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Does anyone know a good general approximation (within 5 percent) for the sum of the products of two numbers such that the sum of one number of one product and another of another product is equal two the other two numbers MENTALLY. Format: $$a\times{b}+(k-b)\times(k-a)$$ where $k$ is the sum of $a$ and $(k-a)$ or $b$ and $(k-b)$ So I am practicing for this number sense competition and I have stumbled upon this question type several times.

A quick example $75\times53+57\times35$ Notice how 75 and 35 or 53 and 57 add up to 110. The acceptance range is $5672-6282$

I have tried to manipulate the general form in several ways but there is no convenient way to do mental math with. Help.

Ian L
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Simplify this first:$$ab+(k-a)(k-b)=2ab-ak-bk+k^2$$This means we need a mental path to get either near or exactly that total.

Obviously, as k gets large, dealing with this via mental math becomes tougher. The method described below works well, assuming $a$ and $b$ are 2-digit numbers.

First, we need to establish a starting point. Knowing $k$, here's a series of mental steps. Calculate: $$\frac{1}{2}k$$ Next, square this: $$(\frac{1}{2}k)^{2}$$ Double this: $$2((\frac{1}{2}k)^{2})$$ Those mental steps give you a simple path to: $$\frac{1}{2}k^{2}$$ Mental math techniques for squaring 2-digit numbers are easily found on the internet if you don't already know them.


Example: We know that, in your example, $k=110$, so:$$\frac{1}{2}(110)=55$$ Next,$$55^{2}=3,025$$ (There happen to be specific mental math tricks for squaring 2-digit numbers ending in 5, so this is surprisingly easy). Don't forget the doubling step!$$3,025 \times 2=6,050$$


Now that we have a starting point $(\frac{1}{2}k^{2})$, we need to either work out or estimate a distance from that starting point. Calculate both:$$a-\frac{1}{2}k \\ b-\frac{1}{2}k$$ From there, you'll multiply them together: $$(a-\frac{1}{2}k)(b-\frac{1}{2}k)$$ Double this total: $$2((a-\frac{1}{2}k)(b-\frac{1}{2}k))$$What you actually wind up calculating mentally here is:$$2ab-ak-bk+\frac{1}{2}k^2$$ Your final estimate can be given by adding this number to your starting point above:$$\frac{1}{2}k^2+(2ab-ak-bk+\frac{1}{2}k^2)=2ab-ak-bk+k^2$$

$2ab-ak-bk+k^2$ should look familiar. It's the answer for which we've been searching!

If $k$ happens to be even, and both $a-\frac{1}{2}k$ and $b-\frac{1}{2}k$ happen to be small enough to easily multiply and double, then you're on your way to an exact answer!

If multiplying $a-\frac{1}{2}k$ and $b-\frac{1}{2}k$ together seems daunting, simply round both numbers to the nearest power of 10, and multiply those together and double that instead. We're just going for an approximation, after all.


Example I: In your example, $a=75$, $b=53$, and $\frac{1}{2}k=55$. We work out$$75-55=20 \\ 53-55=-2$$ Multiply:$$20(-2)=-40$$Double:$$2(-40)=-80$$ Add this to the starting point, working it out mentally: $$\\ 6,050 + (-80) \\ 6,050 - 80 \\ 6,000 - 30 \\ 5,970$$

In this case, it happens that we can easily calculate an exact answer. $$(75)(53)+(57)(35)=5,970$$


What happens when this isn't the case? Let's try this with an odd $k$ and an $a$ and $b$ that aren't very close to $\frac{1}{2}k$. Let's try the more difficult:$$(109)(31)+(92)(14)$$

Example II: In this example, we can work out that $k=123$. First, we work out:$$\frac{1}{2}k=61.5 \\ 61.5 \approx 62$$Next, square:$$62^{2}=3,844$$Double that for the final step:$$2(3,844) = 7,688$$This is our starting point.

How much do we adjust this?$$109-62=47 \\ 31-62=-31$$If you're good with mentally multiplying numbers like this together, do it! Let's assume you're not, however:$$47 \approx 50 \\ -31 \approx -30$$

Multiplying those numbers together is much easier:$$50(-30) = -1,500$$Double that:$$2(-1,500) = -3,000$$

So, our final estimate would be:$$7,688 + (-3,000) \\ 7,688 - 3,000 \\ 4,688$$

The actual answer to the problem is $(109)(31)+(92)(14) = 4,667$, so our estimate of $4,688$ is quite good! How good?$$\frac{4,688}{4,667} \approx 1.0045$$It's not only within 5%, it's within 0.5%!