Question

the sum of three digits number is 12 and the digits are in AP if digit are reserved then the number is the​

Answer 1

Answer:

I hope 642 is correct answer

Answer 2

The original number for the given condition is 642  .

Step-by-step explanation:

Given as :

The sum of three digits numbers = 12

The digits are in Arithmetic progression

Let The digits in A.P  are  a - d ,  a , a + d

So, Original number = 100 (a - d) + 10 a + 1 (a + d)

Or, Original number = 100 (a - d) + 10 a +  (a + d)

According to question

The sum of digits = 12

i.e a - d +  a + a + d = 12

Or,  3 a - 0 = 12

Or,        3 a = 12

∴              a = $\dfrac{12}{3}$

i.e            a = 4                    .........1

Again

if the digits are reversed, then the number is diminished by 396

Now, Reserved number =  100 (a + d) + 10 a +  (a - d)

So,  100 (a + d) + 10 a +  (a - d) = [ 100 (a - d) + 10 a +  (a + d) ] - 396

Rearranging the equation

     100 (a + d) + 10 a +  (a - d) - [ 100 (a - d) + 10 a +  (a + d) ]  = - 396

     (a + d) (100 - 1) + (10 a - 10 a) + (a - d) (1 - 100 ) = - 396

Or,  99 (a + d) + 0 - 99  (a - d) = - 396

Or,                99 (a + d - a + d) = - 396

Or,                              99 (2 d) = - 396

Or,                                         d = $\dfrac{-396}{198}$

i.e                                          d  = - 2                         ............2

Now, From eq 1 and eq 2

∵ Original number =  100 (a - d) + 10 a + 1 × (a + d)

                           =  100 (4 - (- 2) ) + 10 × 4 + 1 × (4 + (-2) )

                           = 100 (4 + 2) +  10 × 4 + 1 × (4 - 2)

                           = 100 × 6 +  10 × 4 + 1 × 2

                           = 600 + 40 + 2

                           = 642

So, The original number = 642

Hence, The original number for the given condition is 642  . Answer