Question

the product of the digits of a two digit number is 18 and the difference of the digits is 3. If the digits in tens place is bigger than that of the units place find the number​

Answer 1

Answer:

The two-digit number is 63.

Step-by-step explanation:

Let the two-digit number be 10x + y.

Given:

xy = 18

x - y = 3

Compute the value of x + y as follows:

$(x-y)^{2}=x^{2}+y^{2}-2xy\\x^{2}+y^{2}=(x-y)^{2}+2xy\\=3^{2}+(2\times18)\\=9+36\\=45$

Then x + y is:

$(x+y)^{2}=x^{2}+y^{2}+2xy\\(x+y)^{2}=45+(2\times 18)\\(x+y)^{2}=81\\x+y=9$

Compute the values of x and y as follows:

$x+y=9\\x-y=3\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\2x=12\\x=6$

The value of y is:

$x+y=9\\6+y=9\\y=3$

The two-digit number is: 10x + y = (10 × 6) + 3= 60 + 3 = 63.

Thus, the two-digit number is 63.