ABCDE * FGHIJ

Lots of fancy math here. Or you can simply write out the long multiplication problem.

The first two steps are:

114X9*1 = 114X9

114X9*20 = 228(2X+1)80

Since you know the tens place and the ones place are now DONE being calculated, you look at the tens place, which must equal 8. So X+8=8. X must equal 0.

From the digital root method, we know that $123456789 \equiv 0\ (\mod 9)$ . We can apply the digital root method to the LHS as well. $(1 + 1 + 4 + X + 9) (1 + X + 8 + 2 + 1) \equiv 0\ (\mod 9) \\ \Rightarrow (6 + X) (3 + X) \equiv 0\ (\mod 9) \\ \Rightarrow 18 + 9X + X^2 \equiv 0\ (\mod 9) \\ \Rightarrow X^2 \equiv 0\ (\mod 9) \\ \Rightarrow X^2 \in \{0, 3, 6, 9\}$ If $X \geq 3$ then the product would be $> 143000000$ , so clearly $X = 0$ .

Relevant wiki: Modular Arithmetic - Multiplication

$\overline{114X9} \times \overline{1X821} = 123456789$

Now I will use modular arithmetic for the last 2 numbers:

$\overline{114X9} \times \overline{1X821} \equiv 89 \pmod{100}$

So:

$\overline{X9} \times 21 \equiv 89 \pmod{100}$

I can also write this as:

$(X \times 10 + 9)(21) \equiv 89 \pmod{100}$

Multiplying:

$X \times 210 +189 \equiv 89 \pmod{100}$

Using the property of addition: (as $X \times 210 = X \times 200 + X\times 10$ )

$X \times 10 +89 \equiv 89 \pmod{100}$

One way of defining congruence mod(n) is if the difference between them is a multiple of n:

$X \times 10+ 89 - 89$ must be a multiple of 100

$X \times 10$ must be a multiple of 100, so

$X \times 10 = 100 \times Y$ with $Y$ being any integer

$X = 10 \times Y$ with $Y$ being any integer

Alternatively, I can say that $X$ must be a multiple of 10

Remember that: $0 \leq X \leq 9$ as $X$ it is a digit of a number

The only number that is a multiple of 10 AND can be a digit of number is 0

So: $\boxed{X=0}$

11409*10821=123456789

Since there is no carry from the first digit = 9 * 1 = 9, the second digit is the least significant digit of 2 * 9 + 1 * X. This sum ends in 8, and since 2 * 9 = 18, 1 * X ends in 0. The only way for this to be true is if X = 0.

I don't know if this is the best way to do it, but I just found the prime factorisation of 123456789 to be 3²x3607x3803. 3x3607=10821 and 3x3803=11409

Let A = 114X9, and let B = 1X821. Then, in the stacked multiplication of A and B, the one's digit of the result will be the ones digit of A multiplied by the ones digit of B, as shown (ie, 9 $\times$ 1 = 9). The 10's digit of AB will be the 1's digit of A times the 10s digit of B PLUS the 10s digit of A times the 1s digit of B (modulo 10, to dispose of the carry). Hence, $X \times 1 + 2 \times 9 = X + 8 \pmod{10}$

So, it follows that X + 8 = 8, and thus, X = 0.

Represent a decimal number $a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}$ as follows:

$a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0} = \sum_{k=0}^{n} a_{k}10^{k}$

Thus, the decimal numbers $114X9$ and $1X821$ (where $X \in \mathbb{Z}_{10}$ ) can be written as follows:

$\begin{aligned} 114X9 &= 11409 + 10X \\ 1X821 &= 10821 + 1000X \end{aligned}$

So: $\begin{aligned} 114X9 \times 1X821 &= (11409 + 10X)(10821 + 1000X) \\ &= 10000X^2 + 11517210X + 123456789 \end{aligned}$

If $114X9 \times 1X821 = 123456789$ , then $X$ is the zero of the polynomial:

$10000X^2 + 11517210X = 10X(1000X+1151721)$

Since $X \in \mathbb{Z}_{10}$ , the only possible solution is $X = 0$ .

114X9 and 1X821 must be integers. Therefore, 123456789 must be divisible by these two numbers. X can be a digit from 0 to 9. Type 12345679/114X9 or 123456789/1X821 on the calculator, and just replace X by 0,1...9 until you get an integer. You get an integer for X=0, thus 0 is the answer.