Count the numbers.

I'll assume that all of you know the floor function and its applications in number theory problems.

First of all, let us find how many numbers between $0$ to $400$ are divisible by $17$ or $23$ , the two given prime numbers.

No. of numbers divisible by $17$ in the given range $=\left \lfloor \dfrac{400}{17} \right \rfloor = 23$

No. of numbers divisible by $23$ in the given range $=\left \lfloor \dfrac{400}{23} \right \rfloor = 17$

No. of numbers divisible by both $17$ and $23$ in the given range $=\left \lfloor \dfrac{400}{17\times 23} \right \rfloor = \left \lfloor \dfrac{400}{391} \right \rfloor = 1$

So, observe that the question says less than 400. So, we have natural numbers 1 to 399 in our considered range. Now, the numbers which satisfy the given criteria should be the numbers which are divisible neither by $17$ nor by $23$ . So, the no. of numbers fitting the criteria $=399-(23+17-1)=\boxed{360}=n$

Thus, $\frac{n}{10}=\frac{360}{10}=\boxed{36}$

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$\large \textrm{The same solution (Explained using Set theory):}$

Take $A$ as the set of numbers divisible by $17$ and $B$ as the set of numbers divisible by $23$ . Also, the sample space for the problem is $S=\{1,2,3,\ldots,399\}$ , since it is said in the problem "NATURAL NUMBERS LESS THAN 400" . So, we have,

$n(A\cup B)=n(A)+n(B)-n(A\cap B) = 23+17-1=39 \\ \implies n(\overline{A\cup B})=n(S)-n(A\cup B)=399-39=\boxed{360}=n \\ \implies \frac{n}{10}=\frac{360}{10}=\boxed{36}$

The correct answer is 36. There are 22 numbers less than 391 that are divisible by 17 and 16 numbers less than 391 that are divisible by 23. Counting 391, we have a total of 22 + 16 + 1 = 39 numbers that are divisible by either 17 or 23. Thus, there are 399-39 = 260 numbers that aren't and the answer is 36

400/17 = 23.529... , which means there 23 numbers less than 400 that are divisible by 17 Similarly, we find that there are 17 numbers less than 400 that are divisible by 23. Hence, the number of natural numbers less than 400 that are NOT divisible by 400 i,e. n = 400 - (23+17) = 360. Therefore, n/10 = 36

Largest multiple of $17$ less than $400$ is $391$ , which is $17\times 23.$ Largest multiple of $23$ less than $400$ is also $391$ , which is $23 \times 17.$

There are $40$ natural numbers divisible by either, but the number(s) divisible by both $17$ and $23$ were included. The only multiple of $17$ and $23$ less than $400$ is again, $391$ , meaning that we have to subtract that: $40-1=39$ , obviously.

The largest natural number less than $400$ is $399$ , so $399-36=360=n$ and therefore, $\frac{360}{10}=36$

Well first I noticed that 17 x 23 is close to 20 x 20 which = 400.

The obvious move was to take each factor up to the point of 400 and subtract that many from the total 399 under 400. It came out to 391, so I knew there were 17 x 1,2,3... up to 23 and 23 x 1,2,3.. to 17. But I can't count 17x23 twice so 17+23-1=39.

399 natural numbers below 400 minus these 39 came out to 360/10 = 36.

The number of natural nos <400 which are divisible by 17 can be found out by taking the quotient of (400/17) i.e. 23 , similarly for the 23 we will get 17 number of factors . Where we can see 17*23=391 which is common to both numbers so we will take it once out of (17+23) factors . So after discarding (17+23-1) factors or 39 factors out of 400 we get , 361 factors so the ans is (361/10) or 36.1