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UIL Number Sense Mar 2026

UIL Number Sense Mar 2026

I competed in UIL Number Sense (it’s a Texas thing) in junior high and high school, I’ve been doing mental math all my life.

Yet, yesterday I learned a ‘new’ shortcut; that is, I don’t recall that I learned it or used it. I suspect I did though. Both. 😊

As a gift for my granddaughter, I purchased the book Secrets of Mental Math by Arthur Benjamin and Michael Shermer.

The shortcut is explained in the book.

EXAMPLE 1

If we’re asked to calculate $\bold{63 \cdot 67}$ , is there a shortcut?

Yes, my first instinct is to say: $65^2-2^2=4225-4=4221$

Reference: https://www.mathpax.com/engineering-math-nov-2023#squares

This works because, from algebra, $(A-B)\cdot(A+B)=A^2-B^2$

But is there a better way? There must be a better way!, or read this, if you don’t know Raymond Hettinger.

The ‘new’ shortcut is to say $(70 \cdot 60)+(3 \cdot 7)=4200+21=4221$

This works because, from algebra:

$\text{Solve }(A-N)\cdot(B+N)$

$\text{Let }\Delta=A-B$

$\text{Let }N_{c}=\Delta-N \text{}$

$\text{Explanation of notation: I think of }N_{c} \text{ as a complement of }N$

$\therefore (A-N)\cdot(B+N) = A \cdot B +(A-B) \cdot N-N^2$

$\therefore (A-N)\cdot(B+N) = A \cdot B +(A-B-N) \cdot N$

$\therefore (A-N)\cdot(B+N) = A \cdot B +(\Delta-N) \cdot N$

$\text{And finally }(A-N)\cdot(B+N) = A \cdot B +N_{c} \cdot N$

$\therefore \text{In our example, we have: }$

$\Delta=70-60=10$

$N=70-67=3$

$\therefore N_{c}=\Delta-N=10-3=7$

$\therefore (A-N) \cdot (B+N) = A \cdot B + N_{c} \cdot N = 70 \cdot 60 + 3 \cdot 7$

But, that’s just a proof and a verbose example.

The problem can be worked in 2 seconds — in our head.

Just think as follows: $(70 \cdot 60)+(3 \cdot 7)=4200+21=4221$ ✔

EXAMPLE 2

The $\Delta$ value doesn’t have to be 10.

Let’s calculate $\bold{102 \cdot 198}$ .

We can say $(200 \cdot 100)+(2 \cdot 98)=20,000+196=20,196$

$\text{In this example, we have: }$

$\Delta=200-100=100$

$N=200-198=2$

$\therefore N_{c}=\Delta-N=100-2=98$

$\therefore (A-N) \cdot (B+N) = A \cdot B + N_{c} \cdot N$

The answer is: $(200 \cdot 100)+(2 \cdot 98)=20,000+196=20,196$ ✔

EXAMPLE 3

Another example with $\Delta=100$ .

Let’s calculate $\bold{115 \cdot 185}$ .

We can say $(200 \cdot 100)+(15 \cdot 85)=20,000+1275=21,275$

$\text{In this example, we have: }$

$\Delta=200-100=100$

$N=200-185=15$

$\therefore N_{c}=\Delta-N=100-15=85$

$\therefore (A-N) \cdot (B+N) = A \cdot B + N_{c} \cdot N$

The answer is: $(200 \cdot 100)+(15 \cdot 85)=20,000+1275=21,275$ ✔

This example exhibits some self similarity.

The nested calculation is: $15 \cdot 85 = (90 \cdot 10)+(5 \cdot 75)=900+375=1275$ ✔

In summary, I’ve shown two good ways to make a fast mental multiplication. I judge the second method to be more general and faster. The first method involves subtraction of terms, the second involves only addition of terms, therefore easier.

Enjoy!

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