A Two digit number is such that the product of the digit is 18. When 63 subtracted from the number the digit interchange the place.Find the number.
ab=18
10b+a = 10a+b-63
9a-9b = 63
a-b = 7
factors of 18 are 9,2 and they are 7 apart, so the number is 92
check:
29 = 92-63
do this sum more informative manner and correct sense
To find the two-digit number, we need to work through the problem step by step.
Let's assume the two-digit number is represented by "10a + b," where "a" and "b" denote the tens and units digits, respectively.
According to the problem, the product of the digits is 18:
a * b = 18
We are also given that when 63 is subtracted from the number, the digits interchange places. This implies:
(10a + b) - 63 = (10b + a)
Now we can solve the two equations simultaneously. Let's rewrite the second equation in terms of "a" and "b" for easier calculations:
10a + b - 63 = 10b + a
9a - 9b = 63
a - b = 7
Now, we can substitute the value of "a - b" from the equation into the first equation to find the values of "a" and "b":
a * b = 18 ---> Substitute a - b = 7 ---> (b + 7) * b = 18
b^2 + 7b - 18 = 0
Let's solve this quadratic equation to find the possible values of "b". Using factoring, we can rewrite the equation as:
(b + 9)(b - 2) = 0
By setting each factor equal to zero, we find two possible solutions for "b":
b + 9 = 0 ---> b = -9
b - 2 = 0 ---> b = 2
Since we are considering a positive two-digit number, we discard the negative value of "b" (b = -9).
Now, substitute the value of "b = 2" back into the equation for "a - b = 7":
a - 2 = 7
a = 7 + 2
a = 9
Therefore, the two-digit number is 92, as "a = 9" and "b = 2".