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My youngest son is in $6$th grade. He likes to play with numbers. Today he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into this equation. $$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it. $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.
I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.
If I then skip three ones and read the fourth, that one equals my sum!

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Just have to say: your son is awesome. =] – Adriano May 29 '13 at 20:55
A similar one that I noticed when I was in 5th grade is $1000*2^n + 10*2^{n+1}+2^{n+2}=2^{n+10}$ – 0xFE May 30 '13 at 2:51
It's great that he's curious about numbers at his age, but honestly this is nothing to get hyped up about. Keep encouraging him though! – CodyBugstein May 31 '13 at 3:27
Isn't this mathematically equivalent to a card trick? Play an arranged pack in a particular way and the cards seem to behave magically.. – geotheory May 31 '13 at 10:00
This question and its answers have generated an absurd amount of votes. – Olivier Bégassat Jun 9 '13 at 3:31

11 Answers 11

up vote 232 down vote accepted

It works because the number

$$ 128 $$ has two special properties: it is a power of $2$, and every digit is a power of $2$. This means that we can write it in two separate ways:

\begin{align} 128&=2^7\\ 128&=100\times2^0+10\times2^1+1\times2^3 \end{align}

Multiplying both sides by $2^n$ then gives:

$$ 2^{n+7}=(100\times2^n)+(10\times2^{n+1})+2^{n+3} $$

Edit: others have done a better job of generalizing this, but I feel I should point out another obvious related sequence.

$128$ is not the only power of $2$ whose digits are all powers of $2$ (or $0$). For instance:

$$ 1024 = 2^{10} $$

So another, similar kind of relation is given by:

$$ 2^{n+10} = (1000\times2^n)+(10\times2^{n+1})+2^{n+2} $$

In your son's notation, that becomes:

Every one of these numbers is two times the number before it.
$1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.
I pick any one of them, times $1000$. Then I add the next one, times $10$. Then I add the next one.
If I then skip seven (!) ones and read the eighth, that one equals my sum!

See if he likes that one.

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+1 While the first answer points out basically the same thing; you go through the steps of explaining that each digit is a power of 2 which isn't obvious at first. – ryan May 30 '13 at 20:59
It's obvious to any programmer. I know nothing about math but as a programmer I'm shocked that this appears to be a big deal. – Locutus May 31 '13 at 4:08
@Kurian: It's not. (And definitely not a big deal to mathematicians; what makes you think that?) – ShreevatsaR Jun 3 '13 at 5:26

Factor out the $2^n$ and you get: $2^n (100+20+8) = 2^n 128 = 2^{n+7}$ since $2^7 = 128$

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Here is another formula similar to your son's, using powers of $5$:


We can derive identities of a similar form more generally. Consider any $m$th degree polynomial $f(x)\in\mathbb{Z}[x]$ of the following form


Notice that all rational roots of $f(x)$ are integers. Suppose $f$ has a rational root $k$. Then we have the following equality:


Your son's formula involves the $7$th degree polynomial $f(x)=100+10x+x^3-x^7$ and the root $k=2$. Above I've used the polynomial $f(x)=10^6+10^5x+10^4x^2+10^3x^3+10^2x^4+x^6-x^9$ and the root $k=5$.

Here are other examples for $k=3$. Consider the polynomials $f(x)=3+8x+80x^3-x^7,$ $g(x)=2130+10x+x^3-x^7$, and $h(x)=99+687x+x^3-x^7$, and notice that $f(3)=g(3)=h(3)=0$. These give us the following identities:

$$3\cdot3^n+8\cdot 3^{n+1}+80\cdot3^{n+3}=3^{n+7}\\2130\cdot3^n+10\cdot3^{n+1}+3^{n+3}=3^{n+7}\\99\cdot3^n+687\cdot3^{n+1}+3^{n+3}=3^{n+7}$$

Try using this method to find identities for other values of $k$ (including negative integers!).

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"Your son's formula involves the 7th degree polynomial..." - not something you hear very often when talking about 6th grade kids... – anaximander May 30 '13 at 10:38
@anaximander: Of course if it were a 6th degree polynomial that would make perfect sense in the 6th grade. – Joren May 30 '13 at 12:21
How do you determine the values of the "a" coefficients? – Dale Jun 6 '13 at 21:27
Did you mean the $a_2 k^{n+1}$ term to be $a_2 k^{n+2}$? – Dale Jun 7 '13 at 16:57
What would $g(x)$ be for the $3^{n+7}$ example? – Dale Jun 7 '13 at 17:00

A proof which, unlike all other suggestions, doesn't require power as a prerequisite.

This is what your son noticed by himself by looking at the first occurences in the sequence:

1 × 100 + 2 × 10 + 8 = 128

Now, let him remark that he can multiply everything by 2 without affecting the equality:

2 × (1 × 100 + 2 × 10 + 8) = 2 × 128

Assuming that he understands that the doubling distributes over the sum, he will obtain:

2 × 1 × 100 + 2 × 2 × 10 + 2 × 8 = 2 × 128

Then remark that when you double any pick you obtain of course the next pick in the sequence:

2 × 100 + 4 × 10 + 16 = 256

Repeat. Multiply both sides by two, distribute and double every pick again, and you obtain one more equality:

4 × 100 + 8 × 10 + 32 = 512


Now he should be pretty much convinced that his statement holds for any pick. (Of course it could be a good opportunity to introduce recurrence!)

