THE FERRY BOATS
Two ferry boats ply back and forth across a river with constant speeds, turning at the banks without loss of time. They leave opposite shores at the same instant, meet for the first rime some 700 feet from one shore, continue on their way to the banks, return and meet for the second time 400 feet from the opposite shore. [Without using pencil and paper] determine the width of the river.
Solution: By the time of their first meeting, the total distance that the two boats have traveled is just the width of the river. It may take one mildly by surprise, however, to realize that, by the time they meet again, the total distance they have traveled is three times the width of the river. Since the speeds are constant, the second meeting occurs after a total time that is three times as long as the time for the first meeting. In getting to the first meeting, ferry A (say) traveled 700 feet. In three times as long, it would go 2100 feet. But, in making the second meeting, A goes all the way across the river and then back 400 feet. Thus the river must be 2100 – 400 = 1700 feet wide.
Source: American Mathematical Monthly, 1940, p. 111, Problem E366, proposed by C.O. Oakley, Haverford College, solved by W.C. Rufus, University of Michigan.
ROLLING A DIE
A normal die bearing the numbers 1, 2, 3, 4, 5, 6 on its faces is thrown repeatedly until the running total first exceeds 12. What is the most likely total that will be obtained?
Solution: Consider the throw before the last one. After the first throw the total must be either 12, 11, 10, 9, 8, or 7. If it is 12, then the final result will be either 13, 14, 15, 16, 17, or 18, with an equal chance for each. Similarly, if the next to last total is 11, the final result is either 13, 14, 15, 16, or 17, with an equal chance for each; and so on. The 13 appears as an equal candidate in every case, and is the only number to do so. Thus the most likely total is 13.
In general, the same argument shows the most likely total that first exceeds the number n (n > 5) is n + 1.
Source: American Mathematical Monthly, 1948, p. 98, Problem E771, proposed by C.C. Carter, Bluffs, Illinois, solved by N.J. Fine, University of Pennsylvania.
RED AND BLUE DOTS
Consider a square array of dots, colored red or blue, with 20 rows and 20 columns. Whenever two dots of the same color are adjacent in the same row or column, they are joined by a segment of their common color; adjacent dots of unlike colors are joined by a black segment. There are 219 red dots, 39 of them on the border of the array, none at the corners. There are 237 black segments. How many blue segments are there?
Solution: There are 19 segments in each of 20 rows, giving 19 x 20 = 380 horizontal segments. There is the same number of vertical segments, giving a total of 760. Since 237 are black, the other 523 are either red or blue.
Let r denote the number of red segments and let us count up the number of times a red dot is the endpoint of a segment. Each black segment has one red end, and each red segment has both ends red, giving a total of 237 + 2r red ends.
But, the 39 red dots on the border are each the end of 3 segments, and each of the remaining 180 red dots in the interior of the array is the end of 4 segments. Thus the total number of times a red dot is the end of a segment is 39(3) + 180(4) = 837. Therefore 237 + 2r = 837, and r = 300.
The number of blue segments, then, is 523 – 300 = 223.
Source: American Mathematical Monthly, 1972, p. 303, Problem E2344, proposed by Jordi Dou, Barcelona, Spain.
A PERFECT 4TH POWER
Prove that the product of 8 consecutive natural numbers is never a perfect fourth power.
Solution: Let x denote the least of 8 consecutive natural numbers. Then their product P may be written
P = [x(x + 7)][(x + 1)(x + 6)][x + 2)(x + 5)][(x + 3)(x + 4)] = (x2 + 7x)(x2 + 7x + 6)(x2 + 7x + 10)(x2 + 7x + 12).
Letting x2 + 7x + 6 = a, we have
P = (a – 6)(a)(a + 4)(a + 6) = (a2 – 36)(a2 + 4a) = a4 + 4a(a + 3)(a – 12).
Since a = x2 + 7x + 6 and x ≥ 1, we have a ≥ 14 and a – 12 is positive.
Hence P > a4.
However, P = a4 + 4a3 – 36a2 – 144a reveals that P is less than (a + 1)4 = a4 + 4a3 + 6a2 + 4a + 1.
Hence a4 < P < (a + 1)4, showing that P always falls between consecutive fourth powers and never coincides with one.
Source: American Mathematical Monthly, 1936, Problem 3703, proposed by Victor Thébault, Le Mans, France, solved by the Mathematics Club of the New Jersey College for Women, New Brunswick, New Jersey.
1 comment:
Re: ROLLING A DIE
Isn't that logic only valid if "the throw before the last one" would have perfectly equal likelihood to be any of those six values (7, 8, 9, 10, 11, and 12)?
I believe the total before the final roll would be more likely to be 12 than 7. Think of it this way: if you're at 7, there's a 1/6 chance you roll a 6 and finish with 13. But a 5/6 chance you roll something else and the penultimate total becomes something other than 7.
Looking at 8, 9, 10, and 11 . . . they all also have a chance at actually becoming 12 as the penultimate total.
Then the possible outcomes for the final roll after 12 (13, 14, 15, 16, 17, and 18) should have more weight than the 13 being rolled after 7. I don't know how much more weight, as the math is a bit much for me right now. But the fact that this wasn't mentioned leaves me wondering if the solution is as simple as it was posited.
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