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?

  • 2
    $\begingroup$ Yes. Since $\gcd(8, 5)=1$, then if $8|k$ and $5|k$, it follows that $40|k$. $\endgroup$ Dec 19, 2014 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 $n$. $\endgroup$ Dec 19, 2014 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, 2014 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, 2014 at 17:09
  • $\begingroup$ "..last three digits..", not "numbers". $\endgroup$
    – user26486
    Dec 19, 2014 at 17:16

3 Answers 3


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 $n$.


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


Yes. Any number whose last $3$ digits are a multiple of $40$ are a multiple of $40$. It also works if the $100$s digit is even and the last $2$ digits are $00$, $40$ or $80$, or if the $100$s digit is odd and the last $2$ digits are $20$ or $60$.


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