Question

A two digit number is 4 times the sum of its digits and twice the product of its digits. The number is _____ .

Answer 1

Hint: First of all, assume a two digit number that is 10b + a where b and a are digit at tens and units place respectively. Now apply the given conditions that are 10b + a = 4(a+b) and 10b + a = 2(ab) to find the value of a and b and hence, find the number.

Complete step-by-step answer:
Here we are given that we have a 2 digit number such that it is 4 times the sum of its digits and twice the product of its digits. We have to find the number.
First of all let us consider a two digit number whose units digit is ‘a’ and tens digit is ‘b’. Therefore, we get the number as,
N = 10b +a .......................(i)
Now we find 4 times the sum of its digits of numbers we get,
M = 4 (a + b)............(ii)
As we are given that the number is equal to the 4 times the sum of its digits, so now we equate equation (i) and equation (ii), we get,
$\Rightarrow 10b+a=4\left( a+b \right)$
By simplifying the above equation, we get,
$10b+a=4a+4b$
By taking the terms containing ‘a’ to one side and ‘b’ to other side, we get,
$\Rightarrow 10b-4b=4a-a$
We get, $6b=3a$
By dividing 3 on both sides, we get,
$\begin{align}
  & \Rightarrow \dfrac{6}{3}b=\dfrac{3}{3}a \\
 & \Rightarrow 2b=a......................\left( iii \right) \\
\end{align}$
Now, we find twice the product of digits of number we get,
$Q=2\left( ab \right)..................\left( iv \right)$
As we are given that is also equal to twice the product of its digits. So now we equate equation (i) and equation (iv), we get,
$\Rightarrow 10b+a=2\left( ab \right)$
By putting a = 2b from equation (iii) in above equation, we get,
$\Rightarrow 10b+2b=2\left[ \left( 2b \right)\times \left( b \right) \right]$
By simplifying the above equation, we get,
$\Rightarrow 12b=4{{b}^{2}}$
By dividing 4 on both sides, we get,
$\Rightarrow \dfrac{12b}{4}=\dfrac{4{{b}^{2}}}{4}$
We get, \[3b={{b}^{2}}\]
Or \[{{b}^{2}}-3b=0\]
By taking out ‘b’ common, we get,
b (b – 3) = 0
As we are given a 2 digit number and we assume ‘b’ to be in tens place, so ‘b’ could not be equal to 0. Therefore we get,
b = 3
By putting the value of ‘b’ in equation (iii), we get
a = 2(3)
So we get, a=6
By putting the value of a and b in equation (i), we get our number as,
N = 10 (3) + 6
N = 30 + 6
N = 36
Hence, we get the two digit number as 36.

Note: Here students must note that any two digit number is of the form 10x + y where x and y are its tens and units digit respectively. Also students must cross check the answer satisfying the given conditions as follows;
First of all, we were given that number is equal 4 times the sum of its digits, that is 10b + a = 4(a +b). As we have got a = 6 and b = 3, by putting these values in above equation, we get,
$\begin{align}
  & \Rightarrow 10\left( 3 \right)+6=4\left( 6+3 \right) \\
 & \Rightarrow 30+6=4\left( 9 \right) \\
 & \Rightarrow 36=36 \\
\end{align}$
LHS = RHS therefore, our answer is correct. In a similar way we can check another condition, that is 10b + a = 2(a.b). By putting a = 6 and b = 3, we get
$\begin{align}
  & \Rightarrow 10\left( 3 \right)+6=2\left( 6\times 3 \right) \\
 & \Rightarrow 30+6=2\left( 18 \right) \\
 & \Rightarrow 36=36 \\
\end{align}$
LHS = RHS, therefore our answer is correct.