# Download Mathematical Recreations and Essays (11th Rev. Ed) by W.W. Rouse, BALL PDF

By W.W. Rouse, BALL

This vintage paintings deals ratings of stimulating, mind-expanding video games and puzzles: arithmetical and geometrical difficulties, chessboard recreations, magic squares, map-coloring difficulties, cryptography and cryptanalysis, even more. "A needs to so as to add for your arithmetic library." — the maths instructor. Index. References for extra learn. contains one hundred fifty black-and-white line illustrations.

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Additional info for Mathematical Recreations and Essays (11th Rev. Ed)

Example text

I content myself with noting the results when n does not exceed 10. When fa = 3, z = 1; when fa = 4, z = 2; when fa = 5, z = 13; when fa = 6, z = 80; when »=7, z=579; when »=8, z=4738; when »=9, z = 43387; and when fa = 10, z = 439792. BACHET'S WEIGHTS PROBLEM: It will be noticed that a considerable number of the easier problems given in the last chapter either are due to Bachet or were collected by him. in his classical Problemu. Among the more difficult problems proposed by him. was the determination of the least number of weights which would serve to weigh any integral number of pounds from 1 lb.

Now, since he goes in the opposite direction to that in which the hands of the watch move, he has to go over (n - m) hours to reach the hour m: also it will make no difference if, in addition, he goes over 12 hours, since the only effect of this is to take him once completely round the circle. Now (n 12 - m) is always positive, since n is positive and m is not greater than 12, and therefore if we make him pass over (n + 12 - m) hours we can give the rule in a form which is equally valid whether m is greater or less than n.

Fry,· who supposes that the bank's wealth is limited, say to £1000000. There is then a probability (1/2)11 that the player will receive £2"-1 at the nth throw, only as long as 2,,-1 < 1000000; thereafter, he will receive merely £1000000. Since JO ~ (,)" 1 • 2,,-1 + ~ (,)" 1000000 = 10 + (1)10 1000000 11 = 10·9536 ... , the player's expectation is about £10 19,. -8 reasonable amount. OrBEB QUESTIONS ON PRoBABILITY Here is a result (due to H. Davenport) which many people find surprising. If you know more than 23 people's birthdays, it is more likely than not that two of them are the same (as to day and month).