An unusual cryptogram

I did some trial and errors.
But definitely I found these conditions:
1. First digit must not be 1 and 5.
2. First digit cannot be 2, since 2 times 6x would yield 12x (double 2), and 2 times 5 would yield either 11x (double 1) or 10x (we did not have 0 here).
3. The last digit cannot be 1 and 5 as well.
4. The fourth digit cannot be larger than 2.
5. The second digit cannot be smaller than 3.

Based on those facts, I tried several combinations and found 3 x 54 = 162.

Trial and Error:

$3 \times 54 = 162$

Looking at last digits only: you can't use 1 as the last digit of any of the numbers. 6 won't work either, or 5. So it has to be 2, 3 and 4. 3 x 4 = 12, so either it's 4 x #3 = ##2 or 3 x #4 = ##2. Clearly the first digit of the last number is 1, as it can't be 5 or 6. Therefore 4 x #3 is ruled out. Since the final number is a multiple of 3, it has to be 162. So the solution is 3 x 54 = 162.

from itertools import permutations

for l in permutations(range(1,7), 6):
    if l[0] * (l[1] * 10 + l[2]) == l[3] * 100 + l[4] * 10 + l[5]:
        print (l)

output is (3, 5, 4, 1, 6, 2)

Well we have 6 positions which i shall label a,b,c,d,e,f

We know that a(10b+c)=100d+10e+f

Therefore 10ab +ac = 100d + 10e +f

as ab is multiplied by 10 it does not affect f. ac mod 10 must therefore be f. I tried this with each number 1-6 and found that the only the following values work:

This reduces the amount of trial and error required significantly with the final answer being 3 X 54 =162

3x54=162

no repeat of any number.

Determine that:

1. First digit cannot be 1 (because then attaining RHS is impossible as a 2 digit number < a 3 digit number).

2. Third digit cannot be 1 either as it will produce a repetition, because digit 6 will equal digit 1.

3. Neither the first or third digit can be 5, because multiplying the two will always give either 5 or 0 as digit -either a repetition, or invalid (0 is not given as an option).

First digit * third digit gives last digit. Listing possibilities:

2*3 gives 6

2*4 gives 8

2*5 gives 0

2*6 gives 2

3*2 gives 6

3*4 gives 2

3*6 gives 8

4*2 gives 8

4*3 gives 2

4*6 gives 4

6*2 gives 2

6*3 gives 8

6*4 gives 4

Bolded are the only viable options. Rest are rejected due to repetition, or invalid digits (8 or 0).

Go through the cases 1 by 1.

2 * _ 3 = _ _6

1, 4, 5 unused.

second digit cannot be 1. (2 * 13 = 26)

second digit cannot be 4. (2 * 43 = 86)

second digit cannot be 5. (2 * 53 = 106)

rejected.

3 * _ 2 = _ _6

1, 4, 5 unused.

second digit cannot be 1. (3 * 12 = 36)

second digit cannot be 4. (3 * 42 = 126)

second digit cannot be 5. (3 * 52 = 156)

rejected.

3 * _ 4 = _ _2

1, 5, 6 unused.

second digit cannot be 1. (3 * 14 = 42)

second digit is 5. (3 * 54 = 162)

solution found.

4 * _ 3 = _ _2

1, 5, 6 unused.

second digit cannot be 1. (4 * 13 = 52)

second digit cannot be 5. (4 * 53 = 212)

second digit cannot be 6. (4 * 63 = 252)

rejected.

It can be a tad more efficient if you, for example, recognize that for second digit to be 1 first digit has to be 6 (because since 100 = 5 * 20, and a two-digit number starting with 1 has to be less than 20, the multiplier has to be greater than 5 to attain a 3-digit number on right hand side) but that kind of takes more time than trying 4 cases.

