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 PLACES. FIND THE NUMBER. (QUADRATIC EQUATION IN ONE VARIABLE)

GUYS PLEASE IF POSSIBLE THEN JUST SEND ME THE PICTURE.

Answer 1

GIVEN :

The product of two digit positive number is such that the product of its digits is 6.

Let us assume that the number is 10x + y

The product of the number = xy = 6 ---(1)

If 9 is added to the number then the digits interchange their values.

10x + y + 9 = 10y + x

10x - x + 9 = 10y - y

9x + 9 = 9y

9x - 9y = -9

x - y = -1

x = -1 + y ----(2)

Substitute x in eq - (1)

(-1 + y )y = 6

-y + y² = 6

y² - y - 6 = 0

y² - 3y - 2y - 6 = 0

y(y - 3) - 2( y - 3 ) = 0

y - 3 = 0 ; y - 2 = 0

y = 3 ; y = 2

Here, we can take either 3 or 2.

Substitute 3 in eq - 1

xy = 6

x(3) = 6

3x = 6

x = 6/3

x = 2

There, the number is 10x + y = 10(2) + 3 = 23.

Answer 2

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

Let ten's digit be M

and One's digit be N.

Product of digits = MN

A.T.Q.

Product of digits = 6

MN = 6

M = $\dfrac{6}{N}$ ________(eq 1)

______________________________

☞ If 9 is added to the number the digit interchange their places.

Original number = 10M + N

On interchanging the digits the number is 10N + M

=> 10M + N + 9 = 10N + M

=> 10M - M + N - 10N = - 9

=> 9M - 9N = - 9

=> M - N = - 1 __________(eq 2)

_____________________________

• Put value of (eq 1) in (eq 2)

=> $\dfrac{6}{N}$ - N = - 1

=> $\dfrac{6 \: - \: {N}^{2} }{N}$ = - 1

=> 6 - N² = - N

=> -N² + N + 6 = 0

=> N² - N - 6 = 0

=> N² - 3N + 2N - 6 = 0

=> N(N - 3) +2(N - 3) = 0

=> (N + 2) (N - 3) = 0

=> N + 2 = 0

=> N = -2 (neglected)

Also

=> N - 3 = 0

=> N = +3 (neglect)

Put value of N = +3 in (eq 1)

M = $\dfrac{6}{3}$

M = +2

________________________________

☞ We have to find the number.

Original Number = 10N + M

=> 10(2) + 3

=> 23

_____________________[ANSWER]