Question

A two digit positive number is such that the product of its digit is 6. If 9 is added to the number, the digits interchange their place. Fing the number to be brilliant...​

Answer 1

Answer:

23

Step-by-step explanation:

Let us consider the two-digit numbers as 'xy'.

Let the digit in ten's place be x and the digit in unit's place be y, So the decimal expansion is 10x + y.

(i)

Product of its digit is 6.

xy = 6

y = (6/x)

(ii)

⇒ 10x + y + 9 = 10y + x

⇒ 9x - 9y = -9

⇒ x - y = -1.

⇒ x - (6/x) = -1

⇒ x² - 6 = -x

⇒ x² + x - 6 = 0

⇒ x² + 3x - 2x - 6 = 0

⇒ x(x + 3) - 2(x + 3) = 0

⇒ (x - 2)(x + 3) = 0

⇒ x = 2, -3{Since, x≠-3)

⇒ x = 2.

Substitute x = 2 in (i), we get

⇒ y = (6/x)

⇒ y = 3.

Now,

⇒ 10x + y = 10(2) + 3

                = 23.

Therefore, the number is 23.

Hope it helps!

Answer 2

Answer: 23

Step-by-step explanation:

Let the no. be 10x + y.

So the no. formed by interchanging the digits is 10y + x.

Here the digits interchange their place when 9 is 'added'. So x < y.

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TO REMEMBER...

The difference between a two-digit number and the no. formed by interchanging the digits is 9 multiplied by the difference of the digits.

$\because\ (10x+y)-(10y+x)=10x+y-10y-x=9x-9y=9(x-y)$

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But here, as x < y, so 9(y - x)=9

Therefore,

$9(y-x)=9 \\ \\ y-x=1=\ Let \ m \ \ \ \ \ \longrightarrow \ \ \ \ \ (1)$

Given that,

$xy=6=\ Let \ n$

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TO REMEMBER...

         

If $y-x=m$

and $xy=n$,

then $y+x=\sqrt{m^2+4n}$

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Here,

$y+x=\sqrt{m^2+4n} \\ \\ y+x=\sqrt{1^2+4\times6} \\ \\ y+x=\sqrt{1+24} \\ \\ y+x=\sqrt{25} \\ \\ y+x=5 \ \ \ \ \ \longrightarrow \ \ \ \ \ (2)$

$(2)+(1) \\ \\ (y+x)+(y-x)=5+1 \\ \\ 2y=6\\ \\ y=\bold{3} \\ \\ \\ (2)-(1) \\ \\ (y+x)-(y-x)=5-1 \\ \\ 2x=4 \\ \\ x=\bold{2} \\ \\ \\ \therefore\ 10x+y=10 \times 2 + 3 = \bold{23}$

So 23 is the answer.

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Hope this may be helpful.

Thank you. Have a nice day. :-))

#adithyasajeevan