How to Prove integers with sum of 10 trick? Stuck on something obvious

Mental Calculation, Mathematics, and Puzzles
1

We “know” that when we multiply two numbers with two terms

a|b * a|c

where

b + c =10, and a is a single-digit number.

that

10a*10(a+1) + bc = (10a+b) (10a+c)

for example,

\begin{align} 625 & = 25 ^ 2\\ & = 20*30 + 25\\ \text{(OR)}\\ 2021 & = 43 * 47\\ & = 40 * 50 + 3 * 7\\ \text{(OR)}\\ 164021 & = 403 * 407\\ & = 400*410 + 3*7 \\ \end{align}

So what is going on here?

Obviously, by expansion:

\begin{align} a|b * a|c & =10a * 10a + 10ac +10ab + bc\\ \end{align}

and my conjecture is

\begin{align} a|b * a|c & = 10a*10(a+1) + b*c \\ \end{align}

Where b+c = 10

\begin{align} 10a * 10a + 10ac +10ab + bc & =10a*10a+10 + bc \\ 10a * 10a + 10ac +10ab &= 10a*10a+10\\ a * a + ac + ab &= = a*a +10\\ ac + ab = 10 \end{align}

This is far closer, still short an a to divide on the right side but I suspect I will find it in the morning.
Thanks Simon… I am happier to accept that my algebra is weak than the other likely alternative.

Second try…
( I suspect there might be a proof down this path but I need to learn so more general facts about squares, sequences, and other bits and pieces )

I know that (a+b)^2 = a^2 +2ab+b^2 and for the case b=5 I can say

(a+b)^2 = a * (a+1) + b^2

\begin{align} (ab)^2 & = 10a^2+2*10ab+b^2 \end{align}

but it is not obvious that I can say the same for a|b * a|c where b+c = 10.
If I assume that I can ( which is silly )

\begin{align} a|b * a|c = 10a^2 + ab + ac + bc\\ \end{align}

if I set these as equivalent

\begin{align} 10a^2+10ac+10ab +bc &= 10a^2 + ab + ac + bc\\ 10ac+10ab &= ab + ac\\ 10b+10a & = b + c\\ \end{align}

My rule was b + c = 10

So that lets me say a probably untrue and useless thing

\begin{align} 10b + 10a & = 10\\ a+b = 1 \\ \end{align}

@kinma Any tips? I assume my algebra has gone off the rails here but I’m not seeing it. Having another coffee .

1 Like
3

I clearly need to go and do a course on mathematical proofs. I suspect it shouldn’t be hard to prove that the trick always works but I’m stuck on how to get there. b+c = 10 should be the result where the equation is true kind of thing.

4

It looks like there’s a mistake in the line after “and my conjecture is”. On the right hand side, the (a + 1) should be multiplied by 10.

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5

Fixed the end point of a+b=c strikes me as equally funny.

6

It’s still wrong… instead of

a|b∗a|c=10a∗(10a+1)+b∗c

it should be:

a|b∗a|c=10a∗10(a+1)+b∗c

Right? E.g. if a = 4, b = 3, c = 7:

43 * 47 = 40 * 50 + 3 * 7

7

@simon merci… still an error on the right hand side but it’s now looking optimistic.

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Anyone else have ideas about the likelihood of an algebraic approach using difference of squares?

By definition; I think any 2 single digit integers that sum to 10 have to be equidistant to 5

\{1, 9\}\{2, 8\}\{3, 7\}\{4, 6\}\{5, 5\}
a|b * a|c ; c>=b, b+c=10; (c+b)/2 = 5

I suspect could be more direct with absolute values

a|b * a|c = ( ab + ac / 2 ) - ( c - b ) ^2\\

Will play down this path after I get my algebra sorted on the first post

9

The right hand side is

10a * 10(a + 1) + bc

which expands to

10a * 10a + 10a * 10 + bc

or

100a^2 + 100a + bc

Your left hand side was:

10a * 10a + 10ac + 10ab + bc

Equating the two sides:

10a * 10a + 10ac + 10ab + bc = 100a^2 + 100a + bc

100a^2 + 10ab + 10ac + bc = 100a^2 + 100a + bc

10ab + 10ac = 100a

Dividing by 10a gives:

b + c = 10

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10

Yes! This is right.
If we have 2 numbers:

x and y

and
x = a|b = 10a + b
and
y = a|c = 10a+c

then indeed:

\begin{align} a|b * a|c & =10a * 10a + 10ac +10ab + bc\\ \end{align}

If we take all the 10a terms together we get:

\begin{align} a|b * a|c & =10a * (10a + c + b) + bc \end{align}

However; we also stated that: b+c = 10. So:

a|b * a|c =10a * (10a + 10) + bc = a*(a+1) | bc

or

a|b * a|c =10a * 10(a + 1) + bc = a*(a+1) | bc

which is your conjecture.

