The most common electrical resistors commercially available are a series of twelve basic resistance values, multiplied by arbitrary powers of ten. This is called the "E12 series", consisting of resistors with the following values (ohms): 10 12 15 18 22 27 33 39 47 56 68 82 100 120 150 180 220 270 330 390 470 560 680 820 1000 1200 1500 1800 2200 2700 3300 3900 4700 5600 6800 8200 etc.... The E12 series extends down into fractional values, i.e., 8.2, 6.8, 5.6, and so on, but we will not consider these. There also exists an "E24 series" that provides 24 resistances per decade, but those are less widely available. Noting that the total resistance of a set of resistors R1, R2, ..., Rn combined in parallel is 1 1 1 1 ---- = ---- + ---- + ... + ---- Rtot R1 R2 Rn we might ask which integer-valued resistances can be produced by combining distinct integer-valued resistances in parallel from the standard E12 series. Of course, if 1/R equals a sum of unit fractions then R must divide the product of the denominators, which implies that R cannot be synthesized by means of 2 or more distinct E12 resistors in parallel if R is divisible by any prime other than those that appear in the E12 sequence, namely, 2, 3, 5, 7, 11, 13, 17, 41, and 43. It's interesting that this covers the first seven primes, even though it's doubtful that the E12 values were chosen with prime factors in mind. According to the commercial literature, the E24 values were defined as the closest integers to 10r^k, k=0,1,2,..,23, where r is the 24th root of 10, although this doesn't match all the values. Even if R is entirely composed of "E12 primes" it may not be possible to synthesize it by just TWO resistors in parallel. It may require three or more. The only "fundamental" integer values of R that are expressible as a sum of just two unit fractions with E12 denominators are those shown below: 1/R = 1/A + 1/B R A B ----- ------ ------- 6 10 15 20 22 220 30 33 330 72 120 180 88 120 330 99 180 220 102 120 680 108 180 270 132 220 330 975 1000 39000 1125 1200 18000 1476 1800 8200 1485 2700 3300 6875 10000 22000 9375 10000 150000 16875 18000 270000 17875 33000 39000 103125 150000 330000 193875 330000 470000 I call these "fundamental" because we can multiply any of them by any power of 10. For example, since 1/99 = 1/180 + 1/220 it's obvious that 1/990 = 1/1800 + 1/2200, and so on. However, although the above solutions are fundamental in this sense, they are not all "primitive", because they may share common E12 factors. For example, the solution [103125, 150000, 330000] is just 15 times the solution [6875, 10000, 22000]. To create other fundamental solutions we must allow more than two resistors in parallel. Here's a summary of the "most economical" solutions I've found for the first several values of R: R denominators --- ------------------------------------------------------------- 3 10 12 15 22 33 220 330 4 10 12 15 5 10 12 100 150 6 10 15 7 10 56 100 120 150 8 10 100 120 150 9 18 22 180 220 10 10 11 12 220 330 12 12 13 15 100 3900 14 33 100 120 150 180 220 330 1000 1500 1800 2200 5600 12000 18000 15 15 16 18 180 1200 1800 17 33 68 120 330 680 1200 6800 18 18 19 impossible 20 22 220 21 33 100 330 560 1000 1200 1500 22 22 23 impossible 24 33 120 330 25 33 150 330 26 39 100 390 3900 27 27 etc. This raises some interesting questions. For example, is the above expression for 14 ohms optimal, or can it be done with fewer than 14 resistors? In general, is it possible to express 1/R as a sum of inverses of distinct "E12 integers" for any R (greater than 1) that is composed entirely of E12 primes? It's easy to see that we can't synthesize 1 ohm from any number of distinct E12 resistors in parallel, because the infinite sum of the inverses of all these resistors implies that the least resistance that can be produced by such a network is 31873589748 ----------- ohms = 1.764 ohms 18062619415 On the other hand, this doesn't rule out the possibility that 2 ohms could be produced. This is the "2 Ohm Problem", i.e., find a combination of distinct E12 resistors that give exactly 2 ohms when connected in parallel. One approach to finding such an expression is to first consider each individual basic resistor, multiplied by powers of 10, focusing on eliminating the large primes from the denominators of these individual sums. For the basic resistors 22, 33, 39, 47, 58, 68, and 82 the smallest sums that eliminate the primes greater than 5 from the denominators are 1 1 1 -- + --- = -- 22 220 20 1 1 1 -- + --- = -- 33 330 30 1 1 77 -- + + ----- = ---- 39 39000 3000 1 1 1 213 -- + ----- + ------ = ----- 47 47000 470000 10000 1 1 143 -- + ----- = ---- 56 56000 8000 1 1 1 1 653 -- + --- + ---- + ------ = ----- 68 680 6800 680000 40000 1 1 1 1 1 271 -- + --- + ---- + ----- + ------ = ----- 82 820 8200 82000 820000 20000 The sum of the seven series listed above is 3561/20000, which differs from 1/2 by 6439/20000. Thus we seek sums k1 to k5 of reciprocal powers of 10 such that / 1 \ / 1 \ / 1 \ / 1 \ / 1 \ 6439 ( --- )k1 + ( --- )k2 + ( --- )k3 + ( --- )k4 + ( --- )k5 = ----- \ 10/ \ 12/ \ 18/ \ 15/ \ 27/ 20000 Since there are only 32 distinct sums of reciprocal powers of 10 up to 10^4, it is feasible to make an exhaustive search. The first solution we find is 1 1 1 1 1 1 k1 = 1 + ---- + ---- + ---- k2 = k3 = 1 + ---- + ---- k4 = 1 + ---- k5 = 0 10^2 10^3 10^4 10^1 10^2 10^4 Thus we can produce 2 ohms by combining the following thirty-two integer resistances from the E12 series: 10 1000 10000 100000 12 120 1200 15 150000 18 180 1800 22 220 33 330 39 39000 47 47000 470000 56 56000 68 680 6800 680000 82 820 8200 82000 820000 This solution makes no use of any resistors from the "27" sequence. There are, of course, many other solutions. Consider, for example, the expression 1.1111 1.1101 1.001 1.0011 1.001 1.1 1.1 1.0111 1.0111 114653 ------ + ------ + ----- + ------ + ----- + --- + --- + ------ + ------ = ------ 82 68 56 47 39 33 22 18 12 360000 To complete a "2 ohm" solution based on this, we need to find series using the resistances 10, 15, and 27 that sum to Q = 65347/360000. Each of the following expressions equals this quantity, so each of them (along with the expression above) gives a distinct solution of the 2 ohm problem: Q = 1.10001/10 + 1.01111/15 + 0.11100/27 = 1.10101/10 + 1.01011/15 + 0.11010/27 = 1.10111/10 + 1.01001/15 + 0.11001/27 = 1.11101/10 + 1.00011/15 + 0.10110/27 = 1.11111/10 + 1.00001/15 + 0.10101/27 One peculiar feature of these solutions can be seen if we interpret the numerators of these terms as binary numbers. For example, the numerator of 10 in the first of these expressions is 1.10001, which we interpret as the binary number 110001 = 49. Similarly we determine the base 2 corresponding to each of the other numerators, adding a zero to each numerator of 27 so that it has the same number of digits as the other numerators. This gives the results 49 47 56 53 43 52 55 41 50 61 35 44 63 33 42 The differences between consecutive numbers in the first column are 4, 2, 6, 2, and these are the negatives of the differences of consecutive numbers in the second and third columns. If we reverse the digits of the numerators, but still interpret them as binary numbers, we have the values 35 61 14 43 53 22 59 37 38 47 49 26 63 33 42 Now the differences between consecutive numbers in the first column are 8, 16, -12, 16, and these are the same as the differences of the third column, and the negatives of the differences of the second column. We can simplify the situation by removing all the terms from the numerators that are common to all rows, leading to the denominators for five equal expressions (swapping the third and fourth sets to put them in numerical order) .0000 .1110 .1100 .0100 .1010 .1010 .0110 .1000 .1001 .1100 .0010 .0110 .1110 .0000 .0101 Taking these as binary digits in forward order or reverse order, we get the sequences forward reverse 0 14 12 0 7 3 4 10 10 2 5 5 6 8 9 6 1 9 12 2 6 3 4 6 14 0 5 7 0 10 In the forward case, the differences between consequtive numbers in the first column are equal and opposite the differences in the second, which are twice the differences of the third column, whereas in the reverse case the differences in the first and third columns are equal and opposite the differences in the second column. The forward case occurs because the component-wise difference between any two sets of numerators, interpreted as decimal numbers, is proportional to 1, -1, -0.9, which implies that each set gives the same quantity when divided by their respective denominators, because of the basic equality 1 1 / 1 1 \ 1 1 0.9 --- - --- - ( --- - --- ) = --- - --- - --- = 0 10 15 \ 27 270/ 10 15 27 The decimal differences between the numerators of consecutive sets (multiplied by 10000) are shown below. .0000 .1110 .1100 100 -100 -90 .0100 .1010 .1010 10 -10 -9 .0110 .1000 .1001 990 -990 -891 .1100 .0010 .0110 10 -10 -9 .1110 .0000 .0101 On the other hand, when we interpret the numbers in binary, the decimal difference given by 1 - .1 = .9 becomes 1 - .1 = .1 (which is 1/2), so the differences are proportional to 1, -1, -0.5 instead of 1, -1, -0.9. It would be interesting to know if there exist larger sets of numbers with this same property, i.e., the component-wise differences are in fixed proportions in two different bases.

Return to MathPages Main Menu