25*101 solve by using distributive property

To solve \( 25 \times 101 \) using the distributive property, we can break down the number 101 into a sum of two simpler numbers and then distribute the multiplication.

Let's express 101 as \( 100 + 1 \):

\[
25 \times 101 = 25 \times (100 + 1)
\]

Using the distributive property:

\[
25 \times 101 = 25 \times 100 + 25 \times 1
\]

Now, multiply:

\[
25 \times 100 = 2500
\]
\[
25 \times 1 = 25
\]

Add these two results together:

\[
2500 + 25 = 2525
\]

So, \( 25 \times 101 = 2525 \).

Would you like more details or have any questions? 

Here are some related questions:

1. How would you use the distributive property to solve \( 12 \times 103 \)?
2. What is the result of \( 47 \times 102 \) using the distributive property?
3. Can you solve \( 19 \times 99 \) using the distributive property?
4. How does the distributive property help in mental math calculations?
5. What happens if we decompose the number differently, for example \( 101 = 50 + 50 + 1 \)?

**Tip:** The distributive property is especially useful for simplifying multiplication problems by breaking them down into easier parts.

Learn how to solve the multiplication problem 25 times 101 using the distributive property with a step-by-step explanation. This problem demonstrates the application of the distributive property in multiplication, suitable for students in Grades 4-6.

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