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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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    $\begingroup$ Wow! 4 answers all posted within 41 seconds of each other! $\endgroup$ – davidlowryduda Jun 21 '12 at 22:48
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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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    $\begingroup$ $4$ answers in $40$ seconds! $\endgroup$ – user17762 Jun 21 '12 at 22:48
  • $\begingroup$ Yeah, I was just noticing that too! $\endgroup$ – davidlowryduda Jun 21 '12 at 22:48
  • $\begingroup$ +1. To nitpick, "primes have the characteristic that..." should be "the definition of a prime is that...". $\endgroup$ – user17762 Jun 21 '12 at 22:55
  • $\begingroup$ 3 and 8 are relatively prime ? could you explain that a bit ? $\endgroup$ – Rajeshwar Jun 21 '12 at 22:56
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    $\begingroup$ @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.' $\endgroup$ – davidlowryduda Jun 21 '12 at 22:57
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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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  • $\begingroup$ Thanks for the great answer $\endgroup$ – Rajeshwar Jun 21 '12 at 22:52
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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 tinyurl.com/mlxk8pw

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If LCM ( a,b) = n, then any number which is divisible by n is also divisible by a & b because n is least common multiple of a & b.

Now if we want to check divisibility of 24 then we have to follow these steps.

  1. We know that there is no predetermined divisibility rule for 24 as in the case of 2,3,4,5,6,7,8,9,11. So we need to find factors of 24 between 2 to 11.

  2. Now we have factors of 24 between ( 2 to 11) is 2,3,4,6,8

  3. Now find factors whose LCM is 24 ( see above rule). LCM ( 8, 3) = 24 , from above rule any number divisible by 24 also divisible by 8 & 3 so check the divisibility of that number from 8 & 3 .

We can't consider (4,6) , ( 8,2) or (2,3,4) etc as their LCM is not 24.

Further, there is one more trick to solve this, see below

The LCM of co-prime numbers is product of these numbers because co-prime number has no common factor. Suppose p & q are co-prime numbers then LCM (p,q) = p*q

Now we can easily find co-prime numbers from factors of 24 ( 2,3,4,6,8) whose multiplication is 24 i.e 8 & 3

8 & 3 are co-prime and its product is 24.

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