Question

A two-digit number has a tens digit greater than the unit’s digit. If the sum of its digits is equal to twice the difference, how many such numbers are possible?

Answer 1

Hint: We will first assume the one’s digit to be x and the ten’s to be y for this question. Hence, the numbers which will follow such a condition will be given as 10x+y. Then we will form equations in x and y from the given conditions. Then we will use the hit and trial method for finding the values of y for every x. Then we will count the number of values of both x and y which follow all the given conditions and hence, we will get our answer.

Complete step-by-step solution:
Let us first assume the one’s digit to be ‘x’ and the tens digit to be ‘y’.
Hence, the required number will be given by $10y+x.$
Now, we have been given that the ten’s digit is greater than the one’s digit. Thus, we can say that:
$y >x$
Now, we also have been given that the sum of the digits is twice the difference between the digits. Thus, we can say that:
$y+x=2\left( y-x \right)$
Simplifying this equation, we get:
$\begin{align}
  & y+x=2\left( y-x \right) \\
 & \Rightarrow y+x=2y-2x \\
\end{align}$
$\Rightarrow 3x=y$ ………………..(i)
Hence, the digit at ten’s place is thrice the number at one’s place.
Now, we will use hit and trial and find the values of y for every x.
Thus, we will get:
Case-1: x=0
Keeping x=0 in equation (i) we get:
$\begin{align}
  & 3x=y \\
 & \Rightarrow 3\left( 0 \right)=y \\
 & \therefore y=0 \\
\end{align}$
Thus, the number will be:
$\begin{align}
  & 10\left( 0 \right)+0 \\
 & \Rightarrow 0 \\
\end{align}$
But the number cannot be 0 as it is a two digit number. Hence, it x=0 is not a solution.
Case-2: x=1
Keeping x=1 in equation (i) we get:
$\begin{align}
  & 3x=y \\
 & \Rightarrow 3\left( 1 \right)=y \\
 & \therefore y=3 \\
\end{align}$
Thus, the number will be:
$\begin{align}
  & 10\left( 3 \right)+1 \\
 & \Rightarrow 31 \\
\end{align}$
Thus, the first number is 31.
Case-3: x=2
Keeping x=2 in equation (i) we get:
$\begin{align}
  & 3x=y \\
 & \Rightarrow 3\left( 2 \right)=y \\
 & \therefore y=6 \\
\end{align}$
Thus, the number is:
$\begin{align}
  & 10\left( 6 \right)+2 \\
 & \Rightarrow 62 \\
\end{align}$
Thus, the second number is 62.
Case-3: x=3
Keeping x=3 in equation (i) we get:
$\begin{align}
& 3x=y \\
& \therefore 3\left( 3 \right)=y \\
& \Rightarrow y=9 \\
\end{align}$
Thus, the number is:
 $\begin{align}
  & 10\left( 9 \right)+3 \\
 & \Rightarrow 93 \\
\end{align}$
Thus, the third number is 93.
Case-4: x=4
Now, if we keep x greater than or equal to 4, all values of y will come out to be 2 digits as $3\left( 4 \right)=12$. Thus, the maximum value of x will be 3 and so will be the respective values of y.
Thus, we will get three numbers, which follow such a condition which are 31, 62, and 93.
Hence, 3 such numbers are possible.

Note: We here have only mentioned the positive numbers because it is not mentioned in the question if we have to consider the negative numbers as well. So we have taken only positive numbers by default. But we notice that if there is any mention in the question for negative numbers, then the number of negative numbers will be exactly equal to the number of positive numbers and hence the answer will get doubled. So read the question carefully and answer accordingly.