Marilyn Lost in the River

Marilyn is Wrong Copyright © 2000 Herb Weiner. All rights reserved.

Ask Marilyn ® by Marilyn vos Savant is a column in Parade Magazine, published by PARADE, 711 Third Avenue, New York, NY 10017, USA. According to Parade, Marilyn vos Savant is listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."

In her Parade Magazine column of October 1, 2000, a reader poses a puzzle that Marilyn correctly solves using seven steps.

In her Parade Magazine column of November 19, 2000, a different reader congratulates Marilyn for solving with "pure logic" a problem that is "not solvable in a straightforward mathematical way" because it "appears to have five unknowns and only four equations."

Rather than point out that the problem is in fact solvable in a straightforward mathematical way using five equations of five unknowns, Marilyn neglects to challenge the reader's claims, and points out that she's heard from 462 readers who believe her answer is wrong.

Sorry Marilyn

John Mallinckrodt <> provided the following solution:

As much as I truly enjoyed the motorboat problem and as much as I also prefer conceptual (what you seem to call "logical") to strictly mathematical solutions, there is certainly no mystery about the mathematical solution to this problem. Are you truly under the impression that it requires solving a set of four equations for five unknowns and that, somehow, "logic" allows you to succeed where math fails? Or was your answer simply designed to lure more incorrect solutions from "mathematicians and other professionals"?

If the former, please allow me to disabuse you of that silly notion ...

The first time the boats pass we have

  dA1 + dB1 = w,            (Eq 1)

where dA1 and dB1 are the distances traveled by the two boats and where w is the width of the river.

We can assume (without loss of generality) that boat A is the one that is heading for the shore that is 700 yards away. Thus,

  w - dA1 = 700.            (Eq 2)

Since you specified that both boats "continue to the opposite shore, then turn around and start moving toward each other again," then, when they pass the second time we have

  dA2 + dB2 = 3 w,          (Eq 3)

where dA2 and dB2 are the new (longer) distances traveled by the two boats.

Because of the previous assumption it now will be boat A that is 300 yards from the shore. So

  2 w - dA2 = 300.          (Eq 4)

Finally, since the boats travel at constant speeds,

  dA2/dA1=dB2/dB1.          (Eq 5)

Thus, we have five equations in five unknowns.

Because of the nonlinearity introduced by the constant speed constraint, there are two mathematical solutions:

  w   =    0
  dA1 = -700
  dB1 =  700
  dA2 = -300
  dB2 =  300


  w   = 1800
  dA1 = 1100
  dB1 =  700
  dA2 = 3300
  dB2 = 2100.

However, we easily rule out the first mathematical solution on physical grounds and are left with the second solution. Q.E.D.

It may be interesting to note that, if you had not specified that each boat reaches its opposite shore before the second passing, Eq 3 could just as easily read

  dA2 - dB2 = w,            (Eq 3A)

implying that boat A moves faster and catches boat B before B reaches B's opposite shore, or

  dB2 - dA2 = w,            (Eq 3B)

implying that boat B moves faster and catches boat A before A reaches A's opposite shore.

In the first case the physical solution would have involved the number R1 = 12 + sqrt(102) = 22.0995... and would have been

  w   = 100 R1         = 2209.95...
  dA1 = 100 R1 - 700   = 1509.95...
  dB1                  =  700
  dA2 = 200 R1 - 300   = 4119.90...
  dB2 = 100 R1 - 300   = 1909.95...

and in the second case the physical solution would have involved the number R2 = [19 + sqrt(235)]/3 11.4432 ... and would have been

  w   = 100 R2         = 1144.32...
  dA1 = 100 R2 - 700   =  444.32...
  dB1                  =  700
  dA2 = 200 R2 - 300   = 1988.64...
  dB2 = 300 R2 - 300   = 3132.97... last updated November 25, 2000 by