To cleverly calculate $A = 57 \times 968 + 43 \times 968$, we can apply the distributive law, which allows us to write it as:
$$A = (57 + 43) × 968$$
$$A = 100 × 968$$
$$A = 96,800$$
We then get the answer $\boxed{96,800}$.
What are the different distributive laws, and when should we use them?
I. Distributive laws
In the following two paragraphs, $\color{red} a$, $\color{red} b$, and $\color{red} k$ designate any numbers.
A. Distribution of multiplication with respect to addition
Assume you’re multiplying $\color{red}k$ to a sum of two or more integers ($a$ and $b$):
$$\color{red}k \color{black}\times (a+b)$$
$\color{red}k$ gets distributed to both $a$ and $b$ as a factor:
$$(\color{red}k \color{black}\times a) + (\color{red}k \color{black} \times b) = \color{red}k \color{black} (a) + \color{red}k \color{black} (b)$$
Example:
Use the distributive law to calculate $2 \times (3 + 4)$:
Answer:
$$2(3) + 2(4) = 6 + 8 = \boxed{14}$$
Alternatively, you may use the order of operations to add $3$ and $4$ first (summing to $7$), then multiplying that sum by $2$:
$$2 \times(7) = \boxed{14}$$
Note
In case you have $(a + b) \times k$, the distributive law may also be applied like this:
$$(a \times k) + (b \times k)$$
Notice how $k$ get distributed the same way demonstrated above. Therefore:
$$(a + b)k$$
$$ak + bk \Longleftrightarrow ka + kb$$
$$k(a + b)$$
B. Distribution of multiplication with respect to subtraction
As with distributing a factor $\color{red}k$ to any sum of terms, the same can be said about subtraction. Assume you’re multiplying $\color{red}k$ to a difference of two or more integers ($a$ and $b$):
$$\color{red}k \color{black}\times (a-b)$$
$\color{red}k$ gets distributed to both $a$ and $b$ as a factor:
$$(\color{red}k \color{black}\times a) – (\color{red}k \color{black} \times b) = \color{red}k \color{black} (a) – \color{red}k \color{black} (b)$$
Example:
Use the distributive law to calculate $3 \times (5 – 2)$:
Answer:
$$3(5)-3(2)=15-6= \boxed{9}$$
C. Generalization
The preceding formulas can be generalized to any number of terms inside the parentheses.
Example:
Evaluate $2 \times (3 + 4 – 5)$:
Answer:
$$2(3)+2(4)-2(5)=6+8-10=\boxed{4}$$
II. Examples of application
A. Mental calculation
Example 1.
If we want to calculate:
- $25 \times 11$
- $25 \times 21$
- $25 \times 31$
We can proceed like this:
$$25 \times 11\newline$$ $$25 \times (10 + 1)$$ $$(25 \times 10) + (25 \times 1)$$ $$250 + 25 = \boxed{275}$$
In the same way:
$$25 \times 21 = 25 \times (20 + 1) = 500 + 25 = \boxed{525}$$ $$25 \times 31 = 25 \times (30 + 1) = 750 + 25 = \boxed{775}$$
Example 2.
If we want to calculate:
- $29\times9$
- $24\times19$
- $24\times29$
We can proceed like this:
$$24 \times 9$$ $$24 \times (10 – 1)$$ $$(24\times10)-(24\times1)$$ $$240 – 24 = \boxed{216}$$
In the same way:
$$24 \times 19 = 24 \times (20 – 1) = 480 – 24 = \boxed{456}$$ $$24 \times 29 = 24 \times (30 – 1) = 720 – 24 = \boxed{696}$$
B. Calculating the lateral surface area of a right prism

Figure 1 shows the net of the lateral surface of a right triangular prism. The unit of length is centimeters.
To calculate the area of this surface, we can carry out the following calculation:
$$(24 \times 14) + (24 \times 19) + (24 \times 12)$$
This would amount to adding together the areas of the three rectangles.
It is, however, simpler to carry out this calculation:
$$24 \times (14+19+12) = 24 \times 45 = 1,080$$
Therefore, we find that the area of this surface is equal to $\boxed{1,080\;\mathrm{cm}^2}$.
Note
$(14 + 19 + 12)$ is the size in centimeters of the perimeter of a base of the prism.
C. Calculating the area of a circular ring
Say we want to calculate the area of the circular ring shaded in figure 2. The unit of length is centimeters. The unit of area will be centimeters squared $(\mathrm{cm}^2)$.

The area of this ring (shaded region) will be equal to the difference between the area of the entire circle with a radius of $3\;\mathrm{cm}$ ($2 + 1$) and the area of the small circle with a radius of $2\;\mathrm{cm}$. The formula to calculate the area of a circle is:
$$\boxed{A=\pi \times r^2}$$
We have:
- Area of the large circle: $\pi \times 3^2 = \pi \times 3 \times 3 = 9\pi$
- Area of the small circle: $\pi \times 2^2 = \pi \times 2 \times 2 = 4\pi$
- Area of the ring: $9\pi – 4\pi = (9 – 4)\pi = 5\pi$.
Therefore, the area of the ring is equal to $5\pi\;\mathrm{cm}^2$. Recall that $\pi$ is roughly equivalent to $\approx 3.14159$. By taking $3.14159$, we find the area is around $\boxed{15.7\;\mathrm{cm}^2}$.
Note
Generally, the area of a circular ring bordered by two concentric circles with respective radii of $R$ and $r$ is equal to $\pi R^2\;-\;\pi r^2$, which is $\pi (R^2\;-\;r^2)$.