Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is stating that a number $x$ is divisible by 15 the same as stating that $x$ is divisible by 5 and $x$ is divisible by 3?

share|improve this question
3  
The two assertions are equivalent. –  André Nicolas Feb 1 '12 at 23:38
    
may be worth noting for somebody who might be new to this, that it happens to work in both directions here, because 3 and 5 are co-prime. i.e. if you ask the same question for a different example you may not get the converse implication. –  Beltrame Feb 1 '12 at 23:49
1  
If you want to figure out the general facts at work here, you could read this Keith Conrad handout. –  Dylan Moreland Feb 1 '12 at 23:55

2 Answers 2

up vote 20 down vote accepted

Yes, if a number $n$ is divisible by $15$, this means $n=15k$ for some integer $k$. So $n=5(3k)=3(5k)$, so it is also divisible by $3$ and $5$.

Conversely, if $n$ is divisible by $3$ and $5$, it is a simple lemma that it is divisible by the least common multiple of $3$ and $5$. Since $3$ and $5$ are coprime, their lcm is just $15$.

share|improve this answer
    
I feel that this answer is incomplete. It does not prove that x|n and y|n implies that lcm(x,y)|n; it just claims it's "a simple lemma". A better answer would show why this is true. –  user22805 Feb 2 '12 at 8:32
    
Suppose there's a number n such that x|n and y|n, but lcm(x,y) does not divide n. Then write m = lcm(x,y) and n = pm+q, where 0 < q < m and p is an integer. Then x and y must both divide q, so m is not the lcm of x and y - contradiction. –  user22805 Feb 2 '12 at 8:35

Yes. $15|n \implies 3*5|n \implies 3|n \text{ and } 5|n$. Conversely, $3|n \text{ and } 5|n \implies 15|n$. This is because if $x|n$ and $y|n$, then $\text{lcm}(x,y)|n$ where $\text{lcm}(xy)$ is the smallest number that is divisible by both $x$ and $y$. In this case, $x=3$ and $y=5$, so $\text{lcm}(3,5) = 15$, therefore $15|n$.

(Note: $a|b$, read as "$a$ divides $b$", means that $b$ is divisible by $a$.)

share|improve this answer
1  
You should also prove the other direction as Dedede has done below. –  user38268 Feb 1 '12 at 23:51
    
@BenjaminLim: Good point. Editing that in now. –  El'endia Starman Feb 1 '12 at 23:53
2  
@El'endiaStarman: You seem to have done nothing except write the OP's question as an assertion using mathematical language. A proof or justification as to why this works would be helpful. –  JavaMan Feb 1 '12 at 23:59
    
I don't know enough math to pick the "most correct" answer. @El'endiaStarman answered first, I understood it, and it looked good to me! Dedede's has more upvotes though, so I went with that one. –  Adam Monsen Feb 2 '12 at 0:02
    
@JavaMan: At this point, is there a reason I shouldn't just delete my answer? Dedede has already answered it perfectly. –  El'endia Starman Feb 2 '12 at 0:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.