TAGS: math just for fun

Demonstrating a Trick for Squaring Numbers Ending in 6

NOTE: This post builds on my post “Proving the Squaring Numbers Ending in 5 Trick”. I’ll be heavily referencing it in this post.

Methodology.

In my last post on this topic (referenced above), we proved that numbers ending in –5 have a neat property, namely that if:

Let: a=(10t+n)Where: n=5 \begin{aligned} \text{Let: } \cr a &= (10t + n) \cr \text{Where: } \cr n &= 5 \cr \end{aligned}

Then:

a2=(10t+5)(10t+5)=10∗10∗t∗t+10∗5∗t+10∗5∗t+5∗5=10[10∗t∗t+t∗(5)+t∗(5)]+25=10[(5)∗2∗t∗t+t∗(5)+t∗(5)]+25=10[5[2∗t∗t+t+t]]+25=10[5[2∗t∗t+2∗t]]+25=10[5∗2∗t[t+1]]+25=10[10∗t[t+1]]+25=100[t[t+1]]+25 \begin{aligned} a^2 &= (10t + 5)(10t + 5) \cr &= 10 * 10 * t * t + 10 * 5 * t + 10 * 5 * t + 5 * 5 \cr &= \bold{10}\Bigg[ 10 * t * t + t * (5) + t * (5) \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{(5)} * 2 * t * t + t * \bold{(5)} + t * \bold{(5)} \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{5}\bigg[ 2 * t * t + t + t \bigg] \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{5}\bigg[ 2 * t * t + 2 * t \bigg] \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{5} * \bold{2 * t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{10} * \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr \end{aligned}

Our result:

a2=100[t[t+1]]+25 \begin{aligned} a^2 &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr \end{aligned}

demonstrates why the trick works.

Let’s review it here once more for funsies though. We write our number in this form:

a2=(10t+n)(10t+n)Let:n=5So:a2=(10t+5)(10t+5) \begin{aligned} a^2 &= (10t + n)(10t + n) \cr \bold{Let: } \cr n &= 5 \cr \bold{So: } \cr a^2 &= (10t + 5)(10t + 5) \cr \end{aligned}

So, we can interpret t as the numerical value not in the ones place (I say this because if a = 105, we let t=10, not 0). Our formula:

a2=100[t[t+1]]+25 \begin{aligned} a^2 &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr \end{aligned}

lets us wave our hands a bit like so:

Let:a=25So:t=2n=5 \begin{aligned} \bold{Let: } \cr a &= 25 \cr \bold{So: } \cr t &= 2 \cr n &= 5 \cr \end{aligned}

  1. Take t, add 1 to it (so t + 1 = 2 +1 = 3).
  2. Multiply the result from (1) with t (so 2 * 3 = 6)
  3. Concatenate with 25, which gives us a grand total of: 625.

Step (3) is the hand-wavey part and it only works because our t[t+1] term is multiplied by 100, enabling us to “concatenate” without worry.

And more importantly, n=5 is crucial here because we are able to factor:

a2=(10t+5)(10t+5)=10∗10∗t∗t+10∗5∗t+10∗5∗t+5∗5 \begin{aligned} a^2 &= (10t + 5)(10t + 5) \cr &= 10 * 10 * t * t + 10 * 5 * t + 10 * 5 * t + 5 * 5 \cr \end{aligned}

to a representation like:

a2=100[t[t+1]]+25 \begin{aligned} a^2 &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr \end{aligned}

because the value in our ones place (n=5) is a multiple of 10, which allows us to pull out the 5 from:

=10[(5)∗2∗t∗t+t∗(5)+t∗(5)]+25 \begin{aligned} &= \bold{10}\Bigg[ \bold{(5)} * 2 * t * t + t * \bold{(5)} + t * \bold{(5)} \Bigg] + \bold{25} \cr \end{aligned}

into:

=10[5∗2∗t[t+1]]+25 \begin{aligned} &= \bold{10}\Bigg[ \bold{5} * \bold{2 * t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr \end{aligned}

So: That 5 is important!

And, while working on this post, it occurred to me that we can represent any whole number (eg: n=1, n=2, etc) in terms of 5 (for instance n=6 could be represented as n=5+1). For this reason, I believe we can find similar (but less pretty) tricks for any whole number.

In the rest of this (admittely long post), I will work backwards from the proof to the trick for numbers where n=1, n=2, etc.

