Answer
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Hint: In this problem, to find the required value we will use the left-distributive property of multiplication over addition. For every real numbers $a,b$ and $c$, we can write $a \times \left( {b + c} \right) = \left( {a \times b} \right) + \left( {a \times c} \right)$. This is called left-distributive property. Note that multiplication can be distributed over parenthesis (brackets).
Complete step by step solution: In this problem, it is given that $893 \times 78 = p\; \cdots \cdots \left( 1 \right)$. We need to find the value of $893 \times 79$. For this, first we will rewrite this multiplication as $893 \times 79 = 893 \times \left( {78 + 1} \right)$. Now we can see that there is addition in one group of two numbers and we need to multiply the number $893$ with that group.
Let us compare $893 \times \left( {78 + 1} \right)$ with $a \times \left( {b + c} \right)$ then we can say that $a = 893,\;b = 78$ and $c = 1$.
Now we are going to use the left-distributive property of multiplication over addition which is given by $a \times \left( {b + c} \right) = \left( {a \times b} \right) + \left( {a \times c} \right)$ where $a,b$ and $c$ are real numbers. Therefore, $893 \times \left( {78 + 1} \right) = \left( {893 \times 78} \right) + \left( {893 \times 1} \right)\; \cdots \cdots \left( 2 \right)$
From $\left( 1 \right)$, we have $893 \times 78 = p$. Now we are using equation $\left( 1 \right)$ in $\left( 2 \right)$, we get $893 \times 79 = p + \left( {893 \times 1} \right)$
$ \Rightarrow 893 \times 79 = p + 893$
Therefore, if $893 \times 78 = p$ then $893 \times 79 = p + 893$.
Therefore, option D is correct.
Note: For every real numbers $a,b$ and $c$, we can write $\left( {a + b} \right) \times c = \left( {a \times c} \right) + \left( {b \times c} \right)$. This is called right-distributive property. The distributive property is valid for matrix multiplication also. In this type of problem, we can use the basic knowledge of multiplication to get required value. Multiplication is repeated addition. $a \times b$ means we need to add the number $a$ itself $b$ times. For example, to find $5 \times 10$ we need to add the number $5$ itself $10$ times. The result will be $50$. Similarly, in a given problem we have to find $893 \times 79$ that means we have to add the number $893$ itself $79$ times. In this problem it is given that $893 \times 78 = p$ that means if we add the number $893$ itself $78$ times then the result is $p$. Therefore, when we add the number $893$ itself $79$ times then the result will be $p + 893$ because one more time we have to add the number $893$.
Complete step by step solution: In this problem, it is given that $893 \times 78 = p\; \cdots \cdots \left( 1 \right)$. We need to find the value of $893 \times 79$. For this, first we will rewrite this multiplication as $893 \times 79 = 893 \times \left( {78 + 1} \right)$. Now we can see that there is addition in one group of two numbers and we need to multiply the number $893$ with that group.
Let us compare $893 \times \left( {78 + 1} \right)$ with $a \times \left( {b + c} \right)$ then we can say that $a = 893,\;b = 78$ and $c = 1$.
Now we are going to use the left-distributive property of multiplication over addition which is given by $a \times \left( {b + c} \right) = \left( {a \times b} \right) + \left( {a \times c} \right)$ where $a,b$ and $c$ are real numbers. Therefore, $893 \times \left( {78 + 1} \right) = \left( {893 \times 78} \right) + \left( {893 \times 1} \right)\; \cdots \cdots \left( 2 \right)$
From $\left( 1 \right)$, we have $893 \times 78 = p$. Now we are using equation $\left( 1 \right)$ in $\left( 2 \right)$, we get $893 \times 79 = p + \left( {893 \times 1} \right)$
$ \Rightarrow 893 \times 79 = p + 893$
Therefore, if $893 \times 78 = p$ then $893 \times 79 = p + 893$.
Therefore, option D is correct.
Note: For every real numbers $a,b$ and $c$, we can write $\left( {a + b} \right) \times c = \left( {a \times c} \right) + \left( {b \times c} \right)$. This is called right-distributive property. The distributive property is valid for matrix multiplication also. In this type of problem, we can use the basic knowledge of multiplication to get required value. Multiplication is repeated addition. $a \times b$ means we need to add the number $a$ itself $b$ times. For example, to find $5 \times 10$ we need to add the number $5$ itself $10$ times. The result will be $50$. Similarly, in a given problem we have to find $893 \times 79$ that means we have to add the number $893$ itself $79$ times. In this problem it is given that $893 \times 78 = p$ that means if we add the number $893$ itself $78$ times then the result is $p$. Therefore, when we add the number $893$ itself $79$ times then the result will be $p + 893$ because one more time we have to add the number $893$.
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