Question

A two-digit number is thrice as large as the sum of its digits and the square of that sum is equal to the tripled required number. Find the number.

Answer 1

Hint: Here we will be using the given information on the required number and forming the equations based on that. Then we will be solving those equations and hence would obtain the required number.

Complete step-by-step answer:
A two-digit number is thrice as large as the sum of its digits and the square of that sum is equal to the tripled required number.
Let us assume that the required number is \[10x + y\] (since it is given in the problem statement that the number is of two-digit) where x and y belongs to the set of Natural numbers and x is the digit of the tenths place and y is the digit of the unit place.
Here we will be using the given information on the required number and forming the equations based on that.
Since it is given that the two-digit number is thrice as large as the sum of its digits, therefore the equation is
\[\Rightarrow 10x + y = 3\left( {x + y} \right)\]
Again, it is given that the square of the sum is equal to the tripled required number.
Hence, we get the equation as \[{\left( {x + y} \right)^2} = 3\left( {10x + y} \right)\].
From the equation \[10x + y = 3\left( {x + y} \right)\], we get that
\[\left( {x + y} \right) = \dfrac{{10x + y}}{3}\].
From the equation\[{\left( {x + y} \right)^2} = 3\left( {10x + y} \right)\], we get that
\[\left( {x + y} \right) = \sqrt {3\left( {10x + y} \right)} \].
Comparing these two equations, we get that,
\[\Rightarrow \dfrac{{10x + y}}{3} = \sqrt {3\left( {10x + y} \right)} \].
Simplifying the equation we get that,
\[\Rightarrow \sqrt {10x + y} = 3\sqrt 3 \].
Squaring both sides, we get that
\[\Rightarrow 10x + y = 27\]
Hence, the required two-digit number is $ 27 $ .
So, the correct answer is “27”.

Note: In problems like these, always remember to convert the given word statement in the correct mathematical expression. Then simplify accordingly and find the value of unknowns. Rest goes well in these types of questions.