Question

a two digit number is such that the product of the digits is 20. If 9 is added to the number the digits interchange their places. find the number

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

Unit's Place be b

Also,

Ten's Place be a

Number = (10a + b)

Situation :-

⇒ Original Number + 9 = Interchange Number

⇒ (10a + b) + 9 = (10b + a)

⇒ 9 = 10b + a - 10a - b

⇒ 9 = 9b - 9a

⇒ 9 = 9(b - a)

⇒ 1 = b - a

⇒ b = a + 1 ............... (1)

Product of Digits :

⇒ ab = 20 ................. (2)

$\Large{\boxed{\sf\:{Substitute\;the\;value\;of\;b\;in\; (2)}}}$

⇒ a(a + 1) = 20

⇒ a² + a = 20

⇒ a² + a - 20 = 0

⇒ a² + 5a - 4a – 20 = 0

⇒ a(a + 5) - 4(a + 5) = 0

⇒ (a - 4)(a + 5) = 0

⇒ a = 4 and, a = - 5

a = 4, as per it is a positive integer

$\Large{\boxed{\sf\:{Substitute\;the\;value\;of\;a\;in\; (2)}}}$

⇒ ab = 20

⇒ 4 × b = 20

$\tt{\rightarrow b=\dfrac{20}{4}}$

⇒ b = 5

New Number :-

⇒ Number = (10a + b)

⇒ Number = 10(4) + 5

⇒ Number = 40 + 5

⇒ Number = 45

Hence,

$\Large{\boxed{\sf\:{Number=45}}}$