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One of the "shortcuts" for determining if a number is divisible by 8 is to see if the last three digits are divisible by 8. One of the "shortcuts" for determining if a number is divisible by 5 is to see whether the last digit is a 5 or a 0. If I have a number of arbitrary length, is it acceptable to say that the number is divisible by 40 if the number passes both shortcuts?

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    $\begingroup$ Yes. Since $\gcd(8, 5)=1$, then if $8|k$ and $5|k$, it follows that $40|k$. $\endgroup$ – Alex Wertheim Dec 19 '14 at 16:48
  • $\begingroup$ Actually the gcd is only needed if you want to state a sufficient condition. If $a$ and $b$ divide $n$, then $ab$ divides $n$. $\endgroup$ – Henrik - stop hurting Monica Dec 19 '14 at 16:52
  • $\begingroup$ @Henrik, this is not always true, unless I am understanding you wrong. Both 6 and 2 divide 18, however 2*6=12 does not divide 18. $\endgroup$ – Dasherman Dec 19 '14 at 16:56
  • $\begingroup$ But if 6 and 2 were coprime (I realize they are not), then Henrick's statement would be correct? $\endgroup$ – Kenneth K. Dec 19 '14 at 17:09
  • $\begingroup$ "..last three digits..", not "numbers". $\endgroup$ – user26486 Dec 19 '14 at 17:16
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Yes, since $8$ and $5$ are coprime $(\gcd(8, 5)=1)$. This means that such an integer can be written as $8\times 5\times n=40\times n$, for some integer $n$.

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To offer some clarity towards Henrik's point:

It is true that if a|n and b|n then if gcd(a,b)=1 then (ab)|n.

However, we may generalize somewhat, and note that lcm(a,b) = (ab)/gcd(a,b) will always divide n, provided that both a|n and b|n

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