Question

1. The sum of the digits of a three-digit number is 21. If the digits are reversed, the new
number is 198 greater than the original. The sum of the first and third digits is one
more than three times the second digit. Find the number.​

Answer 1

Let the digits at hundred's,ten's and one's place be x,y and z respectively.Then,

Original Number=100x+10y+z.

Sum of the digits is 21.

⇒x+y+z=21 ...(1)

Reversing the digits,

New number=100z+10y+x.

Given that the new number is greater than the original by 198.

⇒100z+10y+x=100x+10y+z+198

⇒100z+x=100x+z+198

⇒99z-99x=198

⇒99(z-x)=198

⇒z-x=2 ...(2)

The sum of first and third digit is one more than thrice the second digit.

⇒x+z=3y+1 ...(3)

Substituting eq.(3) in eq.(1), we get

3y+1+y=21

⇒4y=20

⇒y=5

Adding eq.(2) and eq.(3)

z-x+x+z=2+3y+1

⇒2z=3+3y

Putting y=5,

2z=3+3(5)=18

⇒z=9

Putting z=9 in eq.(2), we get

9-x=2

⇒x=9-2=7

The original number=100x+10y+z

=100(7)+10(5)+9

=759.

Answer 2

Answer:

original number is 759

Step-by-step explanation:

let the original number be A(100x+10y+z) ----eqn(1)

the number formed by reversing be B(100z+10y+x) =198+A ---eqn(2)

given,

x+y+z=21 ---eqn(3)

x+z=1+3y ---eqn(4)

therefore,

4 in 3

1+3y+y=21

4y=20 ===> y=5

sub y value in eqn 2,

100z+50+x=198+100x+50+z

99x-99z=198

x-z=2 ---eqn(5) (dividing both sides by 99)

sub y value in eqn (4)

x+z=16 ---(6)

solve eqn (5)&(6)

x=9 and z= 7

the num B is 957 and A is 759