac{b-a+ a pmod{n}+ ndelta(amod n)}{n}rightrfloor$$
where $delta(t) = 1$ when $t=0$, and $delta(t)=0$ when $tne 0$.
It seems to be working but it looks a bit nasty to me, so I'm wondering if it can be simplified or if you know any other formula that's a bit simpler...
elementary-number-theory
$endgroup$
add a comment |
$begingroup$
Given an interval $[a,b]$ and $n$,
find the number of multiples of $n$ in the interval.
($a,b,n$ are natural numbers)
Examples :
1) There are $2$ multiples of $3$ in the interval $[0,3]$
2) There are $1$ multiples of $3$ in the interval $[1,3]$
3) There are $0$ multiples of $5$ in the interval $[1,4]$
By trial and error I manged to cookup below formula
$$leftlfloor dfrac{b-a+ a pmod{n}+ ndelta(amod n)}{n}rightrfloor$$
where $delta(t) = 1$ when $t=0$, and $delta(t)=0$ when $tne 0$.
It seems to be working but it looks a bit nasty to me, so I'm wondering if it can be simplified or if you know any other formula that's a bit simpler...
elementary-number-theory
$endgroup$
add a comment |
$begingroup$
Given an interval $[a,b]$ and $n$,
find the number of multiples of $n$ in the interval.
($a,b,n$ are natural numbers)
Examples :
1) There are $2$ multiples of $3$ in the interval $[0,3]$
2) There are $1$ multiples of $3$ in the interval $[1,3]$
3) There are $0$ multiples of $5$ in the interval $[1,4]$
By trial and error I manged to cookup below formula
$$leftlfloor dfrac{b-a+ a pmod{n}+ ndelta(amod n)}{n}rightrfloor$$
where $delta(t) = 1$ when $t=0$, and $delta(t)=0$ when $tne 0$.
It seems to be working but it looks a bit nasty to me, so I'm wondering if it can be simplified or if you know any other formula that's a bit simpler...
elementary-number-theory
$endgroup$
Given an interval $[a,b]$ and $n$,
find the number of multiples of $n$ in the interval.
($a,b,n$ are natural numbers)
Examples :
1) There are $2$ multiples of $3$ in the interval $[0,3]$
2) There are $1$ multiples of $3$ in the interval $[1,3]$
3) There are $0$ multiples of $5$ in the interval $[1,4]$
By trial and error I manged to cookup below formula
$$leftlfloor dfrac{b-a+ a pmod{n}+ ndelta(amod n)}{n}rightrfloor$$
where $delta(t) = 1$ when $t=0$, and $delta(t)=0$ when $tne 0$.
It seems to be working but it looks a bit nasty to me, so I'm wondering if it can be simplified or if you know any other formula that's a bit simpler...
elementary-number-theory
elementary-number-theory
asked Dec 8 '18 at 22:54
rsadhvikarsadhvika
1,7081228
1,7081228
add a comment |
add a comment |
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