6 SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS

DID. Because EVE must be smallerthan 303, E is 1 or 2. Of the 14possibilities (121, 141, … , 292) only242 produces a decimal fitting .TALKTALK…, in which all the digitsdiffer from those in EVE and DID.

The unique answer is 242/303 = .798679867986… If EVE/DIDis not assumed to be in lowest terms,there is one other solution, 212/606 =.349834983498…, proving as JosephMadachy has remarked, that EVEdouble-talked.

Three SquaresThere are many ways to prove thatangle C in the figure is the sum ofangles A and B. Here is one (see Figure2). Construct the squares indicated byred lines. Angle B equals angle Dbecause they are corresponding anglesof similar right triangles. Since angles Aand D add to angle C, B can be substi-tuted for D, and it follows immediatelythat C is the sum of A and B.

This little problem produced a flood ofletters from readers who sent dozens ofother proofs. Scores of correspondentsavoided construction lines by makingthe diagonals equal to the square rootsof 2, 5, and 10, then using ratios to findtwo similar triangles from which thedesired proof would follow. Othersgeneralized the problem in unusualways.

You have six weights. One pair is red,one pair white, one pair blue. In eachpair, one weight is a trifle heavier thanthe other but otherwise appears to beexactly like its mate. The three heavierweights (one of each color) all weighthe same. This is also true of the threelighter weights.

In two separate weighings on a balancescale, how can you identify which is theheavier weight of each pair?

ANSWERSTalkative EveAs stated earlier, to obtain the simplestfraction equal to a decimal of n repeat-ed digits, put the repeating period overn 9’s and reduce to its lowest terms. Inthis instance TALK/9,999, reduced toits lowest terms, must equal EVE/DID.Consequently, DID is a factor of 9,999.Only three such factors fit DID: 101,303, 909.

If DID = 101, then EVE/101 =TALK/9,999, and EVE = TALK/99. Rearranging terms, TALK = (99)(EVE).EVE cannot be 101 (since we assumed101 to be DID), and anything largerthan 101, when multiplied by 99, has afive-digit product. And so DID = 101 isruled out.

If DID = 909, then EVE/909 =TALK/9,999, and EVE = TALK/11.Rearranging terms, TALK = (11)(EVE). Inthat case, the lastdigit of TALK wouldhave to be E. Sinceit is not E, 909 isalso ruled out.Only 303 remainsas a possibility for

Talkative Eve

This cryptarithm (or alphametic,as some puzzlers prefer to callthem) is an old one of unknown

origin, surely one of the best and, Ihope, unfamiliar to most readers:

The same letters stand for the samedigits, zero included. The fractionEVE/DID has been reduced to itslowest terms. Its decimal form has arepeating period of four digits. Thesolution is unique. To solve it, recall thatthe standard way to obtain the simplestfraction equivalent to a decimal of nrepeating digits is to put the repeatingperiod over n 9’s and reduce thefraction to its lowest terms.

Three SquaresUsing only elementary geometry (noteven trigonometry), prove that angle Cin Figure 1 equals the sum of angles Aand B.

I am grateful to Lyber Katz for thischarmingly simple problem. He writesthat as a child he went to school inMoscow, where the problem was givento his fourth-grade geometry class forextra credit to those who solved it.“The number of blind alleys theproblem leads to,” he adds, “isextraordinary.”

Red, White, and Blue WeightsProblems involving weights andbalance scales have been popularduring the past few decades. Here is anunusual one invented by Paul Curry,who is well known in conjuring circlesas an amateur magician.

EVE

DIDTALKTALKTALK= . ...

Talkative EveMartin Gardner

This article first appeared in the April 1996 issue of Math Horizons.

Figure 1. Prove that angle A plus angle B equals angle C.

“That’s me,” I said.

She smiled and held out a hand. “I’mThelma O’Keefe. We were in the samealgebra 101 class.”

We shook hands.

“You won’t remember me,” she said. “Iwas fat in those days, and shy, and notvery pretty.”

“That’s hard to believe,” I said. “Youlook gorgeous now.”

“Well, thank you, kind sir,” she said,

SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS 7

Charles Trigg published 54 differentproofs in the Journal of RecreationalMathematics Vol. 4, April 1971, pages90–99. A proof using paper cutting, byAli R. Amir-Moéz, appeared in the samejournal, Vol. 5, Winter 1973, pages 8–9.For other proofs, see Roger North’scontribution to The MathematicalGazette, December 1973, pages 334–36, and its continuation in thesame journal, October 1974, pages212–15. For a generalization of theproblem to a row of n squares, seeTrigg’s “Geometrical Proof of a Resultof Lehmer’s,” in The FibonacciQuarterly, Vol. 11, December 1973,pages 539-40.

Red, White, and Blue WeightsOne way to solve the problem of sixweights—two red, two white, and twoblue—is first to balance a red and awhite weight against a blue and a whiteweight.

If the scales balance, you know thereare a heavy and a light weight on eachpan. Remove both colored weights,leaving the white weights, one on eachside. This establishes which white

weight is the heavier. At thesame time it tells you whichof the other two weightsused before (one red, oneblue) is heavy and which islight. This in turn tells youwhich is heavy and which islight in the red-blue pair notyet used.

If the scales do not balanceon the first weighing, youknow that the white weighton the side that went downmust be the heavier of thetwo whites, but you are stillin the dark about the redand the blue. Weigh theoriginal red against the mate of the original blue (orthe original blue against the mate of theoriginal red). As C. B. Chandler (whosent this simple solution) put it, theresult of the second weighing, plus thememory of which side was heavier inthe first weighing, is now sufficient toidentify the six weights.

For readers who liked working on thisproblem, Ben Braude, a New York City

dentist and amateur magician, devisedthe following variation. The six weightsare alike in all respects (including color)except that three are heavy and threelight. The heavy weights weigh thesame and the light weights weigh thesame. Identify each in three separateweighings on a balance scale.

DOI: 10.4169/194762110X524684

Figure 2. Construction for proof of the three-squaretheorem.

Several years ago I was agraduate student at theUniversity of Chicago. I was

working on my doctorate in physics,about possible ways to test superstringtheory, when my brother in Tulsa diedsuddenly from a heart attack. Bothparents had earlier passed away. Afterthe funeral I drove past my past,marveling at the enormous changesthat had taken place since I grew upthere. I drove by the red brick building,now an enormous warehouse, that had

once been Tulsa Central High. Mygrades in history, Latin, and English litwere low, but I was good in math andhad a great physics teacher. He wasmainly responsible for my majoring inphysics after a scholarship took me tothe University of Chicago.

While I was having dinner at a popularrestaurant on the corner of Main andSixth streets, the waitress stared at mewith a look of surprise. “Are youMichael Brown?”

Short Fiction

Superstrings and ThelmaMartin Gardner