Question

A 3 digit number is given such that sum of its digit is 9 and the digits are in a.p.the number formed by reversing the digits is 198greater than the original number. find the original number​

Answer 1

Step-by-step explanation:

Let the digit at the hundredth place of the number be (a + d).

Digit at the tens place be a.

Digit at the ones place be (a – d).

Sum of the digits = 15

(a + d) + a + (a – d) = 15

3a = 15

⇒ a = 5

The number formed by the digits

= 100 (5 + d) + 10 (5) + (5 – d)

= 555 + 99d

The number formed by reversing the digits

= 100 (5 - d) + 10 (5) + (5 + d)

= 555 – 99d

Given that the number formed by reversing the digits is 594 less

the original number.

⇒ (555 + 99d) – (555 - 99d) = 594

⇒ 198 d = 594

⇒ d = 3

(a + d) = (5 + 3) = 8,

a = 5,

(a – d) = (5 – 3) = 2.

Hence, the number formed by the digits = 100(8) + 10(5) + 1(2) = 852.

Answer 2

Step-by-step explanation:

Given that -

sum of digits are =9

To find - the original digit

solution :

Lets the digits be -

Hunderth - a+d

tens - a

ones be - a-d

The original digit be -

100(a+d) +10a + 1(a-d)=

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

111a + 99d

The reversed digit be -

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

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

111a - 99d

According to question -

reversed - original = 198

then,

(111a - 99d) - (111a + 99d) = 198

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

-198d = 198

d = 198/-198

d = -1

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

Tens digit = a = 3

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

The number will be = 234