Question

a two digit no. is such that the product of its digits is 18.when 63 is subtracted from the no, the digits interchange their places. find the numbers

Answer 1

let x and y be the digit in unit's place and ten's place of the number respectively

then the number = 10y+x

And xy = 18
=> y = 18/x --- (1)

from the given condition, we get,
10y+x-63 = 10x+y
=> 10(18/x)+x-63 = 10x+(18/x)
=> (180+x^2-63x)/x = (10x^2+18)/x
=> 180+x^2-63x = 10x^2+18
=> -9x^2-63x+162 = 0
=> -x^2-7x+18 = 0 [ dividing both sides by 9]
=> -x^2-9x+2x+18 = 0
=> -x(x+9)+2(x+9) = 0
=> (-x+2)(x+9) = 0
either -x+2=0
=> -x = -2
=> x = 2

or (x+9) = 0
=> x = -9 [Rejected]

therefore x = 2
putting x = 2 in (1)
y = 18/x
=> y = 18/2
=> y = 9

therefore the number = 10×9+2
=92

Answer 2

Answer:

Step-by-step explanation:

Product of two no.s = 18

Find the factors of 18 by prime factorization

Its coming out to be 2*3*3

Now from the above factors multiply and find two digits whose product is 18

There are two possible options of getting 18

Which are 6*3=18 (63)

9*2=18 (92)

Since the no. Is suppose to be subtracted by 63, therefore 63 - 63 = 0 so the required no. Is 92

CHECK:

92 - 63 = 29 (digits interchange their places)