Question

If we divide a two-digit number by a number consisting of the same digits written in the reverse order, we get 4 as a quotient and 15 as a remainder; now if we subtract 9 from that given number we get the sum of the squares of the digits constituting that number. Find that number.

Answer 1

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Answer 2

Solution :-

Let us Assume that, the Given two digit number is (10x + y).

Than,

→ Reverse of the given Number is = (10y + x) .

Now, we have given that, if we divide given number by its reverse number we get 4 as Quotient and 15 as remainder .

So,

→ Given Number = 4(Reverse Number) + 15

→ 10x + y = 4(10y + x) + 15

→ 10x + y = 40y + 4x + 15

→ 10x - 4x + y - 40y = 15

→ 6x - 39y = 15

→ 3(2x - 13y) = 15

→ (2x - 13y) = 5

→ 2x = 5 + 13y

→ x = (5 + 13y)/2 -------------------- Eqn.(1)

We also have given that, if we subtract 9 from that given number we get the sum of the squares of the digits constituting that number.

So,

→ (10x + y) - 9 = x² + y²

Putting value of x From Eqn.(1) Now,

→ 10{(5 + 13y)/2} + y - 9 = {(5 + 13y)/2}² + y²

→ 5(5 + 13y) + y - 9 = (25 + 169y + 130y + 4y²)/4

→ 4(25 + 66y - 9) = 4y² + 299y + 25

→ 64 + 269y = 4y² + 299y + 25

→ 4y² + 299y - 264y + 25 - 64 = 0

→ 4y² + 35y - 39 = 0

→ 4y² - 4y + 39y - 39 = 0

→ 4y(y - 1) + 39(y - 1) = 0

→ (y - 1)(4y + 39) = 0

→ y = 1 or, (-39/4). { Negative value ≠ } .

Putting value of y = 1 in Eqn.(1) Now, we get,

→ x = (5 + 13*1)/2

→ x = (5 + 13)/2

→ x = 18/2

→ x = 9 .

Therefore,

→ Given Number = (10x + y) = (10*9 + 1) = 90 + 1 = 91 . (Ans.)

Hence, The Given Number is 91.