Question

two digit number is such that the product of its digits is 16when 54 is subtracted from the number,the digits interchange their places.find the number.

Answer 1

let the digits of a number be x and y where x is ten's digit
so number= 10x+y---(i)
reverse of the number = 10y+x....(ii)
Given
(i)-54=(ii)
10x+y-54=10y+x
9x-9y=54
x-y=6....(3)

Given xy=16
so y=16/x....(4)

from 3 and 4 we get
x-16/x=6
x∧2-16=6x
x∧2-6x-16=0
x∧2-8x+2x-16=0
(x-8)(x+2)=0
x=8 since x=-2 isn't possible
so y=2
number =82


Verify
82-54=28

Answer 2

SOLUTION:

Let the two digit number be 10x + y

Given : product of its digits(xy) = 16

xy = 16...................(1)

When 54 is subtracted from the number, the digits interchange their places

10x + y - 54  = 10y + x

10x + y - 10y - x = 54

9x - 9y = 54

9(x - y) = 54

x - y = 54/9

x - y = 6

x = 6 + y……………….(2)

Put this value of x in eq 1.

xy = 16

(6 + y)y = 16

6y + y² = 16

y²  + 6y - 16 = 0

y² +  8y - 2y - 16 = 0

[By middle term splitting]

y(y + 8) - 2(y + 8) = 0

(y - 2 ) ( y + 8) = 0

(y - 2 ) = 0   or ( y + 8) = 0

y = 2  or y = - 8

Since, a digit can't be negative, so y ≠ - 8.

Therefore , y = 2

Put this value of y in eq 1,

xy =16

x× 2 = 16

x = 16/2 = 8

x = 8

Required number = 10x + y  

= 10(8) + 2

= 80 + 2

Required number = 82

Hence, the Required two digit number is 82.

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