In the world of arithmetic, two of the most important concepts are the Highest Common Factor (H.C.F.) and the Least Common Multiple (L.C.M.). While they represent different ways of looking at numbers, they share a very specific and useful relationship.
The most important rule to remember is:
The product of two numbers is always equal to the product of their H.C.F. and L.C.M.
In mathematical terms, if you have two numbers, $A$ and $B$:
$A \times B = \text{H.C.F.} \times \text{L.C.M.}$
Let’s look at an example using the numbers 15 and 24.
To find the L.C.M., we can use the division method:
Divide 15 and 24 by their common prime factor (3).
$15 \div 3 = 5$
$24 \div 3 = 8$
$\text{L.C.M.} = 3 \times 5 \times 8 = \mathbf{120}$
Using the long division method:
Divide 24 by 15 (Remainder is 9).
Divide 15 by 9 (Remainder is 6).
Divide 9 by 6 (Remainder is 3).
Divide 6 by 3 (Remainder is 0).
The last divisor is the H.C.F., which is 3.
Now, let’s test our rule:
Product of the two numbers: $15 \times 24 = \mathbf{360}$
Product of H.C.F. and L.C.M.: $3 \times 120 = \mathbf{360}$
As you can see, both calculations result in 360, proving that the product of the numbers equals the product of their H.C.F. and L.C.M.
What happens when you need to find the H.C.F. or L.C.M. of fractions like $\frac{a}{b}$ and $\frac{c}{d}$? The formula changes slightly:
To find the H.C.F. of two or more fractions, use this formula:
Conversely, to find the L.C.M. of fractions:
In this second part of our series on the Primary Number System, we will explore how to apply Lowest Common Multiple (L.C.M.) and Highest Common Factor (H.C.F.) to fractions and real-world scenarios.
When working with fractions, we use specific formulas that involve calculating both the L.C.M. and H.C.F. (also known as the Greatest Common Divisor or G.C.D.) of the numerators and denominators.
To find the L.C.M. of fractions, find the L.C.M. of the numerators and divide it by the H.C.F. of the denominators:
Example: Find the L.C.M. of $\frac{1}{2}$ and $\frac{3}{4}$.
Numerator L.C.M.: L.C.M. of $\{1, 3\} = 3$.
Denominator H.C.F.: G.C.D. of $\{2, 4\} = 2$.
Result: $\frac{3}{2}$.
To find the H.C.F., find the H.C.F. of the numerators and divide it by the L.C.M. of the denominators:
Example: Find the H.C.F. of $\frac{8}{15}$ and $\frac{16}{125}$.
Numerator H.C.F.: H.C.F. of $\{8, 16\} = 8$.
Denominator L.C.M.: L.C.M. of $\{15, 125\} = 375$.
Result: $\frac{8}{375}$.
H.C.F. and L.C.M. are powerful tools for solving mathematical puzzles and real-world timing problems.
Problem: Find the largest number which can exactly divide 10, 20, and 30.
Solution: This requires finding the H.C.F. using prime factorization:
$10 = 2^1 \times 5^1 \times 3^0$
$20 = 2^2 \times 5^1 \times 3^0$
$30 = 2^1 \times 3^1 \times 5^1$
G.C.D. (or H.C.F.): $2 \times 3^0 \times 5 = 10$.
Problem: The time periods of three planets are 100 days, 250 days, and 300 days respectively. After how many days will the planets return to their present positions simultaneously?
Solution: To find when multiple cycles synchronize, we calculate the L.C.M.:
Required number of days: L.C.M. $\{100, 250, 300\}$.
Result: 1500 days.
Have you ever come across a math problem that asks you to find a specific large number divisible by a set of different factors? While it might look intimidating at first, the process is actually quite logical once you break it down.
Today, we are going to solve a classic example: Find the largest 5-digit number which is divisible by 2, 3, 5, and 11.
If a number needs to be divisible by 2, 3, 5, and 11 simultaneously, it must be a multiple of their Least Common Multiple (LCM).
Since all these numbers (2, 3, 5, and 11) are prime numbers, we simply multiply them together:
Any number divisible by 330 will automatically be divisible by 2, 3, 5, and 11.
The largest 5-digit number in existence is 99,999. However, we don’t know yet if 99,999 is divisible by 330. To find out, we perform a simple long division.
When we divide 99,999 by 330, we want to see how much “extra” is left over.
Calculation: $99999 \div 330 = 303$ with a remainder of 9.
The remainder of 9 tells us that 99,999 is exactly 9 units “too high” to be perfectly divisible by 330.
To get our final answer, we subtract that extra remainder from our starting number:
The largest 5-digit number divisible by 2, 3, 5, and 11 is 99,990.
Quick Tip: You can double-check this easily!
It ends in 0, so it’s divisible by 2 and 5.
The digits add up to 27, so it’s divisible by 3.
The alternating sum of digits $(9-9+9-9+0)$ is 0, so it’s divisible by 11.