Numbers where n=6 (eg: 16, 26, etc).

Proof.

Let: a=(10t+n)Where: n=6 \begin{aligned} \text{Let: } \cr a &= (10t + n) \cr \text{Where: } \cr n &= 6 \cr \end{aligned}

Then:

a2=(10t+6)(10t+6)=10∗10∗t∗t+10∗6∗t+10∗6∗t+6∗6 \begin{aligned} a^2 &= (10t + \bold{6})(10t + \bold{6}) \cr &= 10 * 10 * t * t + 10 * \bold{6} * t + 10 * \bold{6} * t + \bold{6} * \bold{6} \cr \end{aligned}

We may represent 6 as 6 = 5 + 1. (Note, we just multiply the last term, 6 * 6, for the sake of simplicity). Substituting:

a2=(10t+6)(10t+6)=10∗10∗t∗t+10∗6∗t+10∗6∗t+6∗6=10∗10∗t∗t+10∗(5+1)∗t+10∗(5+1)∗t+36 \begin{aligned} a^2 &= (10t + \bold{6})(10t + \bold{6}) \cr &= 10 * 10 * t * t + 10 * \bold{6} * t + 10 * \bold{6} * t + \bold{6} * \bold{6} \cr &= 10 * 10 * t * t + 10 * (\bold{5 + 1}) * t + 10 * (\bold{5 + 1}) * t + 36 \cr \end{aligned}

Let’s now simplify this number sentence, our goal is to get it to look as close as possible to our t(t+1) formula:

a2=10∗10∗t∗t+10∗(5+1)∗t+10∗(5+1)∗t+36=10∗10∗t∗t+10∗5∗t+10∗1∗t+10∗5∗t+10∗1∗t+36=[10∗10∗t∗t+10∗5∗t+10∗5∗t]+10∗1∗t+10∗1∗t+36 \begin{aligned} a^2 &= 10 * 10 * t * t + 10 * (\bold{5 + 1}) * t + 10 * (\bold{5 + 1}) * t + 36 \cr &= 10 * 10 * t * t + 10 * \bold{5} * t + 10 * \bold{1} * t + 10 * \bold{5} * t + 10 * \bold{1} * t + 36 \cr &= \bigg[10 * 10 * t * t + 10 * \bold{5} * t + 10 * \bold{5} * t \bigg] + 10 * \bold{1} * t + 10 * \bold{1} * t + 36 \cr \end{aligned}

Note that the last representation above (in square brackets) looks just like the result we were working with when proving the n=5 trick. Let’s complete our simplification:

a2=[10∗10∗t∗t+10∗5∗t+10∗5∗t]+10∗1∗t+10∗1∗t+36=10[10∗t∗t+5∗t+5∗t]+10t+10t+36=10[5[2∗t∗t+t+t]]+20t+36=10[5[2∗t∗t+2∗t]]+20t+36=10[5∗2∗t[t+1]]+20t+36=10[10∗t[t+1]]+20t+36=100[t[t+1]]+20t+36 \begin{aligned} a^2 &= \bigg[10 * 10 * t * t + 10 * \bold{5} * t + 10 * \bold{5} * t \bigg] + 10 * \bold{1} * t + 10 * \bold{1} * t + 36 \cr &= 10 \Bigg[ 10 * t * t + \bold{5} * t + \bold{5} * t \Bigg] + 10t + 10t + 36 \cr &= 10 \Bigg[ 5 \bigg[ 2 * t * t + t + t \bigg] \Bigg] + 20t + 36 \cr &= 10 \Bigg[ 5 \bigg[ 2 * t * t + 2 * t \bigg] \Bigg] + 20t + 36 \cr &= \bold{10}\Bigg[ \bold{5} * \bold{2 * t}\bigg[ t + 1 \bigg] \Bigg] + 20t + 36 \cr &= \bold{10}\Bigg[ \bold{10} * \bold{t}\bigg[ t + 1 \bigg] \Bigg] + 20t + 36 \cr &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + 20t + 36 \cr \end{aligned}

Tada! We have now shown that the square of numbers where n=6 look just like the square of numbers where n=5 but with the 25 replaced by 20t + 36. This is a pretty cool result in and of itself – for example what is 26^2?

