Animated Solution for Mathematics - Sequence and Series: The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is

We are tasked with finding the number of integers $n$ in the range $1000\le n\le 9999$ such that $\gcd (n,18)=3$.

First, we perform the prime factorization of $18$:

For $\gcd (n,18)=3$ to hold, $n$ must be a multiple of $3$, but it must satisfy two specific constraints to prevent the GCD from being larger than $3$: 1. $n$ must not be divisible by $2$ (it must be odd). 2. $n$ must not be divisible by $9$ (it must contain exactly one factor of $3$).

We define the set of odd multiples of $3$ in the range $[1000,9999]$. The first such number is $1005$, and the last is $9999$.

Since we require odd multiples, we skip even multiples of $3$ (which are multiples of $6$). Thus, the common difference $d$ for this arithmetic progression is $6$.

Using the formula for the $k$-th term, $l=a+(k-1)d$:

We must now subtract the multiples of $9$ that are also odd, as these violate the condition $\gcd (n,18)=3$. These are numbers of the form $9\times (\text{odd number})$.

The first such number is $9\times 113=1017$, and the last is $9\times 1111=9999$. The common difference for this sequence is $18$.

Using the formula $l=a+(m-1)d$:

The number of valid integers is the total count of odd multiples of $3$ minus the count of odd multiples of $9$:

The number of such integers $n$ is 1000.