Now, we use alphabets to represent the equation for easy explanation:

$C \times \overline{AB} = \overline{DEF}$

$\quad \;\; AB\\ \underline{\times\;\; \quad C}\\ \underline{\;\;\;DEF}$

Now, we start by picking digits for $B \times C = F$

Since we cannot repeat the numbers, we cannot put $1$ in $B$ or $C$ , because it would result in $1 \times C = C$ or $B \times 1 = B$ .

Next, we cannot put $5$ in $B$ or $C$ , because that will result in $F$ being $5$ or $0$ , which is not possible.

$2,3,4$ and $6$ are left. Do a bit of trial and error, and you will find 2 possibilities:

$4 \times 3 = 12$ , where $F=2$ or $2 \times 3 = 6$ , where $F=6$ . These gives us 4 possible combinations:

Combination 1

$\quad \;\; A2\\ \underline{\times\;\; \quad 3}\\ \underline{\;\;\;DE6}$

Since the product is a 3-digit number, $A$ can only take $4$ or $5$ . However, this does not give us the answer we want, because

$42 \times 3 = 126\\ 52 \times 3 = 156$

Combination 2

$\quad \;\; A3\\ \underline{\times\;\; \quad 2}\\ \underline{\;\;\;DE6}$

For the product to have 3 digits, $A$ can only be $5$ . However, $53 \times 2 = 106$ , which is not the answer we want.

Combination 3

$\quad \;\; A3\\ \underline{\times\;\; \quad 4}\\ \underline{\;\;\;DE2}$

For the product to have 3 digits, $A$ must be $5$ or $6$ . However, we won't get the answer we want:

$53 \times 4 = 212\\ 63 \times 4 = 252$

Combination 4

$\quad \;\; A4\\ \underline{\times\;\; \quad 3}\\ \underline{\;\;\;DE2}$

For the product to have 3 digits, $A$ must be $5$ or $6$ . We get:

$64 \times 3 = 192\\ 54 \times 3 = 162$

The final one uses the digits $1$ to $6$ , which is the answer we want.

$\quad \;\;\, 54\\ \underline{\times\;\; \quad 3}\\ \underline{\;\;\;\;\,162}$

$3 \times 54 = 162$

Therefore, the answer to submit is $\boxed{354162}$

It took too much time to solve it

The unit's place digit in both the multiplied numbers must have different parity.(which can be checked) Then 3 × 54 = 162 follows immediately

Numbers range between and including both 1 and 6. With only limited odd digits, that is just the three of them and included a 5 at that, we can't have an odd 3 digits answer or we'd need more than one 5.
{ 1 , 2 , 3 , 4 , 5 , 6 }
10n + 2 = 3 x 4 or
10n + 6 = 2 x 3 are the only possibilities. Anyway, 2 & 3 will be taken out of the pool and be put at the rear of the numbers.
Considering the largest two of the digits, multiplying them together only gives an output of 30, so even when the two be let to head the multipliers, it's impossible to reach over 400 (literally, 54 x 6 = 324). With 2 & 3 made unavailable (by their rear placements), then the hundreds place must be 1.
Let's have the boxes be substituted by A x BC = DEF for easier reference. D = 1 is already established. 5 can only be either B or E.
Possibility 1 : If A = 3, then 1 + E + F is a multiple of 3. With F either 2 or 6, E must be 6 or 5 respectively. Considering 6 = 2 x 3 has no carry over, a 5 as E is rendered impossible by the same argument in the first paragraph. 162 / 3 = 54 is one solution found.
---> 354162

Possibility 2 : If A = 4, then EF = 52 (F can only be 2, and alongside it E must be odd, with 5 the only available odd). Division attempt of 152 / 4 = 38 is unsuccessful to reach a feasible result.

Possibility 3 : If A = 2, then EF = 46 (F can only be 6, and alongside it E must be even since 2 x 3 has no carry over, with 4 the only available even number). Division attempt of 146 / 2 = 73 is unsuccessful to reach a feasible result.

I've done it through trial and error. My final answer is 3 \times 54 = 162