the 2 digit no is such that the productof its digits is 21 .If 36 is subtracted from the nos ,the digitsvare interchanged . find the nos. x:5(3z-5/x-3)-33(x-3/3x+5)=-52
Answers
Answer:
Let the unit digit of the number be x
and the tens digit of the number be y.
Thus,
The required number will be (10y + x)
Also,
It is given that, the product of the digits is 21.
=> x•y = 21 ----------(1)
Now,
New number formed after interchanging the digits will be (10x + y).
As per the question,
If 36 is subtracted from the required number then the digits are interchanged.
ie;
=> (10y + x) - 36 = 10x + y
=> 10x + y - 10y - x + 36 = 0
=> 9x - 9y + 36 = 0
=> 9(x - y + 4) = 0
=> x - y + 4 = 0
=> y = x + 4 --------(2)
Now,
Putting y = x + 4 in eq-(1) ,we get;
=> x•y = 21
=> x(x + 4) = 21
=> x^2 + 4x = 21
=> x^2 + 4x - 21 = 0
=> x^2 + 7x - 3x - 21 = 0
=> x(x + 7) - 3(x + 7) = 0
=> (x + 7)(x - 3) = 0
=> x = -7 , 3
Here, x = -7 will be rejected (as digit of a number is never negative) .
Thus, the appropriate value of x is 3.
{ Note; In mathematics, there are just ten digits , ie; 0,1,2,3,4,5,6,7,8,9. }
Now,
Putting x = 3 in eq-(2) , we get;
=> y = x + 4
=> y = 3 + 4
=> y = 7
Thus,
The unit digit of required number is 3
and the tens digit is 7.
Hence, the required number is 73.
Trick to solve;
Here , in the first statement of the question ,it is given that, the product of digits of a two digits number is 21.
21 can be written as 3×7 or 7×3.
Thus,
If unit digit is 3, then tens digit must be 7.
And ,if unit digit is 7 , then tens digit must be 3.
Thus, as per first statement of the question, two numbers can be taken under consideration, ie; 73 and 37.
Now, in the second statement of the question it is given that , if 36 is subtracted from the number, the digits are interchanged.
So, let's check, which number (37 or 73) fulfill the second condition.
When 36 is subtracted from 37;
37 - 36 = 1 (condition not fulfilled)
When 36 is subtracted from 73;
73 - 36 = 37 ( condition fulfilled)
Thus, required number is 73.