explain why the second partial product
is always greater than the first partial product when you multiply a
3-digit number by a 2-digit number.
The digit from the 2-digit number is in the tens column?
the partial product of 29x47
Ah, the mystical realm of partial products! Let me entertain you with a little mathematical jest.
You see, when you multiply a 3-digit number by a 2-digit number, a marvelous dance of digits takes place. The second partial product, my friend, is always greater because it's like the special guest who arrives fashionably late to the multiplication party!
The first partial product, like a diligent math student, jumps in with gusto and enthusiasm. It multiplies the 3-digit number by the least significant digit of the 2-digit number. But alas, it's just a warm-up act, setting the stage for the real fun to come.
Now enters the second partial product, strutting its stuff with an air of superiority. It multiplies the 3-digit number by the tens place digit of the 2-digit number. And oh boy, it knows how to make an entrance! Being associated with the tens place digit and all, it naturally brings a larger contribution to the overall product.
So, my curious friend, the second partial product always steals the spotlight and outshines the first. It's a thrilling performance of multiplying digits, where the numbers play their roles and the second partial product claims its rightful place as the grand finale of the multiplication show!
Remember, in the world of mathematics, even the numbers crave a bit of drama and excitement.
When multiplying a 3-digit number by a 2-digit number, we usually use the traditional method called "long multiplication." In this method, we multiply the 3-digit number (the multiplicand) by each digit of the 2-digit number (the multiplier), one digit at a time, and then add the partial products.
Let's illustrate this with an example using a 3-digit number 123 and a 2-digit number 45:
Step 1: We start by multiplying the ones place digit of the 2-digit number (5) with the 3-digit number:
123 x 5 = 615
Step 2: We multiply the tens place digit of the 2-digit number (4) with the 3-digit number:
123 x 4 = 492
Step 3: Finally, we add the two partial products together:
615 + 492 = 1107
In this example, the first partial product (615) is smaller than the second partial product (492), which can be a bit misleading. However, this is not always the case.
To understand why the second partial product is often greater than the first one, let's consider the place value values of the digits involved in the multiplication. In our example, the ones place digit of the 2-digit number (5) is multiplied by each digit of the 3-digit number (123). The resulting partial product will be in the hundreds place value (5 x 3 = 15, but it is written as 150).
When we multiply the tens place digit of the 2-digit number (4) by each digit of the 3-digit number (123), the resulting partial product will be in the thousands place value (4 x 3 = 12, but it is written as 1200).
Since the thousands place value is greater than the hundreds place value, the second partial product tends to be larger than the first partial product in these cases.
It's important to note that this pattern may not always hold true, especially if the digits involved in the multiplication have unique properties or specific relationships. However, in general, the place values of the digits will determine the relative size of the partial products.