# Is a Number Divisible by 40

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?

• Yes. Since $\gcd(8, 5)=1$, then if $8|k$ and $5|k$, it follows that $40|k$. Dec 19, 2014 at 16:48
• Actually the gcd is only needed if you want to state a sufficient condition. If $a$ and $b$ divide $n$, then $ab$ divides $n$. Dec 19, 2014 at 16:52
• @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. Dec 19, 2014 at 16:56
• But if 6 and 2 were coprime (I realize they are not), then Henrick's statement would be correct? Dec 19, 2014 at 17:09
• "..last three digits..", not "numbers". Dec 19, 2014 at 17:16

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 $n$.
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 $$60$$.