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I know that the number of integers between 1 and 1000 that are divisible by 30 is 33, and the number of integers between 1 and 1000 that are divisible by 16 is 62, but I do not know how to calculate the number of integers that is divisible by both 30 and 16, especially because they are not primes. Could anyone help me please?

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It does not matter if they are primes.

So here we are trying to calculate the number of multiples of $30$ which are not multiples of $16$ below thousand. For this, we take all multiples of $30$ below thousand first, and then subtract from this, all numbers that are multiples of $30$ and $16$ which are below thousand.

Any number is a multiple of $30$ and $16$ if and only if it is a multiple of their least common multiple, which in our case is $240$, which I computed by prime factorization, if you wanted to know how that is done.

So the answer is the number of multiples of $30$ minus the number of multiples of $240$ which are less than thousand. This is then $33-4 = 29$.

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$$N=2×3×5k=2^4K$$

$$\implies K=3×5K'$$ by Gauss Theorem.

thus $N $ is a multiple of $$3×5×16=240$$

hence $$N=240,480,720,960$$

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  • $\begingroup$ your answer is not clear for me at all, could u provide some more details please? $\endgroup$ – Intuition Aug 1 '17 at 10:04
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Eliminate E. Notice that $90=3\times30$. Eliminate D. We know that if we multiple $90$ by $2$ we have $180$ which is divisible by $30$ but not $16$. Eliminate C. Multiply $180$ by $2$ and we have $360$ is divisible by $30$ but not by $16$, so eliminate B. The answer is A.

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  • $\begingroup$ your answer is not clear for me at all. $\endgroup$ – Intuition Aug 1 '17 at 9:47
  • $\begingroup$ We can eliminate E right away because in your original post you said only 33 integers between 1 and 1000 are divisible by 30. Next we can eliminate D because we know 90 is divisible by 30 but not 16. After that we can also eliminate C because we know 180 is divisible by 30 but not 16. Then we can eliminate B because we know that 360 is divisible by 30 but not by 16. $\endgroup$ – JohnColtraneisJC Aug 1 '17 at 16:16

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