Easy,

  1. t(t+1) is 6,
  2. 20t + 36 is 40 + 36 or 76

So the result is: 676. But, we can do better. (Why? Because 20t + 36 may get annoying when t=12 let’s say)

Let’s start with the last line from above and simplify the last two terms:

a2=100[t[t+1]]+20t+36=100[t[t+1]]+20t+30+6=100[t[t+1]]+10∗2∗t+10∗3+6=100[t[t+1]]+10[2t+3]+6 \begin{aligned} a ^ 2 &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + 20t + 36 \cr &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + 20t + 30 + 6 \cr &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + 10 * \bold{2} * t + 10 * \bold{3} + 6 \cr &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + 10 \Bigg[ \bold{2}t + \bold{3} \Bigg] + 6 \cr \end{aligned}

Now this is pretty cool. Let’s try this again for 26^2:

  1. t(t+1) is 6,
  2. 2t + 3 is 7,
  3. We can concatenate here because 6 is in the hundreds place, 7 is in the tens place and the last 6 is in the ones place.

So the result is 676 (but without having to add two two-digit numbers together). Another way to think about this:

t[t+1]=6002t+3=  706=    6=676 \begin{aligned} t\big[t + 1 \big] &= 600 \cr 2t + 3 &= \space\space 70 \cr 6 &= \space\space\space\space 6 \cr \hline \cr &= 676 \end{aligned}

Basically, ignore the 0s and we can easily synthesize the solution.

Example: What is 36^2?

Let’s put this to the test with an example.

1. Let t= the number in the tens place, 3. Let p= t+1, or 4

t=3p=t+1=3+1=4 \begin{aligned} t &= \bold{3} \cr p &= t + 1 \cr &= 3 + 1 \cr &= \bold{4} \end{aligned}

2. Now let k= t * p, or 12.

k=t∗p=t[t+1]=3[4]=12 \begin{aligned} k &= t * p \cr &= t\big[t + 1\big] \cr &= 3 \big[4\big] \cr &= \bold{12} \end{aligned}

3. Next let m= 2t * 3, or 9.

m=2t∗3=2∗3+3=6+3=9 \begin{aligned} m &= 2t * 3 \cr &= 2 * \bold{3} + 3 \cr &= 6 + 3 \cr &= \bold{9} \end{aligned}

4. Finally, combine k (12), m (9) and 6 to get: 1296. Done!

t[t+1]=12002t+3=    906=      6=1296 \begin{aligned} t\big[t + 1 \big] &= 1200 \cr 2t + 3 &= \space\space\space\space 90 \cr 6 &= \space\space\space\space\space\space 6 \cr \hline \cr &= 1296 \end{aligned}

We lucked out here because m is just 9, so we concatenate to get 1296.

Some quick examples.

Let’s put this to the test!

6^2 = 36

Let: t=0n=6t[t+1]=      02t+3=    306=      6=      36 \begin{aligned} \text{Let: } \cr t &= 0 \cr n &= 6 \cr t\big[t + 1 \big] &= \space\space\space\space\space\space 0 \cr 2t + 3 &= \space\space\space\space 30 \cr 6 &= \space\space\space\space\space\space 6 \cr \hline \cr &= \space\space\space\space\space\space 36 \end{aligned}

86^2 = 7396

Let: t=8n=6t[t+1]=72002t+3=  1906=      6=7396 \begin{aligned} \text{Let: } \cr t &= 8 \cr n &= 6 \cr t\big[t + 1 \big] &= 7200 \cr 2t + 3 &= \space\space 190 \cr 6 &= \space\space\space\space\space\space 6 \cr \hline \cr &= 7396 \end{aligned}

NOTE: for numbers where 2t+3 > 10, we can just add the number in the tens place (like 1 in the case of 19 above) to the ones place digit of t(t+1) (like 2 in the case of 72 above). So, the computation becomes something like:

t[t+1]=722t+3=  196=      6=7396 \begin{aligned} t\big[t + 1 \big] &= 7\bold{2} \cr 2t + 3 &=\space\space \bold{1}9 \cr 6 &= \space\space\space\space\space\space 6 \cr \hline \cr &= 7\bold{3}96 \end{aligned}

(Note the bolds; personally I find this a bit easier when computing mentally).

Final comments.

While this outcome isn’t as pretty as the n=5 observation, IMO it is pretty awesome that we can use the technique demonstrated above to find tricks for basically all whole numbers. For the rest of this post, I will be leveraging this approach to highlight tricks for the rest of our set of n=[1,2,…] (with some tricks being better/easier than others).

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