Question

A two-digit number is such that the product of its digits is 20. If 9 is subtracted from the number, the digits interchange their place. Find the number.

Answer 1

Hint: At first, we have to assume that the number is “ab” and then using the conditions given in the question that ab = 20 and $\left( 10a+b \right)-9=\left( 10b+a \right)$, we can find the value of the two digits a and b. We must also keep in mind that the digits can not be positive and we are only interested in positive numbers.

Complete step-by-step solution:
 We are given that the required number is a two-digit number. So, let us assume that the required number is N = ”ab”, that is, the digit at one place is b, and the digit at tens place is a.
Thus, we can write
Required number, N = 10 a + b.
It is also given that the product of digits is 20. Thus, ab = 20…(i)
We are also given that when the number is subtracted from 9, its digits are interchanged.
Hence, we can write
$\left( 10a+b \right)-9=\left( 10b+a \right)$,
where (10b + a) is the reversed number.
On simplification, we can write
$9a-9b=9$.
We can take 9 as common from both sides, and cancel them to get
$a-b=1$
Hence, we can also write that $a=b+1$.
Now, putting the value of a in equation (i), we get
$\left( b+1 \right)b=20$.
Hence, we have the quadratic equation
${{b}^{2}}+b-20=0$.
We can also write this quadratic equation as
${{b}^{2}}+5b-4b-20=0$.
On taking factors as common, we get
$b\left( b+5 \right)-4\left( b+5 \right)=0$.
Hence, we can write
$\left( b+5 \right)\left( b-4 \right)=0$.
Thus, we get
$b=-5,4$.
The value of an integer can never be negative. And so, b = 4.
And thus, a = 4 + 1 = 5.
Hence, the required number is 54.

Note: We must understand that the phrase ‘digits interchange their place’ is also sometimes written as ‘the number is reversed’. They both have the same meanings, and convey that the digit at one's place moves to the tens place, and the digit at tens place, moves to the ones place.