Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am currently familiar with the method of checking if a number is divisible by $2, 3, 4, 5, 6, 8, 9, 10, 11$. While Checking for divisibility for $24$ (online). I found out that the number has to satisfy the divisibility criteria of $3$ and $8$. I agree this gives the answer. But why cant I check the divisibility using the divisibility criteria of $6$ and $4$ ? Is there a rule to this criteria ?

share|improve this question
1  
Wow! 4 answers all posted within 41 seconds of each other! –  mixedmath Jun 21 '12 at 22:48

6 Answers 6

up vote 17 down vote accepted

The problem here might be something like $12$. You see, we have that $12$ is divisible by both $6$ and $4$, but it's not divisible by $24$. The reason they suggest $3$ and $8$ is because they are relatively prime, meaning that you can't have the sort of overlap in the case of $6$ and $4$.

This all has to do with the Fundamental Theorem of Arithmetic, which says that each number can be written uniquely as a product of primes, and primes have the special characteristic (or as Marvis points out, they are defined to be exactly those numbers with the characteristic) that if $p|ab$, then $p|a$ or $p|b$. So if $3$ and $8$ divide a number, then $24$ divides that number. But $6$ and $4$ dividing a number doesn't even guarantee that $8$ divides that number.

share|improve this answer
    
$4$ answers in $40$ seconds! –  user17762 Jun 21 '12 at 22:48
    
Yeah, I was just noticing that too! –  mixedmath Jun 21 '12 at 22:48
    
+1. To nitpick, "primes have the characteristic that..." should be "the definition of a prime is that...". –  user17762 Jun 21 '12 at 22:55
    
3 and 8 are relatively prime ? could you explain that a bit ? –  Rajeshwar Jun 21 '12 at 22:56
4  
@Rajeshwar: This is saying that no prime that divides $3$ also divides $8$, and vice versa. It's easier to look at $4$ and $6$. You see, $2$ is a prime number, but $2$ divides both $4$ and $6$. Thus they are not 'relatively prime.' –  mixedmath Jun 21 '12 at 22:57

If $3 \mid n$ and $8 \mid n$, then clearly the $\mathrm{lcm}(3,8) = 24 \mid n$, since the least common multiple of $a$ and $b$ clearly divides any number divisible by both $a$ and $b$.

On the other hand, $4 \mid n$ and $6 \mid n$ is only enough to conclude that $\mathrm{lcm}(4,6) = 12 \mid n$.

share|improve this answer
    
Thanks for the great answer –  Rajeshwar Jun 21 '12 at 22:52

If you are divisible by both $6$ and $4$, the number could be 12 which is not divisible by 24.

The reason for using $3$ and $8$ is that the least common multiple is 24. So every number that is divisible by both $3$ and $8$ is divisible by $24$.

share|improve this answer

It is not true that if a number's divided by $\,6\,,\,4\,$ then it is divided by $\,6\cdot 4\,$ , as $\,12\,$ proves. Yet it is true that if a number's divided by $\,3\,,\,8\,\,$ then it i divided by $\,3\cdot 8=24\,$. Why? Because the former pair is not coprime (i.e., its minimal common divisor is not $\,1\,$), whereas the latter pair is coprime...Can you take it from here?

share|improve this answer

For instance, $12$ is divisible by $6$ and $4$ but is not divisible by $24$. The problem with $6$ and $4$ is that the $\gcd(6,4) = 2 \neq 1$. You could split $24$ as $8$ and $3$ and check for divisibility by $8$ and $3$ because $\gcd(3,8) = 1$.

In general, to check for divisibility by $n$, look at the prime decomposition of $n = p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ and check whether the number is divisible by $p_j^{\alpha_j}$ for all $j \in \{1,2,\ldots,k\}$.

Another equvialent way is to write $n = ab$ such that $\gcd(a,b) = 1$. Then a number is divisible by $n$ if and only if the number is divisible by $a$ and $b$.

share|improve this answer

According to cross divisibility test (VJ's universal divisibility test) there are infinite test for any number. For 24 the divisibility test are given as 1) 24 | (10T+U) if and only if 24 | (2T+ 5U ) 2) 24 | (10T+U) if and only if 24 | (2T -7U) 3) 24 | (10T+U) if and only if 24 | (2T-17U ) etc

To discover why it works refer 'Modern Approach to speed math Secret' at tinyurl.com/mlxk8pw

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.