So I'm trying to use four $2$'s to make the number $7$. I can use parenthesis, addition, subtraction, indicies, addition, division, multiplication
edit: also have the same problem with $9$ but I'll try that myself if someone could help with $7$
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Do you know if it's possible? – brick Sep 09 '15 at 20:43
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1@brick it's from a friends sister's homework and she is in grade 8 so I don't know why her teacher would give her an impossible task. I'm assuming yes. – iphone7 Sep 09 '15 at 20:44
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Are you allowed to use powers? – Oussama Boussif Sep 09 '15 at 20:45
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@OussamaBoussif Yes, indices are allowed (so $2^2$ or $2^{2 + 2/2}$ etc) – iphone7 Sep 09 '15 at 20:45
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Ah okay. Using only four 2 makes it challenging – Oussama Boussif Sep 09 '15 at 20:46
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Can you put multiple $2$s together, i.e. can you use two+ digit numbers? – user170231 Sep 09 '15 at 20:47
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1Using five 2s, I got $7=\frac{2^{2^2}-2}{2}$. Don't know how to do it in four, though. – Cataline Sep 09 '15 at 20:48
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Well I think that somewhere you have to add or subtract $1$, which is made by $2$ $2$'s and you are left with only two more to make $6$ or $8$. Don't think this is possible... – brick Sep 09 '15 at 20:49
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Or $\frac{22}{2} - 2-2$..... (with five $2$'s) – Sep 09 '15 at 20:49
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@user170231 Could you give an example? Her teacher only told her to use the things listed in the OP. – iphone7 Sep 09 '15 at 20:49
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@Cataline Yeah that's all I could find too. – iphone7 Sep 09 '15 at 20:50
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3Using five 2 s we can find multiple answers if concatenating isnt allowed – Oussama Boussif Sep 09 '15 at 20:50
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@iphone7 see John Ma's comment (I was thinking $-2^2$ in place of $-2-2$). – user170231 Sep 09 '15 at 20:50
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@user170231 $-2^2$ is the same number of $2$'s as $-2 - 2$ so it doesn't make a difference. Just like $22$ still counts as two $2$'s – iphone7 Sep 09 '15 at 20:52
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2Famous last words but I think this is impossible. The only way I can think to get an odd number is $2/2$ – Karl Sep 09 '15 at 20:52
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2For $9$ one can do $(2 + \frac{2}{2})^2$. – Sep 09 '15 at 20:55
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4$2\times2\times2-2^0$… it is well known that $0$ doesn't count as number in grade 8. :-) – Intelligenti pauca Sep 09 '15 at 20:56
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1And for $5$, $2^2+\frac{2}{2}$. Is it possible to reach the middle ground? – user170231 Sep 09 '15 at 20:57
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1@JohnMa Thanks for Ma. They had an assignment to do it with four threes to make up the numbers from 1-10, and do the same with four twos. I've checked over her work for everything but she's still stuck on $7$. $2^0$ I don't believe counts, but if all else fails I'll just put that down lol. – iphone7 Sep 09 '15 at 21:00
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I now have insomnia. Any hope of a proof of the futility of the task? If the only way of getting odd is $2/2$ that limits options. I'd love to see a proof ( and sleep again) – Karl Sep 09 '15 at 21:08
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1$\frac{1}{2}\sum_{|k|\leq 2, k \in \mathbb{Z}} k^2 + 2 = 7$ =D.. But actually $k\neq 2$... :/ (but again, only the digit 2 appears, and it is pretty much only addition division and exponention) – servabat Sep 09 '15 at 21:26
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2$(.2)^{-2/2}+2$ can be found here. – mathlove Sep 09 '15 at 21:29
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Actually, $\frac{1}{2} \sum_{|k|\leq 2, k \in \mathbb{Z}} 2 + 2 = 7$ also works, where no other digits than $2$ appears. – servabat Sep 09 '15 at 21:43
1 Answers
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I previously posted an answer listing every possible real number obtainable this way (assuming $22$ is allowable by using two $2$'s. In the revision history, one can find a table where I assumed that $22$ was not allowed). $7$ is not on the list, hence cannot be thusly formed, assuming my program works (which it does). As has been noted in comments and is present on the list, $9$ may be formed as $$9=\left(2+\frac{2}2\right)^2.$$
Without resulting to brute force, one may prove that $7$ cannot be formed by noting that the only operations capable of forming anything but even numbers when given even numbers are division and negative exponents. We can rule out negative exponents, because that would leave us with a non-integer and, to create an integer, we would either add that with another non-integer (which would take up all $4$ operations and would be $2^{-2}+2^{-2}\neq 7$), divide by a non-integer (which similarly gives $\frac{2^{-2}}{2^{-2}}\neq 7$), subtract a non-integer (which gives $2^{-2}-2^{-2}\neq 7$) or multiply by an integer (which, seeing that the only possible non-integers with less than $4$ twos are $2^{-2}$ and $2^{-2-2}=(2+2)^{-2}$, doesn't give any useful forms. The casework is left as an exercise). We can similarly eliminate the possibility of division yielding a non-integer. Thus, we can conclude that division occurs somewhere, yielding an odd number. If $\frac{2}2$ occurs, then our numbers are $2$, $2$, and $1$ where the largest number that can occur without an exponent is $2+2+1<5$, which is too small. So exponentiation must occur, but for this to be helpful, either the base or the exponent must be at least $3$, giving that $2^3$ and $3^2$ are the only integers numbers larger than $5$ obtainable if we include $\frac{2}2$. Thus, $\frac{2}2$ may not occur either. One may find that the only other form yielding an odd number is $\frac{2+2}{2+2}$ (or many other expressions evaluating to $\frac{4}4$). This is, more or less, a proof.
Milo Brandt
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In lieu of an answer, I would recommend your write this whole response in the space provided on the homework. – Milo Brandt Sep 09 '15 at 21:19