Let's explore the correct way to solve the multiplication [tex]\(29 \times 31\)[/tex] step-by-step and understand where Lori might have made her mistake.
First, let's write the multiplication vertically:
```
29
× 31
_____
```
The number [tex]\(31\)[/tex] has two digits: [tex]\(1\)[/tex] in the units place and [tex]\(3\)[/tex] in the tens place. We will use these digits to perform partial multiplications and then sum the results.
Step-by-step solution:
1. Multiply the multiplicand [tex]\(29\)[/tex] by the units digit [tex]\(1\)[/tex]:
[tex]\[
29 \times 1 = 29
\][/tex]
This gives us our first partial product.
- Write [tex]\(29\)[/tex] directly below the line:
```
29
× 31
_____
29
```
2. Multiply the multiplicand [tex]\(29\)[/tex] by the tens digit [tex]\(3\)[/tex]:
[tex]\[
29 \times 3 = 87
\][/tex]
Since we are multiplying by [tex]\(3\)[/tex] which is in the tens place, we need to add a zero at the end to account for the place value, making it:
[tex]\[
290
\][/tex]
- Write [tex]\(870\)[/tex] directly below the first partial product:
```
29
× 31
______
29
870
```
3. Sum the partial products [tex]\(29\)[/tex] and [tex]\(870\)[/tex]:
[tex]\[
29 + 870 = 899
\][/tex]
- Write the final sum below the partial products:
```
29
× 31
______
29
870
______
899
```
So, the correct answer is [tex]\(\boxed{899}\)[/tex].
Now, addressing Lori's possible mistakes:
- Not placing a 0 in the ones column before multiplying [tex]\(3 \times 9\)[/tex]: This is the likely mistake Lori made. When multiplying by the tens digit [tex]\(3\)[/tex], it is crucial to add a zero at the end of the partial product to maintain the correct place value.
- The correct partial product for [tex]\(3 \times 29\)[/tex] should be [tex]\(870\)[/tex] (not [tex]\(87\)[/tex]).
- It looks like Lori might have only added the [tex]\(29\)[/tex] and [tex]\(87\)[/tex] directly, giving her [tex]\(116\)[/tex], which is incorrect. The regular errors that can lead to an incorrect result also include not carrying over properly or adding partial products incorrectly.
Hence, by following the correct method, we see that the accurate answer for [tex]\(29 \times 31\)[/tex] is indeed [tex]\(899\)[/tex]. Lori needs to place the zero correctly while performing the calculation involving the tens place digit.
