ANSWERING THE SALLOWS CHALLENGE
by Ross Eckler and Leonard Gordon
Word Ways, 1990
In his February 1990 Word Ways article, "The New Merology", Lee Sallows challenges the reader to find more than 46 self-descriptive number names between 0 and 99 by assigning suitable integral values to the different letters in each name. It's easy! They key idea is to "square" the array as described in the article "38 Self-Descriptive Number Names" in the February 1990 Word Ways.
We attack the problem by first solving a slightly more general one: allow letters to be assigned any real-number values, not just integers. By so doing, we reduce the problem to one encountered in high-school algebra, the solution of 15 linear equations in 15 unknowns. We then show how the solution can be modified to solve the Sallows challenge.
Ideally, in squaring the array, we could like to make each name in the two sets (ONE, TWO, ... NINE) and (TWENTY, THIRTY, ... NINETY) self-descriptive, so that one could form a total of 9+8+(9x8) = 89 self-descriptive names. However, this is clearly impossible; one can select TY to satisfy only one of the pairs (SIX, SIXTY), (SEVEN, SEVENTY) and (NINE, NINETY). To fix ideas, we select (SEVEN, SEVENTY) and plan to include SIXTY and NINETY as well (but not SIX and NINE) in the self-descriptive set. This will give a total of 7+8+(7x8) = 71 self-descriptive number names. Since there are 15 different letters (E,F,G,H,I,N,O,R,S,T,U,V,W,X,Y) to be assigned numbers, the problem reduces to the solution of 15 linear equations in 15 unknowns.The 71 names can be increased to 74 by selecting a suitable value for Z from ZERO, and for L from ELEVEN and TWELVE.
The solution proceeds as follows: SEVENTY - SEVEN = 63 = [T+Y] and EIGHTY - EIGHT = 72 = Y; hence T = -9.Next solve the three equations O + [N+E] = 1, W + O = 2 -T = 11, and W + [N+E] = 20 - T - [T+Y] = -34 for the unknowns W, O and [N+E]. U is determined from FORTY - FOUR = 36 = [T+Y] - U. From THIRTY - THREE = 27 = I + 63 - 2E, one obtains I - 2E = -36. This equation, together with [N+E] = -22, and N + I = 90 - [N+E] - [T+Y] = 49, may be solved for N, E, and I. The solutions for Y,T,W,O,E,I,N and U are given in the (SEVEN, SEVENTY) column in the table below.
The remaining seven letters are assigned numbers that are multiples of one-half. F is determined from FIFTY = 50; with F in hand, V is determined from FIVE = 5 and R from FOUR = 4. Knowing V, S is ascertained from SEVEN = 7, and X in turn from SIXTY = 60. Using R, H is obtained from THREE = 3, and G in turn from EIGHT = 8. The values of these letters are given in the (SEVEN, SEVENTY ) column below.
A number of alternative solutions are possible. One can elect to include SIX instead of SIXTY, or NINE instead of NINETY in the above set. One can even include both SIX and NINE, although this leads to 69 self-descriptive numbers instead of 71. Or, one can select the pair (SIX, SIXTY) or the pair (NINE, NINETY) as a base for another set. In the first three column of the table below we summarize the changes that occur in the assignment of numbers to letters:
SIX, SEVEN, NINE, SEVEN, SIXTY SEVENTY NINETY SEVENTY (integers) Y 72 72 72 72 T -18 -9 9 -9 W 3/2 -12 -39 -12 O 37/2 23 32 23 E 98 107 125 107 I 169 178 196 178 N -231/2 -129 -156 -129 U 18 27 45 27 F -173/2 -191/2 -227/2 -120 V -351/2 -369/2 -405/2 -160 R 54 99/2 81/2 74 H -229 -503/2 -593/2 -276 G -12 -33/2 -51/2 8 S 111 213/2 231/2 82 X -274 -575/2 -665/2 -263 Z -341/2 -359/2 -395/2 -204 L 8 7/2 -11/2 -21None of the alternative solutions leads to additional self-descriptive number names. Note that the (SIX, SIXTY) set has three fewer half-integer assignments than the others.
We now turn to Sallows' challenge. Clearly, we cannot make all 15 of the number names self-descriptive, but it is possible to include 14 of them, leaving out only FIFTY.. This yields the same solution for Y,T,W,O,E,I,N and U. Pick an integral value for S, X, or V, whereupon the other two letters are determined from equations using SEVEN and SIXTY. Next, select an integral value for either H or R, and the other is determined from an equation using THREE. The remaining two letters, F and G, follow at once from FIVE (or FOUR) and EIGHT. This results in 63 self-descriptive number names, which can be increased to 66 by adding Z and L as described previously. Results are shown in the final column above.
A few other discoveries may be of interest. The first six integers can be made self-descriptive using numerical values that include every sixth from 1/6 through 13/6:
ONE = 1/3 + 1/2 + 1/6
TWO = 2/3 + 1 + 1/3
THREE = 2/3 + 7/6 + 5/6 + 1/6 + 1/6
FOUR = 4/3 + 1/3 + 3/2 + 5/6
FIVE = 4/3 + 11/6 + 5/3 + 1/6
SIX = 2 + 11/6 + 13/6If one is interested in "tight" integer sets making ONE through NINE self-descriptive, consider the assignment F = -5, S = -4, H = -3, O = -2, E = -1 W = 1, I = 2, T = 3, N = 4 R = 5, U = 6, G = 7, X = 8, V = 9.
Finally, if one wishes to maximize the number of consecutive integers that are self-descriptive under integer assignment, we cannot improve on the 31 numbers NINETEEN through FORTY--NINE given by Sallows. However, the corresponding number of consecutive numbers possible under non-integer assignment is 51. Those from TWENTY through SIXTY-NINE can be obtained using the arguments given earlier in this article. SEVENTY cannot be added because TY cannot simultaneously satisfy (SIX, SEVEN, SIXTY, SEVENTY). NINETEEN can be added as follows. Since SIX and SIXTY must both be included in this set, T+Y = 54. From TWENTY - TWO + ONE = 19 = 2[E+N] - 54, one concludes that [E+N] = -35/2. From the equation THIRTY - THREE = 27, one concludes that I - 2E = -27. From NINE = 9, [N+I] = 9 - (-35/2). Solving these equations for E and N, one obtains E = 71 and N = -178/2. If NINETEEN is to be self-descriptive, TEEN must equal 10 and T must therefore equal -87/2. Other letters are assigned integer values as described previously.