Question

two digit number such that the product of it's digits is 6 if 9 is added the number the digits interchange their places find the number​

Answer 1

Answer :

$\underline{\boxed{\sf{Number\:=\:23}}}$

Step-by-step Explanation :

Let the ten's digit be x and one's digit be y.

Therefore, Number = 10x + y

Two digit number such that the product of it's digits is 6.

According to question,

$\implies\:\sf{x\:\times\:y\:=\:6}$

$\implies\:\sf{xy\:=\:6}$

$\implies\:\sf{x\:=\:\dfrac{6}{y}}$

If 9 is added the number the digits interchange their places.

According to question,

$\implies\:\sf{10x\:+\:y\:=\:10y\:+\:x\:-\:9}$

$\implies\:\sf{10x\:-\:x\:+\:y\:-\:10y\:=\:-\:9}$

$\implies\:\sf{9x\:-\:9y\:=\:-\:1}$

$\implies\:\sf{x\:-\:y\:=\:-\:1}$

$\implies\:\sf{\frac{6}{y}\:-\:y\:=\:-1}$

$\implies\:\sf{6\:-\:y^2\:=\:-y}$

$\implies\:\sf{y^2\:-\:y\:-\:6\:=\:0}$

The above equation is in the form ax² + bx + c = 0

Now, solve it by splitting the middle term

$\implies\:\sf{y^2\:-\:3y\:+\:2y\:-\:6\:=\:0}$

$\implies\:\sf{y(y-3)\:+2(y-3)\:=\:0}$

$\implies\:\sf{(y+2)(y-3)\:=\:0}$

$\implies\:\sf{y\:=\:-2,\:+3}$

(-2 neglected as number can't be negative)

Substitute value of y in x

$\implies\:\sf{x\:=\:\dfrac{6}{3}}$

$\implies\:\sf{x\:=\:2}$

Therefore,

$\sf{\underline{Number \:= \:10x\:+\:y}}$

$\Rightarrow\:\sf{10(2)\:+\:3}$

$\Rightarrow\:\sf{20\:+\:3}$

$\Rightarrow\:\sf{23}$

Answer 2

Question :--- two digit number such that the product of it's digits is 6 if 9 is added the number the digits interchange their places find the number ?

Solution :---

Let the Given Two digits be = 10x+y, where unit digit is y and ten's digit is x .

Now, it has been said that , product of both the digits is 6.

So,

→ x * y = 6

→ xy = 6 ------------------ Equation (1)

Now, it is also said that, when we add 9 in the original number its digits gets interchanged.

So,

→ After interchanging digits we get, = (10y+x)

A/q,

→ (10x+y) + 9 = (10y+x)

→ 10x - x + y - 10y = (-9)

→ 9x - 9y = (-9)

→ 9(x-y) = (-9)

Dividing both sides by 9 we get,

→ (x-y) = -1 ------------------- Equation (2)

______________________________

Now, here Either we use Hit and trial to Find both numbers as mulitple is a small digit, we can easily say that, x can be 2 and y will be 3. [ 2*3 = 6 and (2-3) = (-1) ] .

But Lets Prove it with Algebra Now,,

→ As we know, (a+b)² = (a-b)² + 4ab

Putting values of Equation (1) and Equation (2) now, we get,

→ (x+y)² = (-1)² + 4*6

→ (x+y)² = 1 + 24

→ (x+y)² = 25

Square - root both sides now,

→ (x+y) = ±5

[ As its a 2 digit number sum cant be a Negative Number].

So,

→ (x+y) = 5 ----------------------- Equation (3) .

__________________________

Now, adding Equation (2) and Equation (3) we get,

→ (x-y) + (x+y) = (-1) + 5

→ 2x = 4

→ x = 2.

Putting this value in Equation (1) now,

→ 2*y = 6

→ y = 3.

_______________________

→ Required Two digits Number = 10x+y = 10*2+3 = 20+3 = 23.

Hence, The Required Two digits Number is 23.

[❂❂ Think Differently , Think Smart. ❂❂]