## An answer to a puzzle

The puzzle I posted a few days ago is derived from a puzzle that’s worked its way around the Internet every so often in the past few years. See this forum in Russian (why don’t I know Russian?), or Misha Lemeshko’s blog, or Daniel Lemire for the version that mine is derived from. The incarnation I saw on Wednesday, on Facebook, which inspired this post, says that “This problem can be solved by pre-school children in 5-10 minutes, by programmers – in 1 hour, by people with higher education… well, check it yourself! :)

It’s then followed by the following list of numbers:

8809 = 6
7111 = 0
2172 = 0
6666 = 4
1111 = 0
3213 = 0
7662 = 2
9312 = 1
0000 = 4
2222 = 0
3333 = 0
5555 = 0
8193 = 3
8096 = 5
7777 = 0
9999 = 4
7756 = 1
6855 = 3
9881 = 5
5531 = 0

2581 = ?

Sort of implicit in the hint is that maybe it has something to do with the digits; Real Mathematicians think that puzzles involving digits are somehow inferior. (In A Mathematician’s Apology Hardy observes of facts such as 8712 = 4 × 2178 and 153 = 13 + 5 3 + 33 that “[t]hese are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals much to a mathematician.”

Perhaps the natural thing to do, if this is a claim about digits, is to then assume that the claim 8809 = 6 encodes the statement $f(8) + f(8) + f(0) + f(9) = 6$, and proceed on that basis. On that basis we have $4f(1) = 0, 4f(0) = 4, 4f(2) = 0, 4f(3) = 0, 4f(5) = 0, 4f(6) = 0, 4f(7) = 0, 4f(9) = 4$ and so $f(0)=f(9) = 1, f(1)=f(2)=f(3)=f(5)=f(6)=f(7)= 0$. We still need $f(4), f(8)$. From the first “equality” we have $2f(8) + f(0) + f(9) = 6$ and so $f(8) = 2$. The answer is $f(2) + f(5) + f(8) + f(1) = 0+0+2+0 = 2$.

As for the version I gave — there are eight equations in nine unknowns. These were derived by removing from the “bloated” version of the puzzle all the equations with four of the same digits on the left side, and all those with a zero on the right side. The system again has equations $f(8) + f(8) + f(0) + f(9) = 6$ and so on. By subtracting equations from each other we get $f(0) = 1 + f(1), f(3) = f(2), f(6) = 1+f(2), f(8) = 2+f(2)$. From “6855 = 3” we have $f(6) + f(8) + 2f(5) = 3$, or $2f(2) + 2f(5) = 0$; if we agree that all values are nonnegative then $f(2) = f(5) = 0$. Then from “7756 = 1” we can get $f(7) = 0$. Also from relations we alrady derived, $f(3) = 0, f(6) = 1, f(8) = 2$.

But what are the values of 0, 1, and 9? It turns out that either $(f(0), f(1), f(9)) = (1,0,1)$ or $(2,1,0)$ works, from an addition standpoint.

So why does 4 never appears on the left hand side, therefore meaning we can never work out $f(4)$. This is a feature, not a bug. $f(n)$ is the number of holes in the numeral n. Some people draw 4 with one hole; some draw it with zero. So we choose $(f(0),f(1),f(9)) = (1,0,1)$ and so the answer is $f(2) + f(5) + f(8) + f(1) = 0 + 0 + 2 + 0 = 2$.

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### 38 Responses to An answer to a puzzle

1. jc says:

For what it’s worth, I didn’t get the puzzle at all until after I saw the hint that pre-schoolers could solve it in 5-10 minutes, and then while scrolling through the list the key jumped out at me immediately even before I got to the rest of your post.

2. Just a side note as someone who’s used LaTeX on WordPress for a while: the {aligned} environment is your best friend for lists of equations like

\begin{aligned}4f(1)=0\\4f(0)=4\\4f(2)=0\\4f(3)=0\\4f(5)=0\\4f(6)=0\\4f(7)=0\\4f(9)=4\end{aligned}

3. ae says:

You used 6666=0 in your solution. The puzzle says 6666=4.

4. Pep95 says:

Just count the circles

5. Pep95 says:

Like 6 has one circle, so 6=1
1 has no circles, so 1=0
8 has two circles, so 8=2
etc.

6. Ahmed Atef says:

just count the number of circles of each No. in each series and you’ll get the answer
i.e 6666=4 circles
8809=6 circles
7111=0 circles

7. kingkong says:

Why would kids look for circles in particular? I might have looked for any other shape. This is nonsense. *Damn, those grapes look sour*

8. nice going says:

dude you are so retarded, just write what the values are and add them… dumbasses

9. nice going says:

king kong you genius

10. Wayne says:

The answer is 2. Just count the number of circles in these numbers… for example, 9 has 1 circle and 8 has 2, 0 has 1, and 7 has no circle….

11. amit khan says:

2

12. Gene says:

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13. Luke says:

My brother just emailed me this, and I found the answer to be 2 also, but had a different reason. I found that numbers that could be rotated 180 degrees and still be numbers are worth 1 and numbers that can be rotated to a side and still be numbers (or mathematical anyway) are worth 2. 8 is the only number worth 2, because it is viewed as the infinity symbol sideways.

14. Luke says:

I just found a third way, I think. When drawing the number, without lifting the pen, the minimum number of times that the pen comes in contact with itself by completion of the number, without lifting the pen and without retracing a part of the number is the value of the number. 8 must contact itself twice when drawn without lifting. 0, 6, 9 must contact once. all other numbers do not contact themselves without retracing.

15. Luke says:

I forgot 4, which doesn’t appear, but would also equal 1 by this third rule,

16. Luke says:

I think 4 is quite clearly, intentionally, left out of the left side of the puzzle. By the circle rule, 4 = 0. By a “closed loop” rule, 4 =1. By a rotational rule, 4 = 0. By a “minimum point of contact when drawing, without retracing” rule, 4=1. Alright, time for sleep.

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