Plan 4 From Outer Space

by Dan Katz
Answer: FOURIER TRANSFORM

Each of the first six sets of equations in the puzzle can be solved independently; in each one, all of the digits (except the digit 4) have been encrypted into letters using a different substitution cipher. The first two equations can be used to solve this code, and then the rest can be used to determine the values of the Greek letters, which should be in encrypted form (i.e. letters, not numbers).

Once the first six sets are solved, the values for the Greek letters may be substituted into Set 7 to form a seventh puzzle of the same type. Once the code is deciphered, calculating the missing right side of the final equation gives the result 4IERTRANS4M, yielding the math-related answer FOURIER TRANSFORM.

Set 1:

In the first equation, the expression inside the log simplifies to (NR/I)[e^(4*I)-e^(I*L)]+O. The only way the log can be something as simple as the right side is if (NR/I)e^(I*L) = O, which implies that L = 0. (If I were equal to 0, there would be no log on the right side.) This reduces the top equation to two conditions, O = NR/I and ln (NR/I) + 4*I = BE + B ln R, which reduces to 4*I = BE, R^B = NR/I = O.

In the second equation, D must be 1 (since it can't be 0). Since R^B is a single digit, R and B are 2 or 3 in some order. For the sine to be 1, the denominator DR must be even, so R = 2 and B = 3. Therefore O = 8, and I = 9, N = 7, E = 6, and finally H = 5.

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LDRB4HENOI

Set 2:

Since (S x R^L) is raised to a power of at least three to get a number no greater than 102, it is 2, 3, or 4. The only way this can happen is if either R = 1 or L = 0 (and S = 2, 3, or 4).

In the second equation, the left side is at most 45, and the product is at least 2, so I is 1, 2, or 3. This reduces us to a number of cases based on the values of I and S; in each case, one can narrow down to very few possible values for A and then very few values for O and T. Only one configuration ends up working, and unfortunately I don't have a more elegant argument than that right now.

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LISA4NERTO

Set 3:

The only term that can give the derivative HTx^4 is Tx^T, which gives T = 5, H = 2. x^L gives Lx^R, which gives R + 1 = L, and the leftover terms must match up, so the derivative of Hx^4 = 2x^4 is Ex^D, giving D = 3 and E = 8. Based on the assignments already used up, the only consecutive numbers left are R = 6, L = 7.

37^2 = 1369. 1369 can't be congruent to a two-digit number mod NDM if N = 9, so N = 1, and immediately M = 0, O = 9.

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MNHD4TRLEO

Set 4:

In the first equation, the left side can only collapse to a single term if T = 2, U = 1. The derivative then expands to e^Rx[(R*1H)x^2+(2*1H-R*12)x+(4*R-12)], which means that R*1H = S4, 2*1H-R*12 = 0, and 4*R-12 = 0, giving R = 3, H = 8.

In the second equation, the only way the left side can resolve to an integer is if O = 0, in which case the left side is a formula for the gamma function and equal to S!. If S is 6 or larger, the result will be too large for the right side, so S = 5. We now have EI x (M - 2) = 480, and trial and error gives M = 7, E = 9, I = 6.

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OUTR4SIMHE

Set 5:

The only way a two-digit number choose a one-digit number can be 17 or less is if E = 9 and AS = 10 or 11; clearly it has to be 10. So R + O = 10. The second equation converts to the form 4R sin VT = UT sin LU cos LU = UT/2 sin (2*LU), so UT = 2*4R and 2*LU = VT. U = 8 by the first of these equations, and so by the second, T = 6. Then R = 3 and O = 7, and V = 2.

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SALR4VTOUE

Set 6:

N! (N factorial, or the product of the numbers from 1 to N) is between 300 and 2000, so N = 6 and Y = 3. (Either E = 8, S = 0 or E = 7, S = 5.)

The top equation reduces to two equations: D + I = 3 mod 4 (since the sin comes around to negative cos) and B x L^D x P^I = I6B. B,L,D,P,I are either 1,2,5,7,9 in some order or 1,2,8,9,0 in some order. If there's a 0, it has to be D, which gives B x P^I = I6B, and trying different cases for I confirms that this doesn't work. Therefore E = 8, S = 0.

Since all of the remaining numbers are odd, 2 is one of the exponents. Its base will not be 1 (otherwise the other exponent is huge) so I6B is a multiple of either 25, 49, or 81. 25 is out since then B would also be equal to 5. 49 and 81 are large enough that the 1 will have to be the other base, so the possiblities for I6B are 49 x 5, 49 x 9, 81 x 5, or 81 x 7. The last of these is the only one with a 6 in the middle, so I6B = 567, and the rest of the assignments follow.

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SPDY4INBEL

Set 7:

R = I + 1 by the second equation, so by the first R = 2, I = 1. T = 3 by the ones place for the first equation, and since all the small numbers are taken, S + E = 14, N + M = 13, and L + S = 13. This means L,M,N,S are 5,6,7,8 in some order, and so E = 9, S = 5, L = 8, and N,M are 6,7 in some order. Returning to the second equation, we must have N = 6 and A = 0.

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AIRT4SNMLE