Question

(ii) 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
extreme digits is 1 less than the middle digit. Find the original number.​

Answer 1

AnswEr :

• Three Digit Number is Equal to 17 times of sum of the digits.
• If we reverse the digits then the new number is 198 more than the original number.
• Sum of Extreme Digits is 1 less than the Middle Digit of number.
• Find the Original Number.

Let's take the Numbers be a, b and c. Original Number will be (100a + 10b + c).

• First Part of the Question :

⇒ Number = 17 times of Sum of Digits

⇒ (100a + 10b + c) = 17 × (a + b + c)

⇒ 100a + 10b + c = 17a + 17b + 17c

⇒ 100a - 17a + 10b - 17b + c - 17c = 0

⇒ 83a - 7b - 16c = 0⠀⠀⠀⠀⠀⠀—eq. ( I )

• Second Part of the Question :

⇒ Reverse No. = Original No. + 198

⇒ (100c + 10b + a) = (100a + 10b + c) + 198

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

⇒ 99c - 99a = 198

⇒ 99(c - a) = 198

• Dividing Both term by 99

⇒ c - a = 2

⇒ c = 2 + a⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq. ( II )

• Third Part of the Question :

⇒ Sum of Extreme Digits = Middle Digit - 1

⇒ a + c = b - 1

• putting the Value of c from eq.( II )

⇒ a + 2 + a = b - 1

⇒ 2a + 2 = b - 1

⇒ b = 2a + 2 + 1

⇒ b = 2a + 3⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq. ( III )

_________________________________

• Putting the value from eq.( I ) & ( III ) :

⇒ 83a - 7b - 16c = 0

⇒ 83a - 7(2a + 3) - 16(2 + a) = 0

⇒ 83a - 14a - 21 - 32 - 16a = 0

⇒ 83a - 30a - 53 = 0

⇒ 53a = 53

• Dividing Both term by 53

⇒ a = 1

◗ b = (2a + 3) = 2(1) + 3 = 2 + 3 = 5

◗ c = (2 + a) = 2 + 1 = 3

→ Original Number will be

→ (100a + 10b + c)

→ (100 × 1 + 10 × 5 + 3)

→ (100 + 50 + 3)

→ 153

∴ Therefore, Original Number will be 153.

Answer 2

$\bold{\underline{\underline{\huge{\rm{AnsWer:}}}}}$

Original Number : 153

$\bold{\underline{\underline{\large{\rm{StEp\:by\:stEp\:explanation:}}}}}$

GiVeN :

• 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 extreme digits is 1 less than the middle digit.

To FiNd :

• The original number

SoLuTioN :

Let the digit at the hundreds place be x.

Let the digit at the units place be y.

$\sf{Given\:that\:-The\:sum\:of\:the\:extreme\:digits\:is\:1\:less\:than\:the\:middle\:digit}$

$\sf{\therefore{Middle\:digit\:=\:x\:+\:y\:+\:1\:(tens\:place)}}$

Original Number :

$\hookrightarrow$$\tt{100x\:+\:10(x\:+\:y\:+\:1)\:+\:y}$

$\hookrightarrow$ $\tt{100x\:+\:10x\:+\:10y\:+\:10\:+\:y}$

$\hookrightarrow$ $\tt{110x\:+\:11y\:+\:10}$

Reversed Number :

$\hookrightarrow$$\tt{100y\:+\:10(x\:+\:y\:+\:1)\:+x}$

$\hookrightarrow$ $\tt{100y\:+\:10x\:+\:10y\:+\:10\:+\:x}$

$\hookrightarrow$ $\tt{110y\:+\:11x\:+\:10}$

$\sf{\underline{\underline{As\:per\:the\:first\:condition:}}}$

• A three-digit number is equal to 17 times the sum of the digits.

$\hookrightarrow$$\tt{110x\:+\:11y\:+\:10\:=\:17\:(x\:+\:x\:+\:y\:+\:1\:+\:y\:)}$

$\hookrightarrow$ $\tt{110x\:+\:11y\:+\:10\:=\:17\:(2x\:+\:2y\:+\:1)}$

$\hookrightarrow$ $\tt{110x\:+\:11y\:+\:10\:=\:34x\:+\:34y\:+\:17}$

$\hookrightarrow$ $\tt{110x\:-\:34x\:+\:11y\:-\:34y\:=\:17\:-\:10}$

$\hookrightarrow$ $\tt{76x\:-\:23y\:=\:7}$ ---> (1)

$\sf{\underline{\underline{As\:per\:the\:second\:condition:}}}$

• If the digits are reversed the new number is 198 more than the original number.

$\hookrightarrow$ $\tt{110x\:+\:11y\:+\:10\:+\:198\:=\:110y\:+\:11x\:+\:10}$

$\hookrightarrow$ $\tt{110x\:-\:11x\:+\:11y\:-\:110y\:=\:-\:208\:+\:10}$

$\hookrightarrow$ $\tt{99x\:-\:99y\:=\:-198}$

$\hookrightarrow$ $\tt{99(x-y)\:=\:-198}$

$\hookrightarrow$ $\tt{x-y=\:{\dfrac{-198}{99}}}$

$\hookrightarrow$$\tt{x\:-\:y\:=\:-2}$ ---> (2)

Multiply equation (2) by 23,

$\hookrightarrow$$\tt{23x\:-\:23y\:=\:-46}$ ---> (3)

Solve equation 1 and equation 3 simultaneously by elimination method.

Subtract equation (3) from equation (1),

$\tt{76x\:-\:23y\:-(\:23x\:-\:23y\:)=\:7\:-\:(-46)}$

$\tt{76x\:-\:23y\:-\:23\:+\:23y\:=\:7\:-\:(-\:46)}$

$\tt{76x\:-\:23y\:-\:23x\:+\:23y\:=\:7\:+\:46}$

$\tt{53x=\:53}$

$\hookrightarrow$ $\tt{x\:=\:{\dfrac{53}{53}}}$

$\hookrightarrow$$\tt{x=1}$

Substituting x = 1 in equation (2),

$\hookrightarrow$$\tt{x\:-\:y\:=\:-\:2}$

$\hookrightarrow$$\tt{1\:-\:y\:=\:-2}$

$\hookrightarrow$ $\tt{-y\:=\:-2\:-1}$

$\hookrightarrow$ $\tt{-y\:=\:-3}$

$\hookrightarrow$ $\tt{y\:=3}$

Original Number :

$\hookrightarrow$ $\tt{110x\:+\:11y\:+\:10}$

$\hookrightarrow$ $\tt{110(1)\:+\:11(3)\:+\:10}$

$\hookrightarrow$ $\tt{110\:+\:33\:+\:10}$

$\hookrightarrow$ $\tt{120\:+\:33}$

$\hookrightarrow$ $\tt{153}$

$\sf{\therefore{Original\:Number\:=\:153}}$