Mathematics and Society - OS X Lion Server

DIDACTICSTEACHING VIA PROBLEM SOLVINGLooking back is often neglected. Certainly students should ask themselves,"Is this answer reasonable?," and apply some combination of common sense (the girlrode her bike 200 km/h!?) and/or number sense and mental arithmetic. Perhaps mostimportant in the looking back phase is recognizing the problem-solving process weused; for example, in Problem 5 we looked at several simpler "problems" to generatea pattern which enabled us to predict the solution. We could also try to think ofsimilar problems which we think we could now solve. This requires some care (doesit matter that all of 5, 7 and 999 were odd? that 5 and 7 are primes?) • Many problemswill not suggest very imaginative ideas to middle schoolers, but asking theqtl~stiondoes remind the students that PERHAPS THINGS OTHER THAN ANSWERS ARE ALSOIMPORTANT. Finally, looking for another solution will emphasize to the studentsthat quite often there is more than one way to solve a problem.Another example. Several illustrations of problem solving processes with geometryproblems are given in the teaching emphasis ProbZem SoZving, in GEOMETRY &VISUALIZATION, Mathematics Resource Project. Let us look at one more numericalproblem here.(\Problem 6Each different letter represents a different digit.scrAPPAcrSx P and x SAAROOAARt'lN Find each letter's value.(Understanding the problem. We want to multiply a four~digit number (ScrAP) by itslast digit (P) to get a 5-digit number (AARON), which is the same number we wouldget if we reversed the digits of the original number (PAOS) and multiplied it by thenew final digit (S). Different letters stand for different digits, and we want tofind their values. (That was a mouthful, but it illustrates the power of symbolismin mathematics!)26Devising and carrying out a plan.With so many letters in the problem, perhaps wemight first try "Solve part of the problem." Which part? Since the multiplicationalgorithm starts with the units' digits, we might look there. Aha! (?) The lastdigit of P x P is the same as the last digit of S x S--the digit N.now help to narrow the possibilities.Looking at the last digits •Number skills1 x I = 1 4 x 4 = 16 7 x 7 = 492 x 2=!!. 5 x 5 = 2;? 8 x 8 643 x 3 = 2- 6 x 6 = 36 9 x 9 = 81 (gives the following possibilities for P and S:I and 9 or 2 and 8 or 3 and 7 or 4 and 6.