Question

1. 6×7 =

(__×) + (×__)
__ + __

Explanation:

Step1: Understand the multiplication

We know that \(6\times7\) can be broken down using the distributive property. Let's split 6 into two parts, say 5 and 1 (or any other combination, here we'll use 5 and 1 for example, but another common way is 3 and 3, or 4 and 2). Wait, looking at the array, there are 6 rows and 7 columns? Wait no, the array has 6 rows? Wait no, the circles: let's count the rows. Wait the first part: let's see, maybe split 6 into 5 and 1? Wait no, maybe 3 and 3? Wait no, let's check the standard distributive property. \(6\times7=(5 + 1)\times7=5\times7+1\times7\), or \(6\times7=(4 + 2)\times7=4\times7+2\times7\), or \(6\times7=(3 + 3)\times7=3\times7+3\times7\). Let's check the array. The array has 6 rows? Wait the circles: let's count the number of rows. Wait the first row has 7 circles, and how many rows? Let's count: first row, second, third, fourth, fifth, sixth? Wait no, the image shows 6 rows? Wait no, let's count the rows: 1,2,3,4,5,6? Wait no, the circles: first row 7, second 7, third 7, fourth 7, fifth 7, sixth 7? Wait no, the user's image: let's see, the array is 6 rows? Wait no, the problem is \(6\times7\), so 6 groups of 7, or 7 groups of 6. Let's use the distributive property. Let's take \(6\times7=(5\times7)+(1\times7)\). Then \(5\times7 = 35\) and \(1\times7=7\), so \(35 + 7=42\), and \(6\times7 = 42\). Alternatively, another split: \(6\times7=(3\times7)+(3\times7)\), \(3\times7=21\), \(21+21 = 42\). Or \(6\times7=(4\times7)+(2\times7)\), \(4\times7 = 28\), \(2\times7=14\), \(28 + 14=42\). Let's fill in the blanks. Let's choose a split, say 5 and 1. So \((5\times7)+(1\times7)\), then \(35+7\). Or 3 and 3: \((3\times7)+(3\times7)\), \(21+21\). Or 4 and 2: \((4\times7)+(2\times7)\), \(28+14\). Let's proceed with one of them. Let's take \(6\times7=(5\times7)+(1\times7)\), then \(35 + 7\). So the first blank for \(6\times7\) is 42. Then the next part: \((5\times7)+(1\times7)\), then \(35+7\).

Step2: Calculate each part

First, \(6\times7 = 42\). Then, using distributive property, let's take \(a\times b + c\times d\) where \(a + c=6\) and \(b = d=7\). Let's choose \(a = 5\), \(c = 1\), \(b = d=7\). So \((5\times7)+(1\times7)\). Then \(5\times7=35\), \(1\times7 = 7\), so \(35+7\).

Answer:

\(6\times7=\underline{42}\)

\((\underline{5}\times\underline{7})+(\underline{1}\times\underline{7})\) (or other valid splits like 3,7,3,7 or 4,7,2,7)

\(\underline{35}+\underline{7}\) (or 21+21, 28+14 depending on split)

(Note: The split can vary, but the final product of \(6\times7\) is 42. For the distributive part, common splits are shown above. The key is that the sum of the two products equals \(6\times7\).)