A product of a two digit number is24 when the number is reverse exceed the original number by 27 find the number
Digits be x and y and number is 10x+y
xy=24 (assuming you meant the product of the digits)
And second equation is
10y+x-(10x+y) = 27
Take over
To find the number that satisfies the given conditions, we can set up an equation based on the problem:
Let's assume the original two-digit number is represented by "10a + b", where 'a' and 'b' are the tens and units digits, respectively.
According to the problem, the product of the number and its reverse (which is "10b + a") is 24:
(10a + b) * (10b + a) = 24
Expanding the equation gives us:
100ab + 10a^2 + 10ab + b^2 = 24
Combining like terms:
100ab + 10ab + 10a^2 + b^2 = 24
Simplifying further:
110ab + 10a^2 + b^2 = 24
The problem also states that the reverse of the number exceeds the original number by 27:
10b + a = 10a + b + 27
Simplifying this equation:
9b - 9a = 27
Dividing both sides of this equation by 9:
b - a = 3
Now we have a system of two equations:
110ab + 10a^2 + b^2 = 24
b - a = 3
There are several methods to solve this system, such as substitution or elimination. By solving this system, we can find the values of 'a' and 'b' to identify the two-digit number that satisfies the given conditions.