Question
Question: A two-digit number is 4 times the sum of its digits the number obtained by interchanging the digits ...
A two-digit number is 4 times the sum of its digits the number obtained by interchanging the digits is less by 9 than twice the original number. Find the original number.
Solution
Hint: Here, first of all, take the original number as 10y + x. Then the number by reversing its digits would be 10x + y. Use the given information to find the value of x and y. Then finally substitute x and y in 10y + x to get the original number.
Complete step-by-step answer:
Here, we are given that a two-digit number is 4 times the sum of its digits and the number obtained by reversing its digits is less by 9 than twice the original number. Here, we have to find the original number. Let us consider the original two-digit number such that its units digit is x and the tens digit is y.
So, we get the original number = 10y + x…..(i)
We are given that this number is equal to 4 times the sum of its digits, so we get 10y + x = 4(x + y)
By simplifying the above equation, we get,
⇒10y+x=4x+4y
⇒10y−4y=4x−x
⇒6y=3x
By dividing 3 on both the sides of the above equation, we get
36y=33x
⇒x=2y....(ii)
Now, by reversing the digits of the original number, we get the units digit of the new number as y and tens digits as x.
So, we get the new number as = 10x + y….(iii)
Now, we are given that this new number is less by 9 than twice the original number. So, we get,
New number = 2 (Original number) – 9
By substituting the values of the original number and the new number from equation (i) and (ii), respectively, we get,
10x+y=2(10y+x)−9
By simplifying the above equation, we get,
10x+y=20y+2x−9
⇒10x−2x=20y−y−9
⇒8x=19y−9
By substituting the values of x = 2y from equation (ii), we get,