Question

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

$\large\underline\mathfrak\red{Solution \: :- }$

Let the three digit number be xyz.

Its numerical value = 100x + 10y + z

According to first information provided in the question,

$</p><p> \sf{100x + 10y + z = 17 ( x + y + z ) }$

$\sf\therefore{100x + 10y + z = 17x + 17y + 17z }$

$\sf\therefore{87x - 7y - 16z = 0 \: \: \: ...(1) }$

Number obtained by reversing the digits : zyx

Its numerical value = 100z + 10y + x

According to second information provided in the question,

$\sf{(100x + 10y + z) + 198 = 100z + 10y + x}$

$\sf\therefore{99z - 99x = 198}$

$\sf\therefore{z - x = 2} \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\therefore{z = x + 2 \: \: \: ...(2)}$

According to third information provided in the question,

$\sf{x + y = y - 1}$

$\sf\therefore{x + x + 2 = y - 1 \: \: ...(from(2))}$

$\sf\therefore{y = 2x + 3}$

Substituting the value of z and y in equation 1

$\sf{83x - 7(2x + 3) - 16(x + 2) = 0}$

$\sf\therefore{83x - 14x - 21 - 16x - 32 = 0}$

$\sf\therefore{53x - 53 = 0}$

$\sf\therefore{x = 1}$

$\sf\therefore{y = 2x + 3 = 2(1) + 3 = 2 + 3 = 5}$

$\sf\therefore{z = x + 2 = 1 + 2 = 3}$

x = 1

y = 5

z = 3

Thus, the three digit number is 153.

Answer 2

Solution:−

Let the three digit number be xyz.

Its numerical value = 100x + 10y + z

According to first information provided in the question,

< /p > < p > \sf{100x + 10y + z = 17 ( x + y + z ) }</p><p>100x+10y+z=17(x+y+z)

\sf\therefore{100x + 10y + z = 17x + 17y + 17z }∴100x+10y+z=17x+17y+17z

\sf\therefore{87x - 7y - 16z = 0 \: \: \: ...(1) }∴87x−7y−16z=0...(1)

Number obtained by reversing the digits : zyx

Its numerical value = 100z + 10y + x

According to second information provided in the question,

\sf{(100x + 10y + z) + 198 = 100z + 10y + x}(100x+10y+z)+198=100z+10y+x

\sf\therefore{99z - 99x = 198}∴99z−99x=198

\begin{gathered}\sf\therefore{z - x = 2} \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\therefore{z = x + 2 \: \: \: ...(2)}\end{gathered}

∴z−x=2

∴z=x+2...(2)

According to third information provided in the question,

\sf{x + y = y - 1}x+y=y−1

\sf\therefore{x + x + 2 = y - 1 \: \: ...(from(2))}∴x+x+2=y−1...(from(2))

\sf\therefore{y = 2x + 3}∴y=2x+3

Substituting the value of z and y in equation 1

\sf{83x - 7(2x + 3) - 16(x + 2) = 0}83x−7(2x+3)−16(x+2)=0

\sf\therefore{83x - 14x - 21 - 16x - 32 = 0}∴83x−14x−21−16x−32=0

\sf\therefore{53x - 53 = 0}∴53x−53=0

\sf\therefore{x = 1}∴x=1

\sf\therefore{y = 2x + 3 = 2(1) + 3 = 2 + 3 = 5}∴y=2x+3=2(1)+3=2+3=5

\sf\therefore{z = x + 2 = 1 + 2 = 3}∴z=x+2=1+2=3

x = 1

y = 5

z = 3

Thus, the three digit number is 153