Question

A two digit number is obtained by either multiplying sum of digits by 8 and then adding 1 to it or by multiplying the difference of digits by 13 and adding 2. Find the number.

Answer 1

$\Large{\underline{\underline{\bf{Given:-}}}}$

• A two digit number is obtained by multiplying sum of digits by 8 and then adding 1 to it.
• A two digit number is obtained by multiplying the difference of digits by 13 and adding 2.

$\Large{\underline{\underline{\bf{To \: Find:-}}}}$

• What is the original number ?

$\Large{\underline{\underline{\bf{Solution:-}}}}$

Let the digit at ten's place be "x" and the digit at one's place be "y"

• NUMBER = 10x + y

CASE:- 1)

✧ A two digit number is obtained by either multiplying sum of digits by 8 and then adding 1 to it.

• Sum of digits = x + y

${\underline{\bf{According \: To \: Question:-}}}$

➜ 10x + y = 8(x + y) + 1

➜ 10x + y = 8x + 8y + 1

➜ 10x + y –8x –8y –1 = 0

➜ 2x –7y –1 = 0

➜ 2x = 7y + 1

➜ x = $\bf{\dfrac{7y + 1}{2}....1)}$

CASE:- 2)

✧ A two digit number is obtained by multiplying the difference of digits by 13 and adding 2.

• Difference of digits = x –y

${\underline{\bf{According \: To \: Question:-}}}$

➜ 10x + y = 13(x –y) + 2

➜ 10x + y = 13x –13y + 2

➜ 0 = 13x –13y + 2 –10x –y

➜ 0 = 3x –14y + 2....2)

Put the value of "x" in equation 2)

➜ 0 = 3$\sf{ \bigg( \dfrac{7y + 1}{2} \bigg) }$–14y + 2

➜ 0 = $\sf{\dfrac{21y + 3}{2}}$ –14y + 2

➜ 0 = $\sf{\dfrac{21y + 3 -28y +4}{2}}$

➜ 0 = $\sf{\dfrac{-7y + 7}{2}}$

➜ 0 = –7y + 7

➜ 7y = 7

➜ y = $\sf{\cancel\dfrac{7}{7}}$

➜ ❮⠀y = 1⠀❯

Put the value of "y" in equation 2)

➜ 3x –14 $\times$ 1 + 2 = 0

➜ 3x –14 + 2 = 0

➜ 3x –12 = 0

➜ 3x = 12

➜ x = $\sf{\cancel\dfrac{12}{3}}$

➜ ❮⠀x = 4⠀❯

⁂ NUMBER = 10x + y

⁂ NUMBER = 10(4) + 1

⁂ NUMBER = 40 + 1

⁂ NUMBER = 41

❝ Hence, the number formed is 41 ❞

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Answer 2

Answer:

$\Large{\underline{\underline{\bf{Given:-}}}}$

A two digit number is obtained by multiplying sum of digits by 8 and then adding 1 to it.

A two digit number is obtained by multiplying the difference of digits by 13 and adding 2.

$\Large{\underline{\underline{\bf{To \: Find:-}}}}$

What is the original number ?

$\Large{\underline{\underline{\bf{Solution:-}}}}$

Let the digit at ten's place be "x" and the digit at one's place be "y"

NUMBER = 10x + y

CASE:- 1)

✧ A two digit number is obtained by either multiplying sum of digits by 8 and then adding 1 to it.

Sum of digits = x + y

${\underline{\bf{According \: To \: Question:-}}}$

➜ 10x + y = 8(x + y) + 1

➜ 10x + y = 8x + 8y + 1

➜ 10x + y –8x –8y –1 = 0

➜ 2x –7y –1 = 0

➜ 2x = 7y + 1

➜ x = $\bf{\dfrac{7y + 1}{2}....1)}$

CASE:- 2)

✧ A two digit number is obtained by multiplying the difference of digits by 13 and adding 2.

Difference of digits = x –y

${\underline{\bf{According \: To \: Question:-}}}$

➜ 10x + y = 13(x –y) + 2

➜ 10x + y = 13x –13y + 2

➜ 0 = 13x –13y + 2 –10x –y

➜ 0 = 3x –14y + 2....2)

Put the value of "x" in equation 2)

➜ 0 = 3$\sf{ \bigg( \dfrac{7y + 1}{2} \bigg) }$–14y + 2

➜ 0 = $\sf{\dfrac{21y + 3}{2}}$ –14y + 2

➜ 0 = $\sf{\dfrac{21y + 3 -28y +4}{2}}$

➜ 0 = $\sf{\dfrac{-7y + 7}{2}}$

➜ 0 = –7y + 7

➜ 7y = 7

➜ y = $\sf{\cancel\dfrac{7}{7}}$

➜ ❮⠀y = 1⠀❯

Put the value of "y" in equation 2)

➜ 3x –14 $\times$ 1 + 2 = 0

➜ 3x –14 + 2 = 0

➜ 3x –12 = 0

➜ 3x = 12

➜ x = $\sf{\cancel\dfrac{12}{3}}$

➜ ❮⠀x = 4⠀❯

⁂ NUMBER = 10x + y

⁂ NUMBER = 10(4) + 1

⁂ NUMBER = 40 + 1

⁂ NUMBER = 41

❝ Hence, the number formed is 41 ❞

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