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.
Answers
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
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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