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Prove that the sum [tex]\( 34 \cdot 85 + 34 \cdot 36 \)[/tex] is divisible by 11.

Sagot :

Sure, we'll start by simplifying the given expression using simple algebraic manipulations and then verify if the result is divisible by 11.

Given expression:
[tex]\[ 34 \cdot 85 + 34 \cdot 36 \][/tex]

First, factor out the common term 34 from both products:
[tex]\[ 34 \cdot 85 + 34 \cdot 36 = 34 \cdot (85 + 36) \][/tex]

Next, add the numbers inside the parentheses:
[tex]\[ 85 + 36 = 121 \][/tex]

Now, the expression simplifies to:
[tex]\[ 34 \cdot 121 \][/tex]

To prove that this is divisible by 11, we need to check the divisibility of the product by 11. We already know the values of each individual calculation:
[tex]\[ 34 \cdot 85 = 2890 \][/tex]
[tex]\[ 34 \cdot 36 = 1224 \][/tex]
Adding these:
[tex]\[ 2890 + 1224 = 4114 \][/tex]

Check if 4114 is divisible by 11. Here it is:
[tex]\[ 4114 \div 11 = 374 \, \text{exactly} \][/tex]

Since the result is an integer with no remainder, the previous statement:
[tex]\[ 34 \cdot 85 + 34 \cdot 36 \][/tex]
i.e., 4114 is divisible by 11.

Therefore, we have proven that the sum [tex]\(34 \cdot 85 + 34 \cdot 36\)[/tex] is indeed divisible by 11.