Question

a three digit number is given such that the sum of its digits is 9 and the digits are an AP. the number formed by reversing the digit is 198 greater than the original number. find the original number. I will mark it brainliest. pl ans​

Answer 1

Answer:

Original number = 234

Step-by-step explanation:

Given :-

A three digit number is given such that the sum of its digits is 9 and the digits are an AP. the number formed by reversing the digit is 198 greater than the original number.

To find:-

Find the original number?

Solution :-

Let the digits in the three digit number be

a-d, a,a+d

Since they are in AP (Given )

Let the digit at ones place be a-d

Let the digit at tens place be a

Let the digit at hundreds place be a+d

Then the number = 100(a+d)+10a+(a-d)

=> 100a+100d+10a+a-d

=> 111a+99d -----------(1)

If the digits are reversed then the new number

=> 100(a-d)+10a+(a+d)

=> 100a-100d+10a+a+d

=> 111a-99d ---------(2)

Given that

The sum of the digits = 9

=> a-d+a+a+d = 9

=> 3a = 9

=> a = 9/3

=> a = 3 ------------(3)

Given that

The number formed by reversing the digit is 198 greater than the original number.

=> 111a-99d = 111a+99d+198

=> 111a-99d-111a-99d = 198

=> -99d-99d = 198

=> -198d = 198

=> d = 198/-198

=>d = -1

We have ,

a = 3 and d = -1 then the digits are

a-d = 3-(-1) = 3+1 = 4

a+d = 3+(-1) = 3-1 = 2

The digits are 4,3,2

Then the three digit number

= 111a+99d

=> 111(3)+99(-1)

=> 333-99

=> 234

Answer:-

The Original number for the given problem is 234

Check :-

The number = 234

The digits 2,3,4 are in the AP

Since 3-2 = 1 and 4-3 = 1

Common difference is same throughout the series

Sum of the digits = 2+3+4 = 9

Number formed by reversing the digits = 432

432 = 234+198

New number = Original number+198

Verified the given relations in the given problem.

Used formulae:-

• The general form of an AP , a,a+d,a+2d,...

• For solving AP related problems we take 3 terms as a-d,a,a+d