\(9992 \times 10008\)
The problem asks us to simplify the multiplication of two large numbers: \(9992 \times 10008\). Direct multiplication can be time-consuming and prone to errors. We can use a mathematical trick involving algebraic identities to solve this efficiently.
Observe the two numbers, \(9992\) and \(10008\). They are very close to \(10000\). We can express them in terms of \(10000\) and a smaller number:
Let \(a = 10000\) and \(b = 8\). The expression \(9992 \times 10008\) can be rewritten using these variables:
\( (a - b) \times (a + b) \)
This form matches the algebraic identity known as the "difference of squares":
\( (a - b)(a + b) = a^2 - b^2 \)
Now, we substitute \(a = 10000\) and \(b = 8\) back into the identity:
\( 9992 \times 10008 = (10000 - 8)(10000 + 8) = (10000)^2 - (8)^2 \)
Let's calculate each part:
Now, subtract the second value from the first:
\( 100,000,000 - 64 \)
Performing the subtraction:
\( 100,000,000 - 64 = 99,999,936 \)
Therefore, the simplified value of the expression \(9992 \times 10008\) is \(99,999,936\).
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