To solve using vertical multiplication, follow these steps:
Therefore, the product of is .
To solve this problem, we'll follow these steps:
Now, let's work through each step:
Step 1: The numbers given are 240 and 110. We can express 110 as 100 + 10.
Step 2: Multiply 240 by 100 and 240 by 10 separately:
-
-
Step 3: Add the partial products to find the total:
-
Therefore, the solution to the problem is .
To solve this problem, we'll use the long multiplication method to multiply 561 by 120.
Therefore, the solution to the problem is .
To solve this multiplication problem, we'll perform the calculation using vertical multiplication as follows:
First, write the numbers in a vertical format:
Step 1: Multiply by the last digit in , which is :
Step 2: Multiply by the second digit in (tens place, also ), and shift one place to the left:
Step 3: Multiply by the first digit in (hundreds place, again ), and shift two places to the left:
Step 4: Add up all the intermediate results:
The sum of these products is . Therefore, the solution to the multiplication is \textbf{}.
To solve this problem, we'll follow these steps:
Now, let's work through each step:
Step 1: We start with the equation . This suggests that divided by equals .
Step 2: To solve for , multiply both sides by , resulting in .
Step 3: Divide both sides of the equation by to solve for . Thus, .
Given the problem structure and the context, this step yields as expected. However, this suggests a different setup where the actual computations offer a different approach: Check normal computation errors or errors in initial projection due to illustration confusion.
Therefore, the representative solution leads to using values or potential errors in reading underlined operations, leading us to finalize , yielding based on division-restoration or oversized multiplication.
Thus, from recomputing and realizations, the chosen correct choice is (Choice 2) with .
To solve this problem through vertical multiplication, we will multiply each digit of the number by each digit of sequentially, considering their place values.
Therefore, the product of and is .
This matches the correct answer choice: .
To solve this problem, we'll follow these steps:
Now, let's work through these steps in detail:
Step 1: We have the numbers 500 and 100. Our task is to compute .
Step 2: Recognizing that both numbers end in zeros makes our multiplication process simpler. This means that we can separate them into significant digits and zeros.
Step 3: Multiply the significant figures: . Count the total number of zeros across both numbers. 500 has 2 zeros, and 100 also has 2 zeros, resulting in a total of 4 zeros.
Combining these results gives us followed by four zeros, so the answer is .
Therefore, the solution to the problem is .
To solve this multiplication problem, we follow these steps:
Thus, the product of and is , matching choice .
To conclude, the solution to the problem is .
To solve the problem of multiplying 611 by 300, we'll apply standard vertical multiplication for a three-digit number by a three-digit number. Here is the step-by-step process:
Step 1: Set up the multiplication with 611 on top and 300 below:
611 × 300
Step 2: Begin by multiplying the digits of 300 with 611, starting from the rightmost digit (units place):
First, multiply 611 by the unit's digit (0). Since any number times 0 is 0, the entire row is 0.
611 × 300 ---- 0 (This is 611 × 0, aligned with the unit's place)
Repeat the process for the number in the middle (tens place):
611 × 300 ---- 0 (This is 611 × 0, aligned with the ten's place) 00 (This is again 611 × 0, each digit moved one position to the left)
Now, move to the hundreds digit, which is 3.
611 × 300 ---- 0 00 1833 (This is 611 × 3, aligned with the hundred's place)
Step 3: Sum all the partial products. Start from the rightmost row to the left:
611 × 300 ---- 0 00 183300
When we sum the products, we get:
Therefore, the correct result of multiplying 611 by 300 is .
Matching this with the provided answer choices, it confirms choice 3 as the correct option.
Therefore, the product of is .
To solve the multiplication problem , follow these steps:
The solution to the problem is .
To solve this multiplication problem involving and , we'll proceed with the following steps:
Let's perform these calculations step-by-step:
Step 1: :
403
× 4
————
1612
Step 2: (this step gives zero so doesn't change total but affects placement):
403
× 0
———
0000 (this is used for tens place, so considered as 000)
Step 3: :
40
× 1
————
40300
Step 4: Now add the resulting products, remembering the alignments:
1612
0000 (for tens column alignment)
+40300 (for hundreds column alignment)
—————
41912
Therefore, the final product of multiplying 403 by 104 is .
Upon examining the choices given, the correct answer is indeed .
We begin by multiplying the two numbers, and , using vertical multiplication:
The multiplication results in , matching choice number 1. Therefore, the solution is:
To solve this problem, we will multiply two numbers, 345 and 211, using the vertical multiplication method. Let's break down the steps clearly:
Step 1: Write the numbers in vertical alignment, with 345 on top and 211 below. We'll handle multiplication digit by digit, starting from the rightmost digit of the bottom number.
Step 5: Add all these partial products together:
This yields:
gives .
results in .
Therefore, the solution to the problem is .
To solve this problem, we will perform step-by-step vertical multiplication of the numbers and .
Step 1: Multiply by the digit (the rightmost digit of ):
- .
Keep this result aligned in the ones place.
Step 2: Multiply by the digit (the tens digit of ):
- .
Align this result starting from the tens place, so it becomes .
Step 3: Multiply by the digit (the hundreds digit of ):
- .
Align this result starting from the hundreds place, so it becomes .
Step 4: Add all these partial products:
Add these numbers together:
.
Therefore, the product of is .
To solve this problem, we will perform vertical multiplication of the numbers 436 and 161. Here are the steps:
Therefore, the multiplication of 436 by 161 results in .
To solve this problem, we'll follow these steps:
Now, let's work through each step:
Step 1: Arrange the numbers as follows:
539
× 119
─────────
Step 2: Multiply each digit of 119 by 539.
539
× 9
─────────
4851
539
× 10
─────────
5390
539
× 100
─────────
53900
Step 3: Sum the partial products, taking care of placing each correctly to represent their value:
4851
5390
+53900
─────────
64141
Therefore, the answer to the problem is .
This corresponds to choice 4 in the provided multiple-choice options.
To solve this problem, we will use the vertical multiplication method:
First, multiply by (units digit of ):
Second, multiply by (tens digit of ) and adjust for place value by shifting one position to the left:
Third, multiply by (hundreds digit of ) and adjust for place value by shifting two positions to the left:
Add all partial products:
______
Total:
Therefore, the solution to the problem is .
To solve this problem, we will multiply the two numbers step by step:
Step 1: Multiply the bottom number, 316, by each digit of the top number, 631, starting from the rightmost digit 1.
Step 2: Align each resulting product according to its place value.
Step 3: Sum all the aligned products to find the final result.
Let's carry out the multiplication:
Now, add these products together:
Therefore, the product of 631 and 316 is .