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I was watching this Japanese game show and came across this question:

Solid composed of many cubes on several layers

The contestants were told that each small cube is 2cm on its side and were asked to find the volume of the above figure.

The answer was 3080 $cm^3$.

While I was counting the number of cubes for the first row, one of the contestants was able to answer this within a few seconds.

I'm curious about how he did it. I assumed the figure was constructed in some sort of pattern and was hoping someone could shed some light on this.

(The game show didn't explain how to solve this unfortunately...)

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    $\begingroup$ The back left face has $77$ cubes in it. By looking at the dark squares you can see that the layer next to it has $7$ cubes fewer, that is $70$; and the next has $9$ cubes fewer, that is $61$; and so on. We get$$77+70+61+\cdots=385$$cubes with volume $385\times8=3090$. But I really don't think I could do this in a few seconds. Obviously Japanese game show contestants are smarter than me... $\endgroup$
    – David
    Feb 16, 2018 at 5:57
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    $\begingroup$ @David - seeing the YouTube video and the suggested related videos, it seems that one particular Japanese game show contestant is fast and has an incredible memory - the others do not seem to get a look in $\endgroup$
    – Henry
    Feb 16, 2018 at 9:30
  • $\begingroup$ Is there a chance that in fact the host did declare that it is a 'even pyramid' and thus is some certain fraction (which I don't know!) of simply the overall cubic shape ?? $\endgroup$
    – Fattie
    Feb 16, 2018 at 19:22

3 Answers 3

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I looked at the horizontal layers.

Top layer has seven, and each layer below shows seven more. So the number of cubes is $$ 7+14+\cdots+70=\frac{77}2\cdot10=385\,. $$

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    $\begingroup$ I couldn't even count to seven in the time it took the contestant to answer, let alone count each row and notice a pattern. $\endgroup$
    – Jack M
    Feb 16, 2018 at 16:27
  • $\begingroup$ Yes, @JackM, and it didn’t help me that I first miscounted the number in the top layer. $\endgroup$
    – Lubin
    Feb 16, 2018 at 17:43
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    $\begingroup$ Maybe that contestant is just exceptionally good at recognizing numbers visually, and is able to immediately visually recognize a group of seven objects in the same way that most people can visually recognize two or three. $\endgroup$
    – Jack M
    Feb 17, 2018 at 7:57
  • $\begingroup$ I think the ability to recognize numbers is exactly the key to that person’s speed, @JackM . $\endgroup$
    – Lubin
    Feb 17, 2018 at 18:20
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    $\begingroup$ If they had shaded the top faces of the cubes instead of the right faces I might even have got that - see my comment above :( $\endgroup$
    – David
    Feb 18, 2018 at 23:32
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Here's my idea for how someone could answer this in a few seconds.

  1. See that there are ten horizontal layers
  2. See that each layer adds seven blocks
  3. Know that the tenth triangular number is $55$.
  4. Know that $2 ^ 3 = 8$
  5. Multiply $8 \cdot 55 \cdot 7$
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    $\begingroup$ You'd need step 3 to know step 2 is relevant. $\endgroup$
    – Pete Kirkham
    Feb 16, 2018 at 11:43
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    $\begingroup$ @PeteKirkham You're right. I put it as step two so it'd be obvious where "tenth" comes from. I'll change the order--hopefully it will still be apparent that step three relies on both previous steps. $\endgroup$
    – GoalBased
    Feb 19, 2018 at 2:14
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So we start on the left, and kind of slice it diagonally, if it makes sense.

The first diagonal layer has TWO columns, one with $2$ blocks and another with $2$ blocks.

The second diagonal layer has THREE columns, one with $4$ blocks, another with $3$ blocks, and another with $3$ blocks.

The third diagonal layer has FIVE columns, with $6$, $4$, $2$, $2$ and $1$ blocks.

If we sum it up to the tenth diagonal layer, we end up with a total of $385$ blocks.

EDIT: Didn't see the pattern.

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    $\begingroup$ Seems unlikely someone could do that in a few seconds? $\endgroup$
    – IntegrateThis
    Feb 16, 2018 at 5:56
  • $\begingroup$ people are crazy, this woman gave the 23rd root of a 201 digit number in 50 seconds $\endgroup$
    – Saketh Malyala
    Feb 16, 2018 at 5:57
  • $\begingroup$ i don't even think i can write 200 numbers in 50 seconds bro $\endgroup$
    – Saketh Malyala
    Feb 16, 2018 at 5:58
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    $\begingroup$ fair enough lol. $\endgroup$
    – IntegrateThis
    Feb 16, 2018 at 5:58

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