Theory:

Result I: If a number can be divided by another number, then it can also be divided by each of its factor.
Example:
Lets take two number \(18\) and \(72\).
 
\(72\div18 = 4\) this show \(72\) is divisible by \(18\),
 
Factors of  \(18 = 1, 2, 3, 6, 9, 18\).
 
Now we divide the factors of \(18\) with \(72\).
 
\(72\div1=72\), \(72\div2=36\), \(72\div3=24\), \(72\div6=12\), \(72\div9=8\) and \(72\div18=4\).
 
Thus, \(72\) is divisible by each of factors of \(18\).
Result II: If a number is divided by two co-prime numbers, it is also divisible by its product
Example:
Let's say \(90\) is divisible by \(5\) and \(9\). As we know \(5\) and \(9\) are co-prime numbers.
 
Thus, the product of co-primes is (\(5\times9 = 45\)).
 
\(90\div45 = 2\).
 
Therefore \(90\) is divisible by the product of co-primes \(45\).
Result III: If two given numbers are divisible by a number, then, their sum is also divisible by that number.
Example:
Let us take \(21\) and \(18\). Both the numbers are divisible by \(3\).
 
\(21\div3 = 7\) and \(18\div3 = 6\).
 
Sum of the two numbers is \(21+18 = 39\). Also \(39\div 3 = 13\).
 
Therefore, if \(24\) and \(18\) are divisible by \(3\), then their sum is \(39\) which is also divisible by \(3\).
Result IV: If two given numbers are divisible by a number, then their difference is also divisible by that number.
Example:
Let us take \(58\) and \(54\) are divisible by \(2\).
 
\(958\div2 = 29\) and \(54\div2 = 27\).
 
Difference of the two numbers that \(58-54 = 4\) and \(4\div2 = 2\).
 
Therefore, if \(54\) and \(58\) are divisible by \(2\), then their difference \(4\) is also divisible by \(2\).
More examples:
 
1. A number is divisible by both \(7\) and \(15\). By which other numbers will that number be always divisible?
 
Note that the number is divisible by \(7\) and \(15\). Let us recall 'Result II'.
Result II: If a number is divided by two co-prime numbers, it is also divisible by its product.
Since \(7\) and \(15\) are co-prime numbers, the number must be divisible by the product \(7\times 15 = 105\).
 
So, the given number will always be divisible by \(105\).
 
2. A number is divisible by \(16\). By what other numbers will that number be divisible?
 
As we need to find the more divisible for the given number, let us recall 'Result I'.
Result I: If a number can be divided by another number, then it can also be divided by each of its factor.
The number is divisible by\(16\).
 
The factors of \(16\) are \(1, 2, 4, 8, 16\).
 
Therefore, the number is also divisible by \(1, 2, 4, 8\).