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If four 2s make eight, and three 2s also make eight, how many 2s do you need to make three eights, when you only have less than eight 2s?

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  • $\begingroup$ Given the answer below, this isn't a logic puzzle. $\endgroup$ – Kendall Frey Jun 24 '14 at 13:30
  • $\begingroup$ Ah, okay! I didn't see the "brainteaser" label. Thanks for the correction! $\endgroup$ – Thassa Jun 24 '14 at 14:39
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I take three eights as twenty-four, and can do $(2+2)(2+2+2)$ to get there with five twos. One or two of the plus signs can become multiplies. There are other solutions, depending on what operations are allowed. If we add division, we can multiply the above by $\frac 22$ and have solutions with seven twos. If concatenation and factorials are allowed, there are $22+2$ with only three, and $(2+2)!$ with just two.

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  • $\begingroup$ Oh wow, I mistyped. It's eight 2s, not eight 8s. Sorry! $\endgroup$ – Thassa Jun 24 '14 at 3:02
  • $\begingroup$ If I take three eights as $24$, I can do $(2+2)\times (2+2+2)$ for five. $\endgroup$ – Ross Millikan Jun 24 '14 at 3:05
  • $\begingroup$ Correct! But there's another answer that also works. Can you guess what it is? $\endgroup$ – Thassa Jun 24 '14 at 3:07
  • $\begingroup$ There are a number of others. One or two of the plus signs can change to times (as long as they aren't both in the second group). $(2+\frac 22)\times 2 \times 2 \times 2$ uses six, but still beats eight. Taking off mine, you can do $(2+2)(2+2+2)\cdot \frac 22$ and still have only seven-I'm sure there are many more. $\endgroup$ – Ross Millikan Jun 24 '14 at 3:13
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    $\begingroup$ It is confusing, since now accepted answer does not correspond to the question... $\endgroup$ – klm123 Jun 24 '14 at 6:00
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Here's another way of producing three eights from five twos:

$$2 \times 2 \times {222} = 888$$

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Using the two initial examples given:

$$(2*2*2*2) + (2*2*2)$$

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Here is another way also assuming $3*8=24$ using $5*2$:

$$(2+2+2)*2*2$$

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Or, three eights is 8 × 8 × 8, which is 512.

8 × 8 × 8 = 2(2×2) × 2(2×2) × 2.

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  • $\begingroup$ Or (using 5 2s like all the other answers): 2<sup>(2×2×2)</sup> × 2 $\endgroup$ – Klas Lindbäck Jun 11 '15 at 14:10

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