Result \(1\): If \(a\) divides \(M\) and \(a\) divides \(N\), then \(a\) divides \(M + N\) and \(a\) divides \(M – N\). In other words, if \(M\) and \(N\) are multiples of \(a\), then \(M + N\) and \(M – N\) will also be multiples of \(a\).
Example:
Consider two numbers \(24\) and \(36\).
Here, \(M = 24\) and \(N = 36\) are divisible by \(12\).
Now, \(M + N = 24 + 36 = 60\) which is divisible by \(12\).
And, \(M - N = 24 - 36 = -12\) which is divisible by \(12\).
Therefore, \(a\) divides \(M + N\) and \(a\) divides \(M – N\).
Thus, if \(a\) divides \(M\) and \(a\) divides \(N\), then \(a\) divides \(M + N\) and \(a\) divides \(M – N\). In other words, if \(M\) and \(N\) are multiples of \(a\), then \(M + N\) and \(M – N\) will also be multiples of \(a\).
Result \(2\): If \(A\) is divisible by \(k\), then all multiples of \(A\) are divisible by \(k\).
Example:
Let \(A = 15\) and \(k = 3\).
\(15\) is divisible by \(3\) because \(\frac{15}{3} = 5\)
Multiples of \(15\) are \(30\), \(45\), \(60\), ....
Here, \(30\), \(45\), \(60\), .... are divisible by \(3\).
Thus, all multiples of \(15\) are divisible by \(3\).
Therefore, if \(A\) is divisible by \(k\), then all multiples of \(A\) are divisible by \(k\).
Result \(3\): If \(A\) is divisible by \(k\), then \(A\) is divisible by all the factors of \(k\).
Example:
Let \(A = 80\) and \(k = 40\).
Here, \(80\) is divisible by \(40\).
Factors of \(40\) are \(1\), \(2\), \(4\), \(5\), \(8\), \(10\), and \(20\). Since \(80\) is divisible by \(40\), then \(80\) must also be divisible by \(1\), \(2\), \(4\), \(5\), \(8\), and \(20\).
Let us verify them.
\(80\) is divisible by \(1\) because \(\frac{80}{1} = 80\)
\(80\) is divisible by \(2\) because \(\frac{80}{2} = 40\)
\(80\) is divisible by \(4\) because \(\frac{80}{4} = 10\)
\(80\) is divisible by \(5\) because \(\frac{80}{5} = 16\)
\(80\) is divisible by \(1\) because \(\frac{80}{10} = 8\)
\(80\) is divisible by \(1\) because \(\frac{80}{20} = 4\)
Therefore, \(80\) is divisible by all factors of \(40\).
Thus, if \(A\) is divisible by \(k\), then \(A\) is divisible by all the factors of \(k\).
Result \(4\): If \(A\) is divisible by \(k\) and \(A\) is also divisible by \(m\), then \(A\) is divisible by the LCM of \(k\) and \(m\).
Example:
Let \(A = 120\), \(k = 12\) and \(k = 15\).
\(120\) is divisible by \(12\) because \(\frac{120}{12} = 10\)
\(120\) is divisible by \(15\) because \(\frac{120}{15} = 8\)
Prime factors of \(12\): \(2 \times 2 \times 3\)
Prime factors of \(15\): \(3 \times 5\)
LCM of \(12\) and \(15 = 2 \times 2 \times 3 \times 5 = 60\)
\(120\) is divisible by \(60\).
Therefore, if \(A\) is divisible by \(k\) and \(A\) is also divisible by \(m\), then \(A\) is divisible by the LCM of \(k\) and \(m\).