product of digits of a two-digit number is 14 when 45 is added to the number then the digits interchange their places find the number
Answers
Answered by
88
Let the digits of the no. be x and y
So according to the question :
x × y = 14
xy = 14
Let the two digit bo. be 10x + y
10x + y + 45 = 10y + x
10x - x + y - 10y = - 45
9x - 9y = - 45
9 ( x - y ) = -45
x - y = - 5
x = -5 + y
x × y = 14
-5 + y ( y ) = 14
-5y + y^2 = 14
y^2 - 5y - 14 = 0
y^2 - 7y + 2y - 14 = 0
y ( y - 7 ) + 2 ( y - 7 )
( y + 2 ) ( y - 7 ) = 0
y = - 2 ( neglected)
or
y = 7
we cannot take a negative value so we must take y = 7
xy = 14
x × 7 = 14
x = 14 / 7
x = 2
So the required no. is 27
Hope it helps you .
So according to the question :
x × y = 14
xy = 14
Let the two digit bo. be 10x + y
10x + y + 45 = 10y + x
10x - x + y - 10y = - 45
9x - 9y = - 45
9 ( x - y ) = -45
x - y = - 5
x = -5 + y
x × y = 14
-5 + y ( y ) = 14
-5y + y^2 = 14
y^2 - 5y - 14 = 0
y^2 - 7y + 2y - 14 = 0
y ( y - 7 ) + 2 ( y - 7 )
( y + 2 ) ( y - 7 ) = 0
y = - 2 ( neglected)
or
y = 7
we cannot take a negative value so we must take y = 7
xy = 14
x × 7 = 14
x = 14 / 7
x = 2
So the required no. is 27
Hope it helps you .
Mounikamoparthi:
hi
Answered by
11
Answer:refer to the attachment below
Step-by-step explanation:
Hope This Helps
Mark As Brainliest
Attachments:
Similar questions