Question

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

Answer 1

Let the original no be 10x+y such that x is in tens place and y in units place

x*y=14.....(1)

45+10x+y=10y+x

=9x-9y+45=0

=x-y+5=0

=y-x=5

y=x+5....(2)

Putting (2) in(1)

x(x+5)=14

x^2+5x-14

=x^2+7x-2x-14

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

Either x=2 or x=-7

y=either 14/2 or 14/-7

either 7 or -2

Number =10x+y

either 10*2+7

=27 or

-7*10-2

=-72

These are the 2 numbers.

Answer 2

let no. be 10x + y.
after interchanging digits, no. is = 10y+X
ATQ
10x + y + 45 = 10y+X
9y-9x = 45
y-x = 5
y = X+5

x*y=14
x(X+5)-14=0
${x}^{2} + 5x - 14 = 0$
$(x + 7)(x - 2) = 0$
=> x= -7, 2
we will take the positive value of X, as the digit of a number can't be negative.

=> x= 2

x*y =14
y = 7

The given no. was 10x + y
= 10*2 +7
= 27