A two digit number is such that the product of its digits is 18 when 6...
Let the original no.=x+10y,
x y=18--(1).
x+10y-63=y+10x,
9y-9x=63,
y-x=7,---(2).
from 1 and 2 -->
put x=18/y in (2).
y-18/y=7,
y^2-18=7y,
y^2-7y-18=0,
y^2-9y+2y-18=0,
y(y-9)+2(y-9)=0,
y=9 because - 2 is not required.
put y=9 in (1).
xy=18,
x=18/9=2,
original number =x+10y=2+10(9)=92
A two digit number is such that the product of its digits is 18 when 6...
Problem: A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Let's assume that the tens digit of the number is x and the ones digit is y. Therefore, the actual number can be represented as 10x + y.
Given, the product of the digits is 18. So, we can write:
x * y = 18
Also, when 63 is subtracted from the number, the digits interchange their places. This means that:
10x + y - 63 = 10y + x
Simplifying this equation, we get:
9x - 9y = 63
Dividing both sides by 9, we get:
x - y = 7
Now, we have two equations:
x * y = 18 (Equation 1)
x - y = 7 (Equation 2)
We can solve these equations simultaneously to find the values of x and y.
From Equation 2, we can write:
x = y + 7
Substituting this value of x in Equation 1, we get:
(y + 7) * y = 18
Simplifying this equation, we get:
y^2 + 7y - 18 = 0
We can solve this quadratic equation using factorization or the quadratic formula to find the value of y. After solving this equation, we get two values of y: -9 and 2. However, the value of y cannot be negative, as we are dealing with a two-digit number. Therefore, the value of y is 2.
Substituting this value of y in x = y + 7, we get:
x = 2 + 7 = 9
Hence, the number is 92.
Therefore, the solution to the problem is that the number is 92.
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