Question

Let A be a two-digit number and B be another two-
digit number formed by reversing the digits of A.
If A + B + (Product of digits of the number A) =
145, then what is the sum of the digits of A?
(a) 12
(b) 10
(C) 13
(d) 11​

Answer 1

Given :- Let A be a two-digit number and B be another two- digit number formed by reversing the digits of A.

If A + B + (Product of digits of the number A) = 145, then what is the sum of the digits of A ?

(a) 12

(b) 10

(C) 13

(d) 11

Solution :-

Let A = 10x + y = Two digit number .

so,

B = 10y + x = Reverse .

A/q,

→ A + B + (Product of digits of the number A) = 145

→ (10x + y) + (10y + x) + xy = 145

→ 11x + 11y + xy = 145

→ 11(x + y) = (145 - xy)

→ (x + y) = (145 - xy)/11

since , for values of x and y , RHS must be divisible by 11 .

→ near to 145 , 143 is divisible by 11 . so, xy must be 2 . then, (x = 1 , y = 2) or vice versa .

• (145 - 2)/11 = 143/11 = 13
• But 1 + 2 ≠ 13

now,

→ 132 ÷ 11 = 12 = (x + y)

→ xy = 145 - 132 = 13 => x,y = (1,13) , (13,1)

But,

• 1 + 13 ≠ 12 .

again,

→ 121 ÷ 11 = 11 = (x + y)

→ xy = 145 - 121 = 24 => x,y = (1,24), (2,12) , (3,8) , (4,6)

out of these ,

• 1 + 24 ≠ 11
• 2 + 12 ≠ 11
• 3 + 8 = 11
• 4 + 6 ≠ 11 .

hence,

→ sum of digits of A = 3 + 8 = 11 (Ans.D) .

Note :- A will be 38 or 83 .

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