Question:
A two digit number is such that the product of its digits is 14. When 45 is added to the number, the digits are reversed. Find the number.
Answer:
Let the two digit number is 10x + y
Given, the product of its digits = 14
=> xy = 14 ............1
Again, when 45 is added from the number, the digits are interchanged.
=> 10x + y + 45 = 10y + x
=> 10x + y - 10y - x = -45
=> 9x - 9y = -45
=> 9y - 9x = 45
=> 9(y - x) = 45
=> y - x = 45/9
=> y - x = 5 .............2
Now, (y + x)2 = (y - x)2 + 4xy
=> (y + x)2 = 52 + 4 * 14
=> (y + x)2 = 25 + 56
=> (y + x)2 = 81
=> y + x = √81
=> y + x = ±9
Case 1. when y - x = 5 and y + x = 9
After solving it, we get
x = 2, y = 7
Case 2. when y - x = 5 and y + x = -9
After solving it, we get
x = -7, y = -2, wehich is not possible.
So, x = 2, y = 7
So, the number = 10*2 + 7
= 20 + 7
= 27
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