Now, this is true for a being a single digit or between 0 and 9.
However; this is also true for a > 9, for example when double digits:

\begin{align} 403 * 407 = 40 * 41 | 3 * 7 = 1640 | 21 = 164021 \end{align}

In the example above, a= 40.

11

Excellent!
Seems we were both writing a reply at the same time.

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12

Thanks for the hands. I’m happy my algebraic weakness could provide you with a bit of entertainment. :slight_smile:

Proving the relationships for shortcuts seems like good practice in its own little way. Will probably continue down this path with other common techniques. Might even improve my algebra along the way.

I was watching a video interesting relationships amongst squares https://www.youtube.com/watch?v=VrkpD2uPOjs that looked like it might bear fruit. I suspect I will watch more of this authors videos as he presents some excellent nuggets that I really haven’t considered.

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13

Not sure which technique you are trying to prove. None of the calculation methods should require more than a couple of lines to demonstrate:

Perhaps it’s this on Same first digit but last digits sum to10?

(10a+d)*(10a+c)    where d+c = 10

(10a+d)(10a+c) = 100a2 +10(d+c)a +dc

but 10(d+c) = 100

=> product = 100a2 + 100a +dc
= 100a(a+1) +dc

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14

For the record, I don’t know if we did this already, but most of these techniques are specializations of the Anchor/Anker/Base method which allows you to fold the multiplication around an arbitrary ‘anchor’ usually convenient to multiply, usually a multiple of 10.

for any M,N MxN can be written (A+x)(A+y) where A is the anchor.

MxN= (A+x)(A+y) = A2 + A(x+y) + xy

That’s the generic Anchor pattern. Notice the middle term is also easy , the sum of two small numbers multiplied by the Anchor, and especially easy if they sum to 10.

eg: 67x61 = (60+7)(60+1) = 602 + 60(1+7) + 1*7
= 3600 + 480 + 7 = 4087

If we restrict our choice of Anchors to multiples of 10, ie A =10a, and our choice of multiplicands to two digits, we can go quite a bit further

MxN = (10a+x)(10a+y) = 100a2 + 10a(x+y) + xy

It’s important to keep track of the signs of x & y

24*17 = (20+4)(20-3) = 400 + 20(4-3) - 12 = 420-12 = 408

And the Anchor doesn’t have to lie between the multiplicands

97*89 = (100-3)(100-11) = 1002 + 100(-3+(-11)) + 33
10000 - 1400 + 33 = 8633

You can see how the squaring method is just a specialization when the Anchor is chosen exactly between - which is only possible if the difference is even. In this case the extra benefit is that the middle term vanishes taking the Anchor with it. So all you have to do is square the anchor.

NxN = (A+d)(A-d) = A2 - d2

89*97 = (93-4)(93+4) = 932 - 42

Anchor type methods work best when the numbers are fairly close and the difference small. Or if far apart, then easy to work with.

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15

The algebra is useful for seeing the underlying patterns and also for laying out the parts in a way that expresses the calculation pattern. With actual numbers one gets overwhelmed with detail. It’s just another kind of language skill and practice will make you facile. I commend you for having a go at it.

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16

You just posted one of the best explanations for these techniques!

I recommend everybody interested in mental multiplication to study this.

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I read this and I find myself wanting to build a VM, install Jupyterlab, learn how to write in Latex and load up Sagemath. It’s a sickness. I may very well start playing with the algebra for each of the little tricks I know. Better off just practicing but there are so many shiney toys to play with. That said, it would be nice to have a little place to keep my toys.

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18

I have a similar idea. I am working on a compendium of all the 2x2 mult & squaring tricks. Hope,like you, to be able to display it in a nice format, I’m hoping to do a website. A pretty presentation is very important, it makes it much easier to memorize the patterns.

I’ve mostly been working on techniques I’ve gathered here and there but there’s a lot to be harvested from previous discussions here, especially Kinma.

Anyway, we should compare notes, from time to time - that way we can fill in holes, catch each other’s mistakes etc. It would be good to develop this out of all the community contributions and make it available as an organized resource

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19

Thank you for your kind words. You probably shouldn’t encourage me :slight_smile: As I mentioned, I’m trying to compile list of these techniques.

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20

I probably should have stopped while ahead but I wanted to include Art Benjamin’s method of squaring 2 digit numbers in the list of Anker techniques. I doubt he invented it but it often goes by his name.

Starting with the result we had before, the identity

(A+d)(A-d) = A2 - d2

Shift terms back and forth across the =

A2= (A+d)(A-d) + d2

Choose A to be the number to be squared, Choose d to bring one of the terms to a multiple of 10

632=(63-3)(63+3) + 32

                   ...=60*66+9   =3969  <-- do not know why the editor impose this styling on this line

272= (27-3)(27+3) + 32 = 24*30+9 = 729

Formatting posts
21

Again; great way of showing this technique!

Anker. We keep the name.

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