Concatenation = Product

Let's try looking at the digits which only have 9's.

100 > 99 > 81

1000 > 999 > 729

10000 > 9999 > 6561

As the digits increase, the powers of 9 cannot catch up with the powers of 10.

And if we use other digits that are not 9, the multiplication result will decrease drastically.

So there are no such positive integers greater than 9.

Consider the $(n+1)$ -digit integer $N=\overline{a_n a_{n-1}\cdots a_1 a_0}$ (where $n>0$ and $a_n\neq 0$ ). Note that $0\leq a_i \leq 9$ for all $i$ . $\begin{aligned} N &= \sum_{i=0}^n 10^i a_i \\ &\geq 10^n a_n \\ &> a_n\prod_{i=0}^{n-1} a_i \\ &= \prod_{i=0}^{n} a_i \end{aligned}$

Thus, the product of the digits will always be strictly less than $N$ and the answer is no .

Let us take digits be a and b such that :-

10a+b= ab.

By algebraic manipulation, b equals 10a/(a-1).

b is a digit so it must be from 0 to 9. Thus 10a/(a-1)€(0,9)

The LHS will be an integer only when a equals to 2,3 and 6 because factors of 10 are 1,2,5,10.

(a is also a digit so it cannot be 11.)

But for these values ,we get 20,15,12 as b that is not possible as b is a digit.

Thus, answer is NO.

The most significant digit times the product of the other single-digit multipliers cannot produce the most significant digit times its power of 10 multiplier. Therefore the answer is NO.

When we get to the ‘tens’(11,10,12, ect), we get 1 x. With the twenties, we get 2 x. The 30s are 3 x. But 30 is 3x10. This also holds with hundreds:1 x*y.This has no effect, and with 200s, we are already above.

Let this integer be 'N', have 'k' digits and the first digit 'a'. N = a×10^(n-1) + b×10^(n-2) + ... . None of the digits can be '0' and all of them are <10. The product of the digits is obviously < a×10(n-1), which is < N always. Answer is no.

Suppose $N > 9$ has $n$ digits - say $N = a_n a_{n-1} \cdots a_2 a_1$ . Then the product of the digits is:

$(a_n)(a_{n-1}) \cdots (a_2)(a_1) = ((a_{n-1}) \cdots (a_2)(a_1))a_n \leq 9^{n-1} a_n$ since $a_i \leq 9$ for $i = 1, \cdots, n-1$

In turn, $9^{n-1} a_n < 10^{n-1} a_n + 10^{n-2} a_{n-1} + \cdots + 10 a_2 + a_1 = N$

Therefore, a positive integer can never equal the product of its digits.

To prove there is no such two digit number, assume $ab = 10a + b$ then $a = b/(b-10)$

Substituting b = 1 to 9 gives $a$ as a negative number.

Expanding this to n+1 digits, let a = 1 to 9 and b = an integer between 1 and $(10^{n}-1)$ and let m be the product of all the digits of b. Hence $m \leq 9^{n}$

Assume $am = 10^{n}a + b$ , then $a = b/(m-10^{n})$ but the denominator $m-10^{n}$ is negative, so again $a$ comes out as a negative number.

To see this more clearly take the example n=3:

Let a = 1 to 9 and b = an integer between 1 and 999 and let m be the product of all the digits of b. Hence $m \leq 9^{3} = 729$

Assume $am = 1000a + b$ , then $a = b/(m-1000)$ but the denominator $m-1000$ is negative, so again $a$ comes out as a negative number.

Let there be an $n$ digit integer $N > 9$ which is equal to the product of its digits ${x_1, x_2, ... x_n}$ where $x_n$ is the right most digit:

$N = \prod_{i=1}^{n} x_i$

It must be the case that $N = x_n \cdot M$ where $M = \prod_{i=1}^{n-1} x_i$ by the transitive property of multiplication.

It must also be the case that $N = 10 \cdot M + x_n$ by the properties of the base 10 counting system, which is a contradiction.

It follows that no positive integer greater than 9 can exist which is equal to the product of its digits.

With 2 digits a and b: ab=10a+b => a(b-10)=b. but b-10 <0 so no solution other than a=b=0.

With 3 digits: abc=100a+10b+c => a(bc-100)=10b+c. Same reasoning, since bc<100: the only solution has a=0.

This reasoning holds for any higher number of digits, so: No.

Let there be $n$ digits in the integer we are trying to find. Since each digit is less than $10$ , the product of the digits will always be less than $10^n$ , which is less than or equal to the number itself.

The question above is a way of asking whether Concatenation of intetgers equals Product of those integers.

Concatenation of two integers. $a$ and. $b$ is given as,

$a||b = ab$ , not. $a\times b$ where, $||$ is the symbol of Concatenation.

Example, $2||3 = 23 \neq 2\times 3$

Now, $\boxed{a||b = a.β^{[log_β b]+1}}$ , is the general formula for Concatenation.

We need, $a||b = a\times b,$ which is only possible if $b$ doesnt exist, i.e, only possible for single digit integers, which are already excluded.

For more than 1 digit integers, $\boxed{a||b||c..... >a\times b\times c...... }$ Always.

So, $ANSWER:\boxed{NO}$

I tried every number greater than 9 and nothing works

Let the Number $a_1a_2a_3.....a_n$ be a solution to the given condition. So

$10^{n-1}a_1 + 10^{n-2}a_2 + .... + a_n = a_1 × a_2 × .... × a_n$

On rearranging the equation we get

$10^{n-2}a_2 + .... + a_n = a_1 ×(a_2 × .... × a_n - 10^{n-1})$

Since $a_i < 10$ we get

$a_2 × .... × a_n < 10^{n-1}$

Therefore,

$a_2 × .... × a_n - 10^{n-1} < 0$

So here our RHS becomes negative but our LHS is positive So equality can not occur for a number which is not a single digit number