Mathematics and Society - OS X Lion Server

DIDACTICSTEACHING VIA PROBLEM SOLVINGNow what? 8ince we have some information about P and S,we might pursue them.The given data tell us that when we multiply 8 x Pand possibly add it to any digit carried from P x 0 or8 x A, we get AA, a double digit number different fromSOAPx PAARtltJeither PP or 88. 8ubstituting the possibilities generated for P and 8:Similarly,(1 x 9) +(2 x 8) += AA:= AA:(1 x 9) + 1 = 11, which can be ruled out sinceP or 8 =1, so A cannot = 1 also.(3 x 7) + 1 = 22 and (4 x 6) + .2. =12.(2 x 8) + ~= 12 (Why can this be ruled out?)AARGI\lThis last possibility can be eliminated (in a one-digit-times-one-digit calculation,what is the greatest number that might be carried?).P = 7, 8 = 3, A = 2 or P = 3, 8 = 7, A = 2.and3 2x2 277980, we haveThe former does not work (try it, keeping in mind that each different letter representsa different digit).8 1'1 A P7 2 3x 32 2 9AARI'1NTherefore,8ince in the first case 3 x 2 = 6, /J must = 6. The rest is easy . . .7623 3267x 3 and x 722869 22869Looking back. We should be sure to check that our result does indeed give correctcalculations and assign different digits to different letters. We might review howwe proceeded in this problem: We tried to solve part of the problem and were able tobuild on the possibilities generated to get a solution; we obtained information byconsidering units digits of products and the possible "carries" in the multiplicationalgorithm; we ruled out some possibilities by the conditions of the problem and othersby trial and error. Anyone of thesemight come in handy in a similar problem.As another form of looking back,BATH 51.\2.8 MAP~ Ll.~ 7 X P3799G AARONwe might try to make up a similar problem.Or we might ask ourselves, Isthere another method of solution?27