But is it a semiprime?

Let the top number be $\overline{a3}$ and the second number be $\overline{3b}$ , where $a$ and $b$ are single-digit non-negative integers. Then the product is:

$\begin{aligned} (10a+3)(30+b) & = 3\square \square \\ {\color{#3D99F6}300a}+10ab+3b + 90 & = 3\square \square & \small \color{#3D99F6} \text{It is obvious that }a \text{ can only be 1, then} \\ 390 + {\color{#3D99F6}13b} & = 3\square \square & \small \color{#3D99F6} \text{It is obvious that }b \text{ can only be 0, then} \\ 390 + 0 & = \boxed{390} \end{aligned}$

Estimating, we have $x \cdot 30 \approx 300$ or $x \approx 10$ . This means the first number is probably $13$ , and since $13 \times 30 = 390$ , the final product is $\boxed{390}$ .

Since $13$ times any integer larger than $30$ is larger than $399$ , and since $23$ times any integer $30$ or above is larger than $399$ , and since $03$ times any integer $39$ or lower is smaller than $300$ , $13 \times 30 = 390$ is the only possible solution to the above equation.

The product can only begin with three if the top left square is $1$ and the square next to 3 is $0$ .

This was found by first putting $1$ in each place mentioned: $\boxed {1}3$ $\times3\boxed {1}$ This gave me $\boxed {403}$ . I figured reducing the top wouldn't help, and that $300<403-13 <400$ , so I changed the $3$ 's partner to $\boxed{0}$ to get $13*30=\boxed{390}$

Let the top number be 3a and the second number be b3. Then rewrite this product as (10a+3)(30+b) = 10ab + 3b + 300a + 90 Note that 300 <= 10ab + 3b + 300a + 90 < 400. If a = 0, then 10ab + 3b + 300a + 90 = 90 + 3b which is <= 117 a contradiction.
If a => 2, then 10ab + 3b + 300a +90 > 600 a contradiction. Thus, a = 1. Then 10ab + 3b + 300a + 90 = 390 + 13b < 400. Then 13b < 10. This is impossible if b => 1. Thus, b = 0. Thus, 10ab + 3b + 300a + 90 = 390.

31 x 13 > 400, so the product can only be 13 x 30. Ed Gray

Let's don't use a "mathy" way to solve this problem. Since the product is 300 or more, it is clear that the first digit of the first multiplier must be 1 (only 1 times 3=3). Now, our answer must be 300 or more. To make sure the answer is smaller than 400, we should put the smallest (0) on the second multiplier's last digit. Now, we've got the answer: 13 times 30=390.

By first looking you realized that the answer is in the range of 300 to 399, however multiplying two digit number with other two digit number the answer must be three digit number or more, since the 100th digit of the answer is 3 that means 3 1 there for the first box is 1, that means the first digit is 13, now the second digit is among 30,31,32,....,39 so we start from the lowest which is 30, and test the answer if it doesn't match then we move to 31 till the end, so 13 30=399 that's the answer .

The result is in any case 300 and something. So, the first square must be 1. If the square was 2 or greater we would get at least 690 (23 * 30), which is greater than 300 and something. If the second square was 1 we would get 13 * 31 = 403, also greater than 300 and something. What we have left is 13 * 30 = 390

i got lucky. i thought of 13 times 30 randomly and, well, it matched so i got it right lol.

I figured it in my mind that if 13 30=390 if we have taken 03 31 then it would not have worked else we could also not use 13*31 as it will cross 400.

On inspection the top number must be 13 as nothing else could result in a product in the 300s. From there acting out a long multiplication with anything 1 or more in the bottom right hand corner would result in carrying over into the 400s as you always have a 9 in the 10s column, so the bottom right blank must be a 0.

$13\times 30 = \boxed{390}$

Let the second number be 30. The first number has to be more than a 10 to reach 300. And less than 14 so you don't go over 399. So 13 * 30 = 390

Oh my gosh I got the answer on my first try!!!

I just kind of worked it out in my head, testing different numbers to see if they fit.

Let assume the first no . Be x3 and lower 3y ...then carrying out simple multiplication we get 3x( hundredth place ) (xy)+9 { tenth place } 3y { ones place }..to be the product.. but given hundredth place = 3 ..=>x =1 ..put x = 1 in tenth place and you get 9 + y shouldn't be equal to 10 else the one adds to hundreds place ..so ..it must be zero...and hence calculate

1x3 = 3 (bilangan dimulai angka 3) 2x3 = 6 (bilangan dimulai angka 6) 3x3 = 9 (bilangan dimulai angka 9) Dst. Dapat dianalogikan ketika di soal diketahui angka yg memulai suatu bilangan adalah 3, maka bilangan "pengali yg diatas haruslah dimulai angka 1 -> 13 Sedangkan untuk bilangan pengaki yg dibawah sudah dimulai dari angka 3, agar hasil nya sesuai dengan qlue anda harus masukkan angka 0

Simple way 3 3 3__ Of course if the hundreds section of the number is three than three should be multiplied by one to get the first number three

Then you can try the multiplicant by replacing ones place from 0to2 asnd you ge15tg your answer.

Very Easy !😂😂