Question

If g.c.d \[(24,20) = 3x + 1\], then \[x = \_\_\_\_\]

Answer 1

Hint: We calculate the greatest common divisor of two given numbers and use that value of greatest common divisor to find the value of ‘x’ using the given equation.
* G.C.D means the greatest common divisor. We can calculate G.C.D of two numbers by writing their prime factorization and calculating the highest multiplicative number existing in both prime factorizations.
* Prime factorization is a process of writing a number in multiple of its factors where all factors are prime numbers.

Complete step-by-step solution:
We have to find the g.c.d of the numbers 24 and 20.
We write the prime factorization of both the numbers separately.
\[ \Rightarrow \]Prime factorization of \[24 = 2 \times 2 \times 2 \times 3\]
And
\[ \Rightarrow \]Prime factorization of \[20 = 2 \times 2 \times 5\]
Since we can see the greatest common factor between both the prime factorizations of 24 and 20 is \[2 \times 2 = 4\].
So, g.c.d \[(24,20) = 4\]
Now we use this value and equate it with the given equation.
\[ \Rightarrow 3x + 1 = 4\]
Shift all constant values to right hand side of the equation
\[ \Rightarrow 3x = 4 - 1\]
Calculate the subtraction on RHS of the equation
\[ \Rightarrow 3x = 3\]
Divide both sides of the equation by 3
\[ \Rightarrow \dfrac{{3x}}{3} = \dfrac{3}{3}\]
Cancel same factors from both numerator and denominator on both sides of the equation
\[ \Rightarrow x = 1\]

\[\therefore \]The value of \[x = 1\]

Note: Students many times get confused while writing the value of g.c.d as they write only the common divisor but we have to find the highest number that divides both the given numbers.
Keep in mind g.c.d of any two or more numbers can never be greater than the maximum of three numbers. Also, many students make mistakes when shifting values from one side of the equation to another, keep in mind we always change sign from positive to negative and vice-versa when shifting values to the opposite side of the equation.