# How many positive integers less than 1000 are multiples of 5 and are equal to 3 times an even number?

Question: How many positive integers less than 1000 are multiples of 5 and are equal to 3 times an even number?

So Multiples of $5$ and $6$

If a number is a multiple of $5$ and $6$ then it is a multiple of $30$ as well. Because of the law,

Multiple of $a$ and $b$ $\implies$ multiples of lcm$(a, b)$

lcm($5, 6$) = $30$

$$1000 \equiv 10 \pmod{30}$$

I am trying to solve this problem using number-theory, any help? The answer is $33$

Also, how does $1000/30$ give the right answer?

Thanks!

As you have stated yourself, the integers we are looking for must be a multiple of $30$. Now it is simply a matter of counting all the multiples of $30$ below $1000$, which is found by dividing $1000$ by $30$ and take the integer part of the answer. That would be $33$.
To be more explicit, the numbers we are looking for are of the form $30k$ with $k \in \mathbb{N}$ such that $30 k <1000$. The numbers $k$ therefore have an upper limit that is governed by $k < 1000/30$. Since $k$ is an integer number, we might just as well rewrite this as $$k \leq \mbox{floor}(1000/30)=33,$$ where floor means to take the integer part of a number. In this case floor$(1000/30) =$floor$(33.333) = 33$. Since $k$ should at least be $1$, we have $1 \leq k \leq 33$, which means there are $33$ numbers satisfying our requirements.
• How are multiples of $30$ given by dividing $1000$ by $30$ – anonymous Jan 10 '15 at 10:27
• Thanks! But doesnt this just say $k$ is one number? – anonymous Jan 10 '15 at 10:41
• Well, no. We are looking for all numbers of the form $30k$, so we want all to count all the allowed values for $k$. The total number of allowed values for $k$ is 33. – Hrodelbert Jan 10 '15 at 10:44
• This says if you divide $1000$ into $30$ equal pieces you get $33$ pieces. But how does say there are $33$ numbers. – anonymous Jan 10 '15 at 10:48