Question

the sum of the digit of a number consisting of three digits is 12 the middle digits is equal to half the sum of the other two if the order of the digits is reverse the number is diminished by 198 find the number ​

Answer 1

Answer: 543

Step-by-step explanation:

Let a be the hundreds digit, b be the tens digit and c be the ones digit

Then a+b+c=12

b=12-a-c -------(1)

b=(a+c)/2 --------(2)

100a+10b+c-198=100c+10b+a

99a-99c = 198

99(a-c) = 198

a-c = 2----(3)

From (1) and (2), 12-a-c=(a+c)/2

24-2a-2c=a+c

3a+3c=24

a+c=8-----(4)

(3) + (4) --->

2a = 10

a = 5

(4)-(3)

2c=6

c=3

From (2), b=(5+3)/2 = 4

The number is 543

Answer 2

Answer:

543

Step-by-step explanation:

Let the digits be x, y and z respectively.

So, the number would be 100x + 10y + z.

and the reverse would be 100z + 10y + x

Given sum of digits = x + y + z = 12

Also, Given that, middle digit is equal to half the sum of other two.

⇒ y = (x + z)/2

So, the number becomes

100x + 10y + z

⇒ 100x + 10(x + z)/2 + z

⇒ 100x + 5(x + z) + z

⇒ 100x + 5x + 5z + z

⇒ 105x + 6z

Now, the reverse of the number would be

100z + 10y + x

⇒ 100z + 10(x + z)/2 + x

⇒ 100z + 5x + 5z + x

⇒ 105z + 6x

Now, given that reverse is diminished by 198

⇒ 105x + 6z = 105z + 6x + 198

⇒ 105x - 6x + 6z - 105z = 198

⇒ 99x - 99z = 198

⇒ 99(x - z) = 198

⇒ x - z = 198/99

⇒ x - z = 2

Also, given that x + y + z = 12

but, y = (x + z)/2

⇒ x + (x + z)/2 + z = 12

⇒ 2x/2 + (x + z)/2 + 2z/2 = 12

⇒ (2x + x + z + 2z)/2 = 12

⇒ (3x + 3z)/2 = 12

⇒ 3x + 3z = 12 × 2

⇒  3(x + z) = 24

⇒ x + z = 24/3

⇒ x + z = 8

And, x - z = 2

so, adding the two equations, we get

2x = 10

⇒ x = 5

and, x + z = 8

⇒ 5 + z = 8

⇒ z = 8 - 5

⇒ z = 3

And, y = (x + z)/2

⇒ y = 8/2

⇒ y = 4

So, the number is 100x + 10y + z

⇒ 100(5) + 10(4) + 3

⇒ 500 + 40 + 3

⇒ 543