# The product of a two-digit natural number and a number consisting of the same digits in the reverse order is 2430. Calculate that two-digit number?

## I know, that the answer is 54 and 45, but how should I calculate it step by step?

Sep 19, 2017

See explanation

#### 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.)

Sep 19, 2017

The numbers are $45 \mathmr{and} 54$

#### Explanation:

The last digit of $243 \textcolor{b l u e}{0}$ is $0$.

This indicates that one of the numbers must end in a $5$, and the other number must be even.

Let the missing digit be $x$.

The number $x 5$ has a value $10 x + 5$

The number $5 x$ has a value $50 + x$

The product is $2430$

$\left(10 x + 5\right) \left(50 + x\right) = 2430$

$500 x + 10 {x}^{2} + 250 + 5 x = 2430 \text{ }$ make a quadratic equal to $0$

$10 {x}^{2} + 505 x - 2180 = 0 \text{ } \leftarrow \div 5$

$2 {x}^{2} + 101 x - 436 = 0$

Using $a c$ method: $2 \times 436 = 872$
Find factors of $872$ which duffer by $101$

$1 , 2 , 4 , \textcolor{b l u e}{8} , \ldots . . \textcolor{b l u e}{109} , 218 , 436 , 872$

$\text{ } 2 \mathmr{and} 436$
$\text{ "darr" } \downarrow$
$\text{ "2color(white)(.xx)+109" } \rightarrow 1 \times + 109 = + 109$
" "1color(white)(xxxx)-4" "rarr " "2 xx -4=ul(" " -8)
$\textcolor{w h i t e}{w w w w w w w w w w w w w w w w w w w w w w} + 101$

$\left(2 x + 109\right) \left(x - 4\right) = 0$

$x = - 54.5 \text{ } \leftarrow$ reject, not an integer.

$x = 4$

If $x = 4$, then the two numbers are $45 \mathmr{and} 54$

Check $45 \times 54 = 2430$