Question

A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Answer 1

Here we have 2 conditions on the both digits of 2 digit number. Every two digit number can be written as a sum of 10 times the tens digit and 1 time the units digit. So now assume 2 variables to represent the units and tens digits, say a and b. Using data given in question, form 2 equations. First one would be ab=20. Now use a substitution method to solve these equations and get the values of 2 variables. Now substituting these variables into the statement given by – “Every two digit number can be written as sum of 10 times the tens digit and 1 time the units digit” we get the value of 2 digit number. This 2 digit number is the required solution.

Complete step-by-step answer :
Let us assume the digits of 2-digit number as a and b, we get:
First digit = a
Second digit = b
First condition given in the equation about the digits:
Product of them $=20$
$\left( a \right)\cdot \left( b \right)=20$ …..(i)
A 2-digit number is said to be written as 10 times the digit in tens place added to 1 times the digit in the units place.
Example : - $21=2\times 10+1\times 1$ ; $71=7\times 10+1\times 1$ ; $25=2\times 10+5\times 1$
Similarly, here we have our number to be written as:
$required\text{ }number=\left( a\times 10 \right)+b$
Given the number will be reversed.
By reversing the digits, we get:
First digit = b
Second digit = a
So, new number is also a 2-digit number called reverse of number it can be written as:
$reverse\text{ }of\text{ }the\text{ }number=\left( b\times 10 \right)+a$
Given condition is if 9 is added to required number it gives the reverse of the number as the result.

Mathematically, we get
Required number + 9 = reverse of the number
By using representations as given by above 2 equations, we get
$\left( a\times 10 \right)+b+9=\left( b\times 10 \right)+a$
By subtracting with a and 10b on both sides, we get
$9a-9b+9=0$
By taking the 9 common and cancelling, we get that:
$a-b+1=0$
From (i) we can substitute the value of a as:
$\dfrac{20}{b}-b+1=0$ $\Rightarrow {{b}^{2}}-b-20=0$
Equation can be written as: $\left( b+4 \right)\left( b-5 \right)=0$
b is digit of number so always positive; $b=5$
By substituting this in equation (i) we get a as $a=4$
The required number $=\left( a\times 10 \right)+b=40+5=45$
45 is the 2 digit number satisfying the given condition.

Note :Be careful while assuming the reverse number because it depends on previous assumptions. Do the assumptions correspondingly. While you solve quadratic in b you get 2 solutions. But we assumed b as a digit of a number, so we know a, b > 0 always.