Question
Question: A digit of a three-digit natural number is in A.P and their sum is 15. The obtained number by revers...
A digit of a three-digit natural number is in A.P and their sum is 15. The obtained number by reversing the digits is 396 less than the original number. Find the number.
Solution
First of all, we will assume an A.P, (a – d), a, (a + d). Then we will find the value of a with the given condition on the sum of these three terms of A.P. After this, we will consider the second condition and apply logics of addition to get an equation to find d. Once we get a and d, we can find the original number.
Complete step by step answer:
It is given that the three digits of the number are in AP
Let the 2nd term of this AP be a and the common difference be d. Therefore, the first digit of the number is (a – d), second digit is a and third digit is (a + d).
The sum of this AP is given as 15.
⇒ first digit + second digit + third digit = 15
⇒ (a – d) + a + (a + d) = 15
⇒ 3a = 15
⇒ a = 5
Thus, we know that the second digit of the original number and the reversed number is 5.
So, the original number = (5 – d) 5 (5 + d)
And the reversed number = (5 + d) 5 (5 – d)
Now, it is given that the reversed number is 396 less than the original number. This means that the sum of the reversed number and 396 is the original number.
⇒(5+d) 5 (5−d) + 3 9 6 (5−d) 5 (5+d)
As we can see, to get the second digit of the original number as 5, 1 should be carried over from the addition of digits at third positions, [(5 – d) + 6]. This means, [(5 – d) + 6] is greater than 10.
If one is carried over in the middle term, the sum will be 15 and again, 1 will be carried over to digits at first positions.