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This is Velleman's exercise 3.4.26 (b):

Prove that it is NOT true that for every integer $n$, 60 divides $n$ iff 6 divides $n$ and 10 divides $n$.

I do understand that a number will be divisible by 6 and 10 if it is divisible by 60 and that it will not necessarily be divisible by 60 if it is divisible by 6 and by 10. 30 is an example of it.

I still have an issue actually discovering and writing up the proof. To illustrate the issue and to put the question in context, I would like to refer to the first part of the question already asked by another user, Velleman's exercise 3.4.26 (a). Thanks in advance.

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    $\begingroup$ There's nothing more to understand. It's all in the counterexample. It is not complicated. $\endgroup$ Aug 20 '15 at 0:49
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Consider the number $30$. It is divisible by $6$ and $10$ and not by $60$. We have found a counterexample and therefore the statement is not true.


The statement $ab$ divides $n$ if and only if $a$ divides $n$ and $b$ divides $n$ is only true if $a$ and $b$ are relatively prime (share no prime factors).

Otherwise we can take the least common multiple of $a$ and $b$, this number will be smaller than $ab$ (since they are not relatively prime) and it will be divisible by $a$ and $b$. The least common multiple will always be a counterexample if $a$ and $b$ are not relatively prime.

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    $\begingroup$ If 30 is a contradiction, your proof is over. Simple as that. But is suppose what you want is a more mathematical justification, ask that in your question. $\endgroup$ Aug 20 '15 at 0:06
  • $\begingroup$ well, I found that 30 contradicts it ( and wrote it down). I need a well written proof that it is not true ... $\endgroup$
    – Eugene
    Aug 20 '15 at 0:11
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    $\begingroup$ "The statement is false since $30$ is a counterexample" This is a well written proof. $\endgroup$
    – Asinomas
    Aug 20 '15 at 0:13
  • $\begingroup$ Thank you, but my problem is of a more general nature. Imagine it would be difficult to discover the counterexample. How would I proceed? $\endgroup$
    – Eugene
    Aug 20 '15 at 0:26
  • $\begingroup$ Here is a link to the first part of the question by another user: math.stackexchange.com/questions/342128/… . That link is Velleman's exercise 3.4.26 (a) $\endgroup$
    – Eugene
    Aug 20 '15 at 0:28

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