Question

(11) A three-digit number is equal to 17 times the sum of the digits. If the digits are
reversed the new number is 198 more than the original number. The sum of the
extreme digits is 1 less than the middle digit. Find the original number.​

Answer 1

Answer:

$\large\boxed{\sf{153}}$

Step-by-step explanation:

Let the digits at ones, tens and hundred places be "x" , "y" and "z" respectively.

.°. Original number = (100z + 10y + x)

Now, according to question,

=> 100z + 10y + x = 17(x + y + z)

=> 100z + 10y + x = 17x + 17y + 17z

=> 83z = 7y + 16x ..........(1)

Now, when the digits are reversed.

New Number = (100x + 10y + z)

According to Question,

=> 100x + 10y + z = 100z + 10y + x + 198

=> 99x = 99z + 198

=> x = z + 2 ...........(2)

Also, from the Question,

=> x + z = y - 1 ..........(3)

=> z + 2 + z = y - 1

=> 2z + 2 = y - 1

=> y = 2z + 3 ..........(4)

Now in eqn (1),

=> 83z = 7(2z + 3) + 16(z + 2)

=> 83z = 14z + 21 + 16 z +32

=> 83z = 30z + 53

=> 53z = 53

=> z = 1

=> y = 2×1 + 3

=> y = 2+3

=> y = 5

=> x = 1 +2

=> x = 3

.°. Original Number = (100+50+3) = 153