Problem: Compute \( 117 \cdot 1001 + 9 \cdot 999 \cdot 13 \).
Ooh… \( 1001 + 999 = 2000 \). And also, if we multiply \( 9 \) by \( 13 \), we get \( 117 \).
Thus, we can factor out \( 117 \) as follows:
\[ 117 \cdot 1001 + 9 \cdot 999 \cdot 13 = 117 \cdot 1001 + 9 \cdot 13 \cdot 999 = 117 \cdot 1001 + 117 \cdot 999 \]
\[ = 117(1001+999) = 117(2000) = 234000. \]
If you don’t get the above steps, write the first line above on a piece of paper a couple times. Notice how we cleverly observed that \( 1001 + 999 = 2000 \), a nice number. And also \( 9 \cdot 13 =117 \), so our reformed problem collapses by simply reversing the distributive property.
Nothing complicated, but this last step was quite clever. Don’t feel discouraged if you couldn’t see these insights at first glance. What you should especially take away from this example is that seemingly complicated arithmetic can be sometimes simplified drastically with a few neat observations about the problem, whether it be something about the structure of the problem (sum of products -> makes us think about distributive property), the numbers (1001+999=2000, and multiples of hundreds or thousands are easy to multiply) or something that may seem coincidental that may be what the intended “clever” solution looks like (\( 9 \times 13 = 117 \), so we can factor out \( 117 \)) from the entire expression).
Try to find another solution to this example. Did we have to factor \( 117 \) out? What if we tried other numbers? What if we broke down each product into the prime factorizations and factored from that step? There are many possibilities waiting for you to discover…