Question

fill in the missing numbers

Explanation:

Step1: Identify the multiplier

We know the bottom number (multiplier) is 5, and the product of the units digit of the top number and 5 ends with 0. The top number's units digit is 0, which fits. Now look at the tens digit: the product of the top number's tens digit and 5 gives a number ending with 2. The only digit that when multiplied by 5 ends with 2 is 4, since $4 \times 5 = 20$, the units digit is 0, but we carry over 2, which matches the tens digit of the partial product 52. So the top number's tens digit is 4.

Step2: Find the top number's hundreds digit

The partial product from multiplying the hundreds digit by 5 plus the carryover 2 gives 13. Let the hundreds digit be $a$. So $5a + 2 = 13$. Solving for $a$: $5a = 13 - 2 = 11$? No, wait, the partial product is 13 (hundreds and tens place), actually $5a$ gives a number where the last digit plus carryover 2 gives 3, so $5a$ ends with 1? No, correction: the partial product is 13 (the first two digits of 130), so $5a = 13 - 0$ (carryover from tens place: 4*5=20, carryover 2, so 5a + 2 = 13, so $5a = 11$ is wrong. Wait, the full product is 130, no, the partial product under the line is 52_, and the total product is 130? No, wait, the multiplication is:

  2_
x   5
-----
130
-----
 52_

Wait no, the correct structure is: the top number is a two-digit number: 2$\boldsymbol{6}$? No, wait $26 \times 5 = 130$, and the partial product (wait, no, the 52_ is the intermediate? No, wait, the missing digit in 52_ is 0, because 2620=520, and 2625=650? No, wait the total product is 130? No, the rightmost digit of the total product is 0, the middle is 2, left is 5, and the top partial is 130. Oh! It's a multiplication of a two-digit number by 25:
Let the top number be $n$. $n \times 5 = 130$, so $n = \frac{130}{5} = 26$. Then $n \times 20 = 26 \times 20 = 520$. So the missing digit in 52_ is 0, and the missing digit in the top number (2_) is 6.

Step3: Verify the full multiplication

$26 \times 25 = (26 \times 20) + (26 \times 5) = 520 + 130 = 650$, which matches the total product's digits (5,2,0 plus 1,3,0 gives 6,5,0, the rightmost 0 is visible).

Answer:

The missing digit in the top number (2$\boldsymbol{\square}$) is 6, and the missing digit in 52$\boldsymbol{\square}$ is 0. So the completed multiplication is:

  26
x 25
-----
 130
 520
-----
 650