Question

A three digit number abc is 459 more than the sum of its digits.What is the sum of the 2 digit number ab and the 1digit number a?

Answer 1

✵Question :

A three digit number abc is 459 more than the sum of its digits.What is the sum of the 2 digit number ab and the 1digit number a?

✵Answer :

Given, abc is an three digit number

☛Then, three digit no. is ,

$\large{ \sf{ = 100a + 10b + c}}$

✵According to question:

☛ The three digit number is greater than

sum of digits by 459, That means

$⇢ \: \large{ \sf{ 100a + 10b + c = a + b + c + 459}} \: \\ \sf{ \underline{ \underline{ \star{ \green{on \: solving \: this \: equation}}}}} \\ \\ ⇢ \large{ \sf{100a + 10b + \cancel{c} = a + b + \cancel{ c }+ 459}} \\ \\ ⇢ \large{ \sf{ 99a + 9b = 459}} \\ \\ ⇢ \large{ \sf{9(11a + b) = 459}} \\ \\ ⇢ \large{ \sf{11a + b = \cancel{ \frac{459}{9} }}} \\ \\ ⇢ \large{ \sf{ \boxed{ \purple{11a + b = 51}}}} \:$

☛ Here, we got an equation 11a+b=51

✵To Find:

⦿Sum of two digit no. ab and one digit no. a

☛Two digit number ab= 10a+b

☛And, one digit number a= a

$⇢ \large{ \sf{ 10a + b + a}} \\ \\⇢ \large{ \sf{11a + b}} \\ \\ ⇢\large{ \sf{ \boxed{ \purple{51}}}}$

✵So final answer:

$\huge{ \boxed{ \bold{ \red{ = 51}}}}$