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Yeah, it becomes almost trivial if you think about it this way. – Donkey_2009 May 31 '13 at 13:23

You may observe that whenever in any geometric progression $a_n=\lambda^na_0$, one term satisfies a linear recurrence ($a_n=c_0a_0+c_1a_1+\cdots+c_{n-1}a_{n-1}$, where some coefficients $c_i$ may be zero and the corresponding terms dropped), then every further term of the sequence satisfies the same recurrence: $a_{i+n}=c_0a_i+c_1a_{i+1}+\cdots+c_{n-1}a_{i+n-1}$. This is because the relation has simply been multiplied by $\lambda^i$. Even every term in another geometric progression with ratio$~\lambda$ will satisfy the recurrence.

What make the example especially attractive is that the nonzero coefficients are all powers of$~10$, which makes the linear combination easy to compute; this is related to the fact that all (nonzero) digits of $2^7=128$ themselves occur as powers of $2$. The same phenomenon happens for $2^{10}=1024$: take $1000$ times some power of$~2$, add $10~$times the next power of two, and once the one after that; the result is present again $8~$places further in the same progression. (For the digits of $2^{11}=2048$ you basically get the same relation.)

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I seriously considered giving up mathematics knowing that there are people who naturally understand such a thing at such a young age. – hyg17 May 30 '13 at 19:14
@hyg17: There is a difference between remarking and understanding. I remember that when I first tried adding $1+2+\cdots+n$ for various $n$, I observed it was always divisible by the odd number among $n,n+1$. This does not imply I understood or would have been able to explain it then; I don't think I was even aware proving such a thing would be possible. – Marc van Leeuwen May 31 '13 at 14:30
@hyg17 read this, it's worth your time!… – raindrop May 31 '13 at 19:52

Nice discovery, truly impressive for a boy. The answers have already show the justification, I just wanted to display three examples of the property, probably this corresponds to how it was discovered:

    1 . .        4 . .          1 6 . .    
  +   2 .      +   8 .        +   3 2 .    
        8          3 2            1 2 8    
    -----       ------         --------    
    1 2 8        5 1 2          2 0 4 8    
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+1 for making it visual. Words don't work for the ADD. Like me... – Chris K Oct 29 '13 at 5:11

A way of beginning understanding it is for example:

Take any number in the series, multiply by $30$, add the next one. You get the one that is five to the right of where you started.

Because multiplying by $30$ then adding the next one is the same as just multiplying by $32$. Which is the same as multiplying by $2$ five times.

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Proof by Induction.
Base case for $n = 1$ is $100 \cdot 2^1 + 10 \cdot 2^2 + 2^4 = 200 + 40 + 16 = 256 = 2^8 = 2^{1+7}$.
Now to prove that it is true for $n+1$, given it is true for $n$.
$f(n+1) = 100 \cdot 2^{n+1} + 10 \cdot 2^{n+2} + 2^{n+4}$ $= 2 \cdot (100 \cdot 2^n + 10 \cdot 2^{n+1} + 2^{n+3})$ $= 2 \cdot 2^{n+7} = 2^{(n+1)+7}$.

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Here's my non-mathematical answer:

  1. It works for elements $1, 2, 4$ and $8$ of your series (whose values are $1, 2, 8$ and $128$): $1 \times 100 + 2 \times 10 + 8 = 128$
  2. If you start anywhere else (say, at term $16$, whose value is $\,\,32768$), then you can just divide all those terms by $32768$ to have the same expression as in step 1.

So I think any expression involving a series of terms in the $2^n$ series would work.

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One could just manipulate the left-hand-side to be multiples of $2^{n+2}$ $:$ $${5^2} \times {2^{n + 2}} + 5 \times {2^{n + 2}} + 2 \times {2^{n + 2}}$$

Factoring, $${2^5}{2^{n + 2}}$$ And therefore, the Right Hand Side $$2^{n+7}$$ q.e.d.

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The result simply holds because $$\begin{align} (100*2^n)+(10*2^{n+1})+2^{n+3} &=(2^6+2^5+2^2)*2^n+(2^3+2)*2^{n+1}+2^{n+3} \\ & = 2^{n+6} +2^{n+5}+2^{n+2} +2^{n+4} + 2^{n+2}+2^{n+3} \\ & =2^{n+6}+2^{n+5}+2^{n+4} + 2^{n+3}+(2^{n+2}+2^{n+2})\\ & =2^{n+6}+2^{n+5}+2^{n+4} +(2^{n+3}+2^{n+3})\\ & =2^{n+6}+2^{n+5}+(2^{n+4}+2^{n+4})\\ & =2^{n+6}+(2^{n+5}+2^{n+5})\\ & =(2^{n+6}+2^{n+6})\\ &=\mathbf{2^{n+7}} \end{align}$$

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protected by Alex Becker May 31 '13 at 18:37

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