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How many natural numbers less than 1000 are divisible by 5 or 7 but NOT by 35? (a) 285 (b) 313 (c)341 (d) 243

i know that numbers upto 1000 that are divisible by 5 are 1000/5=200 and by 7 are 142. so total numbers that are divisible by 5 or 7 =342. then i am blank about how to progress. i guess we have to find area of divisibles of 5 and 7 excluding the area common between them in venn diagram representation.

Manish Kumar Balayan
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    Try using $#(A\cup B) = # A+#B -#(A\cap B)$. – Jacky Chong Jan 29 '20 at 08:06
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    You're well on your way! You then have to figure out how many are divisible by 35 and then, indeed, a Venn diagram should help. – Matti P. Jan 29 '20 at 08:07
  • The answer is $$\bigg\lfloor\frac{1000}5\bigg\rfloor+\bigg\lfloor\frac{1000}7\bigg\rfloor-2\bigg\lfloor\frac{1000}{35}\bigg\rfloor$$I've leave it to you to figure out why. – Rushabh Mehta Jan 29 '20 at 08:07
  • i get it 28 numbers are divisible by 35 upto 1000. then we get union set of a and b =342-28=314. then what to do. oh i get it now. we get union set then we have to subtract the set of 35 from it. – Manish Kumar Balayan Jan 29 '20 at 08:15

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Similarly, the number of numbers divisible by $35$ from $1$ to $999$ is $\left[\frac{999}{35}\right]=28$, so the answer is $199+142-2\times28=285$. Also, you are to use $999$ and not $1000$ (owing to the "less than $1000$" in your question) for getting your sort of Venn diagram areas. So, to put it exactly, it is

$$\left[\frac{999}{5}\right]+\left[\frac{999}{7}\right]-2\times\left[\frac{999}{35}\right]=285$$ Explanation to the Venn-diagram approach:

(i) There are $28$ numbers that are divisible by both $5$ and $35$(numbers divisible by $35$ are also divisible by $5$)

(ii) There are $28$ numbers that are divisible by both $7$ and $35$(numbers divisible by $35$ are also divisible by $7$)

These $28$ numbers each of which are being counted twice (once the first term and once in the second term) are to be removed from the enumeration.

Sameer Baheti
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