Question

Find two consecutive multiples of 3 whose product is 648.

Answer 1

Hint: Here, we will assume one of the multiples to be \[x\] and since they are consecutive multiples of 3, thus the next multiple will be \[\left( {x + 3} \right)\]. We will then multiply these two assumed numbers and equate it to the given product. We will solve the equation further to get the value of \[x\], hence the first number. We will then substitute the value of the first number in the expression for the second number to get the required consecutive multiples of 3.

Complete step by step solution:
Let one of the multiples of 3 be \[x\].
Since, we are required to find two consecutive multiples of 3, thus, the second multiple of three will be \[\left( {x + 3} \right)\].
According to the question,
The product of these two consecutive multiples of 3 is 648.
Thus, we can form a mathematical equation using this information as:
\[x\left( {x + 3} \right) = 648\]
Now multiplying the terms using the distributive property, we get
\[ \Rightarrow {x^2} + 3x = 648\]
\[ \Rightarrow {x^2} + 3x - 648 = 0\]
The obtained equation is a quadratic equation. Now, doing middle term spit in this quadratic equation, we get,
\[ \Rightarrow {x^2} + 27x - 24x - 648 = 0\]
\[ \Rightarrow x\left( {x + 27} \right) - 24\left( {x + 27} \right) = 0\]
Factoring out common terms, we get
\[ \Rightarrow \left( {x - 24} \right)\left( {x + 27} \right) = 0\]
By using zero product property, we get
\[\begin{array}{l} \Rightarrow \left( {x - 24} \right) = 0\\ \Rightarrow x = 24\end{array}\]
 Or
\[\begin{array}{l} \Rightarrow \left( {x + 27} \right) = 0\\ \Rightarrow x = - 27\end{array}\]
As the product is positive, we will neglect the negative value and only consider the positive one.
Thus, The first required multiple of 3 \[ = x = 24\].
And, the second required multiple of 3 \[ = \left( {x + 3} \right) = \left( {24 + 3} \right) = 27\]

Therefore, the two consecutive multiples of 3 whose product is 648 are 24 and 27 respectively.
This is the required answer.

Note: An alternate way of solving this question, is to let the first multiple of 3 as \[3x\] and the second consecutive multiple of 3 as \[3\left( {x + 1} \right)\]. Since, the product is 648, thus we can write the equation as:
\[3x \times 3\left( {x + 1} \right) = 648\]
\[ \Rightarrow 9x\left( {x + 1} \right) = 648\]
Again, opening the brackets,
\[ \Rightarrow 9{x^2} + 9x - 648 = 0\]
Dividing both sides by 9, we get
\[ \Rightarrow {x^2} + x - 72 = 0\]
Now, doing middle term split, we get
\[ \Rightarrow {x^2} + 9x - 8x - 72 = 0\]
\[ \Rightarrow x\left( {x + 9} \right) - 8\left( {x + 9} \right) = 0\]
Now we will factor out the common terms.
\[ \Rightarrow \left( {x - 8} \right)\left( {x + 9} \right) = 0\]
By using zero product property, we get
\[ \Rightarrow x = 8\] or \[x = - 9\]
As the product is positive, we will neglect the negative value and only consider the positive one. Thus,
The first required multiple of 3, \[ = 3x = 3 \times 8 = 24\]
And, the second required multiple of 3, \[ = 3\left( {x + 1} \right) = 3\left( {8 + 1} \right) = 27\]
Therefore, the two consecutive multiples of 3 whose product is 648 are 24 and 27 respectively.
This is the required answer.
Hence, we can use either of the two ways to find the required answer.