with NUMBER SUMS |

Here is an interesting problem that only involves basic arith- metic, but with an unusual twist. Find 5 numbers such that when each number is multiplied by the sum of the remaining 4 numbers, one of the following values will result:44 63 95 108 128 WOW! Did you get that? I hope you didn't get too confused. All it really says is: let's take five numbers -- say 1, 3, 4, 7, and 8 -- add any four of them, then multiply the sum by the remaining number. In this sample, try adding 1, 3, 7, & 8, getting 19; now multiply 19 by 4, getting 76. Now it seems easy enough, right? But that's because we knew the 5 numbers in advance. This is another of those problems I call "JEOPARDY" problems. When you are given the "answer", you must find the "question". You know? Perhaps in this case I'll give you some pretty heavy hints, as it does seem to be a "tough nut to crack" this time. We begin by forming factor T-charts for each of the 5 products. 4463951081281 44 1 63 1 95 1 108 1 128 2 22 3 21 5 19 2 54 2 64 4 11 7 9 3 36 4 32 4 27 8 16 6 18 9 12 Next locate those factor pairs that have the same sum; in this case it should be 24. The smaller members of the required pairs are indicated in red. They are the 5 numbers which solve the original problem. Perhaps you're wondering how I knew the common sum should be 24. It's really quite simple if you look at the chart that has the fewest entries: 95. Since it only has two factor pairs, whose sums are 96 and 24, this shortens our search quite a bit. Then since the first two charts (for 44 and 63) only have factor sums less than 96, this leaves only 24 to be considered. The rest, as they say, is "easy as pie".PROOFThe reason the procedure works out so well can be proved by simple algebra. Assume the five numbers are a, b, c, d, and e. There- fore, we havea(b + c + d + e) = 44 b(a + c + d + e) = 63 etc. Notice that the sum of the two factors in each equation is always the same, namelya + b + c + d + e, the sum of the desired numbers. Hence, the procedure of finding the factor pairs possessing equal sums becomes rather obvious.Here are some problems that can be presented to your math class after the procedure has been explained. Many more can be prepared as needed. 1. 245 297 152 320 360 2. 220 328 468 490 360 3. 111 319 375 204 175 4. 522 742 816 570 882## Teacher Notes:

Variations1. Of course, more (or fewer) numbers than five can serve as the basis. It only depends on student interest, level, time, etc. 2. A problem can be given with one of the products missing. The problem will be considered as solved when that product has been found. (This is easier than it sounds at first.) 3. A competition between 2 students can be set up. Each player chooses, say 6 numbers, and prepares the set of products. A limit on the size of the numbers should be agreed upon in advance. Then they trade number sets and the first to declare the other's 6 numbers is the winner.

Final NotesSince the method of solution is being directly presented to the student from the start, the pedagogical objectives of this activity are these: a) Important practice in finding factor pairs of whole numbers---a skill much needed in algebra courses; b) Experience in following directions in a non-tradi- tional problem setting; and c) Drill in systematic-search techniques for problem solving. d) NOTE: Calculators should be considered as an option, especially if large numbers are employed.

REFERENCE: Maurice Kraitchik, MATHEMATICAL RECREATIONS, Dover Pub., 1953. p. 46. Answers: (1) 4, 7, 9, 10, 12 (2) 5, 8, 9, 13, 14 (3) 3, 5, 6, 11, 15 (4) 9, 10, 14, 16, 18 tt(1/5/86)

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