The product of a 2 digit natural number and number obtained by reversing its digits is 2430. The smaller of the two numbers is
Answers
Explanation:
Well, I didn't know how to solve it absolutely from just what you have in your problem statement. But I made just 1 guess, and the answer flowed from that.
Assume your digits differ by 1. (If this doesn't work, we'll assume they differ by 2, and continue with a similar procedure to what I've done here.)
So, let A be one of the digits. The second will be A + 1.
Then your first number is (A * 10) + (A + 1), and your second is ((A + 1)* 10) + A.
Their product is then:
(10A + A + 1) * (10A + 10 + A) = 2430
...multiply every term in the first set of parens with every term in the second set. Collect terms, and you get:
121
A
2
+
121
A
+
10
=
2430
121
A
2
+
121
A
−
2420
=
0
...which is a quadratic equation with roots 4 and -5.
The digits we want must be "natural" numbers, i.e., positive integers. So disregard the -5, and pick A = 4. The next digit must be 5. (Remember, we made the single assumption that they differed by 1.)
Which would give 45 and 54.
(I went through a similar procedure with the assumption that the 2 digits differed by 2. Your resulting quadratic equation is
122
A
2
−
242
A
−
2390
=
0
, which has real roots, but they are not natural integers. I'm assuming if you tried this with the two digits differing by 3, 4, etc., the same will be true.)