Democratic Sentinel, Volume 19, Number 36, Rensselaer, Jasper County, 13 September 1895 — AN OLD WOMAN’S EGGS. [ARTICLE]

AN OLD WOMAN’S EGGS.

Problem that Is Racking the Braiae of the- Sleepy Philadelphians. Some as- yet unidentified Philadelphian came momentarily out of his accustomed, state of peaceful trance a few days ago and proceeded to utilize the half hour-or so that elapsed before he went to sleep again by evolving the following highly characteristic problem: “An old. woman who was selling eggs, when! asked how many she had in her replied that if she took them out by twos, threes, fours, fives or sixes in each; case there would be left one odd egg la each basket, but if she removed them: by sevens there would be none remainlug. How many eggs did the old woman have?” Since then Philadelphia has been stirred as never before in its history. All the local papers accepted the-problem as one well calculated to test the mathematical knowledge of their readers—id which supposition they seem to have made no mistake—and have been devoting much space to it ever since. Curiously enough, no residenit of the “City of Dead Calm” noticed that the "sum” as stated is insoluble:. After fumbling about in the wilderness of figures for a while, the brighter-minded customers of Mr. Wanamaker and a few of his cash girls hit upon 301 as the right answer. The accuracy of this was at once admitted by all the papers and peace again brooded, on the Schuylkill. A troublesome fellow who vehemently declared that 2,401 is the correct number was sternly silenced by the information that no basket could hold 2,401 eggs, and no .old woman could carry them. A New York schoolboy to whom this enigma, was submitted looked it over for a moment, and then asked, “Why didn’t they say, ‘A woman had some eggs, how many did she have?’ ” An amateur mathematician treated the matter more gravely: “Put in exact terms,” he-said, “the problem is to find a common multiple of 2,3, 4; 5 and 6, which, increased by 1, in divisible by 7. There is not enough paper in the world, to hold all the correct answers to that question. Therefore, strictly speaking,, there is no answer. Any multiple of 60, the least common multiple of 2,3,. 4, 5 and 6, which, by the addition of 1, becomes a multipleof 7, fulfills the conditions. F&r instance, GO, multiplied by 5, 12, 17, 10 and 26, gives 300, 420, 720, 1,020, 1,140 and 1,560, and 1 added to any of these makes a- number that will do. Thereare innumerable millions more of them.” Evidently Philadelphia must try again if she wants to enter the first class in arithmetic.—New York Times,