Is a Number Divisible by 40

Question

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 $40|k$. — Alex Wertheim, Dec 19 '14 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 $n$. — Henrik supports the community, Dec 19 '14 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. — Dasherman, Dec 19 '14 at 16:56
But if 6 and 2 were coprime (I realize they are not), then Henrick's statement would be correct? — Kenneth K., Dec 19 '14 at 17:09
The logical implication $a\mid n, b\mid n\implies ab\mid n$ holds for all $n\in\mathbb N$ if and only if $(a,b)=1$. If $(a,b)\neq 1$, then this implication does not necessarily hold and $\exists n\in\mathbb N(a\mid n \wedge b\mid n\wedge ab\not\mid n)$. — user26486, Dec 19 '14 at 17:20
@Dasherman: Right, but $\gcd = 1$ still isn't necessary ($2$ and $4$ both divide $24$ and so does their product, even though their $\gcd $ is $2$). — Henrik supports the community, Dec 19 '14 at 17:38
@Henrik: You are right. I think I misunderstood your original comment. gcd(a, b)=1 is indeed a sufficient, but not a neccesary condition. — Dasherman, Dec 19 '14 at 17:54

score 5 · Accepted Answer · edited Dec 19 '14 at 16:55

5

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 $n$.

edited Dec 19 '14 at 16:55

Sujaan Kunalan

10,944

answered Dec 19 '14 at 16:49

Dasherman

4,206

score 0 · Answer 2 · answered Dec 19 '14 at 17:37

0

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

answered Dec 19 '14 at 17:37

The Bardbarian

26

score -1 · Answer 3 · edited Feb 13 '24 at 08:29

-1

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 $60$.

edited Feb 13 '24 at 08:29

Dominique

2,130

answered Feb 13 '24 at 08:04

Matias Pereira Miranda

1

Is a Number Divisible by 40

3 Answers3