Question

A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.



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Answer 1

let two digit no be XY

X× Y =14

XY + 45 = YX

10 X + Y + 45 = 10 Y + X

9X - 9 Y + 45 = 0

X - Y + 5 = 0

also X× Y = 14

X = 14/Y

14/Y - Y + 5 = 0

14 - Y.Y + 5 Y = 0

-Y.Y -2 Y + 7 Y + 14 = 0

- Y( Y +2) +7( Y+2) = 0

Y = 7 , -2

SO TAKE 7

X× Y = 14
X = 2


So no is 27

Answer 2

$\huge \bf \pink{Hey \: there \: !! }$

Let the ten's digit of the required number be x.

And, the unit's digit be y .

Then, xy = 14 .

Required number = ( 10x + y ) .

Number obtained on reversing its digit = ( 10y + x ) .

•°• ( 10x + y ) + 45 = ( 10y + x ) .

=> 10y - y + x - 10x = 45 .

=> 9y - 9x = 45 .

=> 9( y - x ) = 45 .

=> y - x = 45/9 .

=> y - x = 5 ..............(1).

▶ Now, using Identity :-)

→ ( y + x )² = ( y - x )² + 4xy .

$= > (y + x) = \sqrt{ {(y - x)}^{2} + 4xy} . \\ \\ = > y + x = \sqrt{ {5}^{2} + 4 \times 14 } . \\ \\ = > y + x = \sqrt{25 + 56} . \\ \\ = > y + x = \sqrt{81} . \\ \\ = > y + x = 9............(2).$

On adding equation (1) and (2), we get

y - x = 5 .

y + x = 9.

+ + +

_________

=> 2y = 14 .

=> y = 14/2 .

•°• y = 7 .

Putting y = 7 in equation (2), we get

=> 7 + x = 9.

=> x = 9 - 7 .

•°• x = 2 .

Therefore, the required number = 10x + y .

= 10 × 2 + 7 .

= 20 + 7 .

$\huge \red{ \boxed{ = 27 . }}$

✔✔ Hence, it is solved ✅✅.

$\huge \green{ \boxed { \mathcal{THANKS}}}$

$\huge \bf \blue{ \#BeBrainly.}$