Wikipedia:Reference desk/Archives/Mathematics/2013 December 12

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December 12

All the 2s

My son was given this homework problem. He's turned it in now so this is no longer a homework question :)

The problem is to use only the number 2 and the operators +, -, x, / to get to a given value of n, say 13. e.g. (22+2+2)/2 = 13. What other ways of getting to 13 (since that was the number he was asked to get to) are there? And is there some kind of equation or something that would simplify the process and enable one to quickly find a few answers for a given value of n? Thanks, --Viennese Waltz 10:14, 12 December 2013 (UTC)

(2 + 2) x (2 + 2) - 2 - (2 / 2)

2 x 2 x 2 x 2 - (2 + 2 + 2) / 2

(2²² + 2^{2 + 2 / 2} + 2) / 2

(2 x 2) x (2 + 2 / 2) + 2 / 2

(2 x 2 x 2 - 2) x 2 + 2 / 2

(22 - 2 x 2) / 2 + 2²

(2 + 2 / 2)² + 2 x 2

((2 + 2)! + 2) / 2

2 x (2² + 2) + 2 / 2

((2 x 2²)² - 2^{2 x 2} + 2²) / 2²

10 versions of 13... EdChem (talk) 11:05, 12 December 2013 (UTC)

Excellent, thanks very much. --Viennese Waltz 11:15, 12 December 2013 (UTC)

(edit conflict)(EdChem beat me to an answer, and found more than I did, so ...) I'll just add an obvious solution without brackets: 2 x 2 x 2 + 2 x 2 + 2 / 2 (using algebraic order of operations). I don't thing there is any formula for answers, but writing n in base 2 will give a quick and easy solution. There is no limit to the number of different answers that could be constructed, given an unlimited number of 2s. Dbfirs 11:17, 12 December 2013 (UTC)

Thanks, and sorry but I don't get the bit about base 2. 13 in base 2 is 1101, how does that help me get to a solution? --Viennese Waltz 11:36, 12 December 2013 (UTC)

1101 base 2 = 2³ + 2² + 1 (base 10) = 2 x 2 x 2 + 2 x 2 + 2 / 2, the "obvious solution" to which Dbfirs referred. Any other number could be approached in the same way. EdChem (talk) 11:41, 12 December 2013 (UTC)

Did the problem limit the number of 2's? ROUND(n/2) * 2 - MOD(n/2)*2. Depending on whether n is odd or even tells whether you need to create a 1 with division with modulus. For n=13, ROUND(13/2)*2 - MOD(13/2)*2 = 7*2 - (1/2)*2=14-1=13. For 7, you use the same technique and reduce it until you have 2's and 2/2's . Division is the operator that lets you bring a 1 into the equation to create odd numbers. Using the above technique to continue intermediate (2*2*2)*2 - 2/2 - 2/2 - 2/2 . This will yield an all 2 solution. --DHeyward (talk) 12:18, 12 December 2013 (UTC)

No it didn't but he's only 9 so wouldn't have had a lot of patience with a lot of 2's anyway. Thanks for the help. --Viennese Waltz 13:28, 12 December 2013 (UTC)

The simple 22/2 + 2 hasn't been suggested. But puzzles of this type often disallow digit concatenation as in making "22" from two 2's. If the formulation said "use only the number 2" as you said, and not "use only the digit 2", then it would usually be implied that "22" is disallowed - at least when the target is adults. At age 9 they are probably not expected to see how base-dependent digit concatenation is different from standard operations like +, -, x, /. PrimeHunter (talk) 13:51, 12 December 2013 (UTC)

It was to use only the digit 2, 22 was allowed. --Viennese Waltz 14:00, 12 December 2013 (UTC)

(ec) Take 2/2 + 2/2 + 2/2 ... and add them as many times as you want, then subtract ... – 2/2 – 2/2 as many times as you need. This way you will get infinite set of solutions. --CiaPan (talk) 13:52, 12 December 2013 (UTC)

Another tricky approach: declare '13' to be in the base 5 and solve 13₅ = 2*2*2 = 2 + 2 + 2 + 2 = ...  :) --CiaPan (talk) 13:55, 12 December 2013 (UTC)

Or base 21? 13₂₁ = (2+2)!--Naraht (talk) 17:08, 12 December 2013 (UTC)

! is not one of the allowed operators. -- Jack of Oz ^{[pleasantries]} 05:22, 14 December 2013 (UTC)

Overkill

This sort of feels like someone has asked us to destroy a shack and we are considering which Nuclear Weapon to use. :)Naraht (talk) 17:29, 12 December 2013 (UTC)

I'm surprised no-one has given 22/2+2 which uses only four 2s. Dmcq (talk) 23:37, 13 December 2013 (UTC)

I wrote "The simple 22/2 + 2 hasn't been suggested". PrimeHunter (talk) 00:39, 14 December 2013 (UTC)

Sorry didn't see that. Dmcq (talk) 17:35, 15 December 2013 (UTC)

Here's one with a 2's, a 22, and a 222:     2^{22/2 - 2 x 2} - 222 / 2 - 2 x 2     :) EdChem (talk) 00:45, 14 December 2013 (UTC)

And with a 2222, 222, 22, and 2:     (2222 + 2²) / 2 + 2 x 2^{2 + 2} - (2 x 2² - 2^{2 - 2} - 2) x 222 - 22     EdChem (talk) 01:09, 14 December 2013 (UTC)

At the other extreme, there's one with 13x2^32 twos that only works on 32-bit digital computers, by exploiting integer overflow. Sławomir Biały (talk) 02:10, 14 December 2013 (UTC)

QED vs QEO vs QEI

In the Principia Mathematica 1729 English version scan on Google Books, most proofs end in Q.E.D. as expected. But then I saw Q.E.I. on p.231 (Prop. LXIV, Problem XL) and p.286 (Exam. 3), and Q.E.O. on p.232-233 (Case 1 & 2), among many other places with varying consistency. Many other statements' justifications end in neither Q.E.D. nor its variants.

Does anyone know what these statements mean (inb4 "Quod Erat Obvious"), and how they are distinct from each other? SamuelRiv (talk) 19:11, 12 December 2013 (UTC)

I think I recall a Q.E.F. for quod erat faciendum, for geometric constructions, but I can't guess what I or O would be. --Trovatore (talk) 19:50, 12 December 2013 (UTC)

Ah, I think Q.E.I. is quod erat inveniendum, "that which was to be discovered". --Trovatore (talk) 22:13, 12 December 2013 (UTC)

By the way, the best rendering of Q.E.D. is quod ego dico, "because I say so". --Trovatore (talk) 22:14, 12 December 2013 (UTC)

quod erat ostendendum DTLHS (talk) 04:50, 13 December 2013 (UTC)

I've also seen Q.E.A. = Quod Est Absurdum = which is absurd, used with indirect proofs. It should be noted that all of these phrases are pretty much obsolete, despite being kept alive in popular culture, for example Q.E.D. (BBC TV series). --RDBury (talk) 05:04, 13 December 2013 (UTC)

Our List of Latin abbreviations only mentions Q.E.C., Q.E.F. and Q.E.I. —Kusma (t·c) 09:53, 13 December 2013 (UTC)