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? Yes, if a number \$n\$ is divisible through \$15\$, this means \$n=15k\$ for some integer \$k\$. For this reason \$n=5(3k)=3(5k)\$, so the is likewise divisible by \$3\$ and \$5\$.

Conversely, if \$n\$ is divisible through \$3\$ and also \$5\$, that is a straightforward lemma that it is divisible through the least common multiple that \$3\$ and also \$5\$. Due to the fact that \$3\$ and \$5\$ space coprime, their lcm is simply \$15\$.

You are watching: Numbers divisible by 3 and 5 \$egingroup\$ ns feel that this price is incomplete. That does no prove that x|n and y|n indicates that lcm(x,y)|n; it just insurance claims it's "a simple lemma". A much better answer would display why this is true. \$endgroup\$
\$egingroup\$ expect there's a number n such that x|n and y|n, but lcm(x,y) does no divide n. Then compose m = lcm(x,y) and n = pm+q, whereby 0 \$endgroup\$
Yes. \$15|n implies 3*5|n implies 3|n ext and also 5|n\$. Vice versa, \$3|n ext and also 5|n implies 15|n\$. This is due to the fact that if \$x|n\$ and also \$y|n\$, then \$ extlcm(x,y)|n\$ whereby \$ extlcm(xy)\$ is the smallest number the is divisible through both \$x\$ and also \$y\$. In this case, \$x=3\$ and \$y=5\$, therefore \$ extlcm(3,5) = 15\$, as such \$15|n\$.

(Note: \$a|b\$, review as "\$a\$ divides \$b\$", means that \$b\$ is divisible by \$a\$.) Thanks for contributing an answer to benidormclubdeportivo.org Stack Exchange!

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