Question

The total number of $4$-digit numbers whose greatest common divisor with $18$ is $3$, is

Answer 1

Find the total number of $4$-digit numbers:

Let $N$ be the four digit number.

Gcd $\left(N,18\right)=3$

Thus $N$ is an odd integer which is divisible by $3$ but not by $9$.

Since $4$-digit odd multiplies of $3$ are,

$1005,1011,...........9999$

We can see that it is an A.P. with $a=1005,d=3,a_n=9999$

Since, $n^{th}$ term of A.P. is given by

$$\begin{gathered} a_n=a+(n-1)d \\ \Rightarrow 9999=1005+(n-1)6 \\ \Rightarrow n=1500 \end{gathered}$$

Now, $4$ digit odd multiplies of $9$ are,

$1017,1035,............,9999$

It is an A.P. with $a=1017,d=9,a_n=9999$

Since, $n^{th}$ term of A.P. is given by

$$\begin{gathered} a_n=a+(n-1)d \\ \Rightarrow 9999=1017+(n-1)18 \\ \Rightarrow n=500 \end{gathered}$$

Therefore, total such numbers $N=1500-500$

$=1000$

Hence, total number of $4$-digit numbers whose greatest common divisor with $18$ is $3$, is $1000$.