When we recently bought honey roasted peanuts at the grocery store, Eli speculated that there were 215 peanuts in the jar.
“Okay, so now we have to count them,” Eli said.
“No,” I said, explaining that I didn’t want them touching food that others would be eating. I then showed them the back of the jar, which said that one serving contained about 39 pieces and the whole jar contained about 16 servings. They knew that 39 × 16 would approximate the number of pieces, and they estimated that the jar contained 40 × 15 = 600 pieces.
But then they wanted the actual value, and I wondered how we could use the estimate to find the exact product. More importantly, I wondered if it was possible to find an algorithm that would allow an easily calculated estimate to be converted to the exact value with some minor corrections.
My sons’ estimate used one more than the larger factor and one less than the smaller factor; that is, they found (m + 1) × (n – 1) to estimate the value of mn. A little algebra should help to help to provide some insight.
The product had a value of 600, so further refinement led to:
This led to an algorithm:
- Find an estimate with nice numbers.
- Add 1.
- Add the larger factor.
- Subtract the smaller factor.
This gives 600 + 1 + 39 – 16 = 624. And sure enough, 39 × 16 = 624.
This method works any time you want to find the exact value of a product when the larger factor is one more than a nice number and the smaller factor is one less than a nice number. Just estimate with the nice numbers, then follow the steps. The method can be modified if the larger factor is one less than a nice number and the smaller factor is one more than a nice number:
- Find the estimate.
- Add 1.
- Subtract the larger factor.
- Add the smaller factor.
So if you want to find the product 41 × 14, then the larger factor is one more than 40 and the smaller factor is one less than 15. The estimate is again 40 × 15 = 600.
Then 600 + 1 – 41 + 14 = 574. And sure enough, 41 × 14 = 574.
The same idea can be extended to numbers that aren’t the same distance from nice numbers. But that’s not the point. The intent was not to find general methods for every combination; instead, the hope was to use an easily calculated estimate as the basis for an exact calculation. I’m not sure this method completely succeeds, but it was fun for an afternoon of mental gymnastics.