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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 ?

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Wow! 4 answers all posted within 41 seconds of each other! – mixedmath Jun 21 '12 at 22:48
up vote 18 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.

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$4$ 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
@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 $\mathrm{lcm}(4,6) = 12 \mid n$.

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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 $24$.

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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?

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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 $b$.

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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

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