Show that $$ 2^{2011} \cdot 5^{2007}=1,6 imes 10^{2008} $$ Hence determine the sum of the digits of $$ 2^{2011} \cdot 5^{2007} $$

Question

Show that $$ 2^{2011} \cdot 5^{2007}=1,6 	imes 10^{2008} $$ Hence determine the sum of the digits of $$ 2^{2011} \cdot 5^{2007} $$

UpStudy AI · Accepted Answer

To solve the problem, let's break it down into two parts:

### 1. Showing that $$ 2^{2011} \cdot 5^{2007} = 1.6 \times 10^{2008} $$

We start by manipulating the expression:

$$
2^{2011} \cdot 5^{2007}
$$

Notice that $$ 2^{2011} $$ can be written as $$ 2^{2007} \cdot 2^4 $$. Therefore:

$$
2^{2011} \cdot 5^{2007} = 2^{2007} \cdot 2^4 \cdot 5^{2007}
$$

Combine $$ 2^{2007} $$ and $$ 5^{2007} $$:

$$
2^{2007} \cdot 5^{2007} = (2 \cdot 5)^{2007} = 10^{2007}
$$

Now, multiply by $$ 2^4 $$ (which is 16):

$$
2^4 \cdot 10^{2007} = 16 \times 10^{2007} = 1.6 \times 10^{2008}
$$

So, we've established that:

$$
2^{2011} \cdot 5^{2007} = 1.6 \times 10^{2008}
$$

### 2. Determining the Sum of the Digits of $$ 2^{2011} \cdot 5^{2007} $$

From the above result, $$ 1.6 \times 10^{2008} $$ is equivalent to:

$$
16 \times 10^{2007}
$$

In decimal form, this is:

$$
16000\ldots0
$$

where there are **2007 zeros** following the digits **1** and **6**.

To find the sum of the digits:

$$
1 + 6 + \underbrace{0 + 0 + \ldots + 0}_{2007 \text{ zeros}} = 7
$$

### **Answer:** 
The sum of the digits of $$ 2^{2011} \cdot 5^{2007} $$ is **7**.
The sum of the digits of $$ 2^{2011} \cdot 5^{2007} $$ is **7**.

Show that \( 2^{2011} \cdot 5^{2007}=1,6 \times 10^{2008} \) Hence determine the sum of the digits of \( 2^{2011} \cdot 5^{2007} \)