The Listener Crossword

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Boxes by Radix

Puzzle explanation

In clue order, the final digits of the heights spelt TOTAL HEIGHT IN BASE TWO-FOUR. Summing all the heights gave a total of 42509, which in base 24 is CASE, a box of sorts. The lengths/widths/heights of the boxes in the grid exhibited 180° symmetry.

Puzzle solution process

Here is one possible logical solution path. As in the puzzle, we use capitals for the clues and boxes, and lower case for the dimensions: for example, Pb represents the width b of box P.

The following addition table along with squares for the base-24 digits (Z = 0 to W = 23) will be useful. Note that odd squares (shown in red below) end in one of {A, I} and even squares (in blue) end in one of {Z, D, L, P}. Since exactly one of a, b, c must be odd, d² is always odd and we have the following possibilities for the final digits of a² + b² + c² = d²: either d² ends in A and the other squares’ last digits correspond to one of the sums {A+Z+Z, A+L+L, I+P+Z, I+D+L} or d² ends in I and the others end in {A+D+D, A+P+P, I+Z+Z, I+L+L}.

Furthermore, a square ending in A must have a root ending in one of {A, E, G, K, M, Q, S, W}, the root of a square ending in I ends in {C, I, O, U}, the root of a square ending in Z ends in {Z, L}, the root of a square ending in D ends in {B, J, N, V}, the root of a square ending in P ends in {D, H, P, T}, and the root of a square ending in L ends in {F, R}.

x²

Given the number of digits and the fact that b is always entered 90° clockwise from a, (and that there are to be no empty cells), every entry can be positioned unambiguously.

Since all of (a, b, c) must be less than d, if d starts with A then any other dimension with the same number of digits must also start with A. Thus Aa, Ca, Ga, La, Ma, Na, Qb, Ra, Va and Xa start with A, so A can be entered in their eight starting cells.

The maximum value for Ha and Hc is WW = 23×24 + 23 = 575, so Hb ≥ √(37339² − 575² − 575²) ≈ 37330.1, so Hb is at least 37331, which is BPSK in base 24 (2×24³ + 16×24² + 19×24 + 11). To be less than Hd = BPSS, Hb must be BPS_, with the last digit in the range [K, R].

From the grid, Aa is now A_S_. Its minimum value is √(15461² − 575² − 575²) ≈ 15439.6, so Aa ≥ 15440 = ABSH. To be less than Ad = ABTE, Aa must be ABS_, with the last digit in the range [H, W].

Ca ≥ √(15477² − 575² − 575²) ≈ 15455.6, so Ca ≥ 15456 = ABTZ. To be less than Cd = ABTU, Ca must be ABT_, with the last digit in the range [Z, T].

Qb ≥ √(13869² − 575² − 575²) ≈ 13845.1, so Qb ≥ 13846 = AZZV. To be less than Qd = AZAU, Qb must be in the range [AZZV, AZAT] so we can enter the Z in second place and mark the third as either Z or A.

Xa ≥ √(15019² − 575² − 575²) ≈ 14996.97, so Xa ≥ 14997 = ABZU. To be less than Xd = ABAS, Xa must be in the range [ABZU, ABAR] and we can enter the B (the third digit is either Z or A, confirmed by the third digit of Qa).

Va ≥ √(19585² − 575² − 575²) ≈ 19568.1, so Va ≥ 19569 = AIWI. Va is less than Vd = AJZA, and AJZZ would make Wb start with Z, so Va must be in the range [AIWI, AIWW] and we can enter the IW.

Wb is W_, in the range [552, 575], so the maximum value for Wa is √(12941² − 552² − 24²) ≈ 12929.2 and the minimum is √(12941² − 575² − 575²) ≈ 12915.4, so Wa is in the range [12916, 12929] = [VJD, VJQ] and we can enter the first two digits.

Ra ≤ √(19021² − 1² − 576²) ≈ 19012.3, so it’s in the range [13824, 19012] = [AZZZ, AIZD].

Ka is between √(10713² − 575² − 23²) ≈ 10697.5 and √(10713² − 24² − 1²) ≈ 10712.97, so it’s in the range [10698, 10712] = [RMR, RNH]. From Ra, we know that the last digit is ≤ I, which reduces the range to [10704, 10712] = [RNZ, RNH] and we can enter the initial RN.

Ob ≤ √(10611² − 1² − 576²) ≈ 10595.4, so it’s in the range [576, 10595] = [AZZ, RIK].

Mb starts with R and its second digit is the first digit of Ob, so it’s in the range [RAZ, RRW] = [10392, 10823]. Ma is then between √(24183² − 10823² − 23²) ≈ 21625.9 and √(24183² − 10392² − 1²) ≈ 21836.3, putting it in the range [21626, 21836] = [AMMB, AMUT], and we can enter the M. We know from the grid that Ma ends in V, so the range must be narrowed to [AMMV, AMTV] = [21646, 21814] (in steps of 24). Mb is then between √(24183² − 21814² − 23²) ≈ 10438.7 and √(24183² − 21646² − 1²) ≈ 10782.8, so its range is narrowed slightly to [10439, 10782] = [RBW, RQF]. That in turn narrows the range for Ma to [21647, 21813] = [AMMW, AMTU], which is narrowed to [AMNV AMSV] = [21670, 21790] to keep the last digit as a V. That narrows Mb to [10489, 10734], which narrows Ma again to [21671, 21789] = [AMNW, AMSU] and then [AMOV, AMRV] = [21694, 21766]. Then Mb is in [10539, 10685], which narrows Ma to [21695, 21765] = [AMOW, AMRU] and then [AMPV AMQV] = [21718, 21742]. Finally, Mb is now in [10588, 10637], so Ma ≤ √(24183² − 10588² − 1²) ≈ 21741.9, which forces Ma = 21718. That forces Mb = 10637, which means Mc = √(24183² − 21718² − 10637²) = 14 (happily a whole number!). So box M is (21718, 10637, 14, 24183) = (AMPV, RKE, N, AQWO).

Kb is between √(10713² − 10712² − 23²) ≈ 144.6 and √(10713² − 10704² − 1²) ≈ 439.04, so it’s in the range [145, 439] = [FA, RG]. Ob is then in the range [KFB, KRB] = [6482, 6770] (in steps of 24). Oc is then between √(10611² − 6770² − 23²) ≈ 8170.7 and √(10611² − 6482² − 1²) ≈ 8401.01, so it’s in the range [8171, 8401]. Then Ob ≤ √(10611² − 8171² − 1²) ≈ 6769.6, so its upper bound must be reduced by 24 to 6746. That then raises the lower bound for Oc to 8191, which reduces Ob’s upper bound again to 6722. This back-and-forth continues until Ob is forced down to 6482 and Oc becomes 8401, so box O is (14, 6482, 8401, 10611) = (N, KFB, NNA, RJC).

Kb is now F_, in the range [FA, FW] = [145, 167]. Then Ka ≥ √(10713² − 167² − 23²) ≈ 10711.7, so its lower and upper bounds coincide at Ka = 10712 = RNH. Then Kb ≤ √(10713² − 10712² − 1²) ≈ 146.4, so its upper bound is 146. Then Kc ≥ √(10713² − 10712² − 146²) ≈ 10.4, so its lower bound is 11. Then Kb ≤ √(10713² − 10712² − 11²) ≈ 145.96, so Kb is forced to 145 and box K is (10712, 145, 20, 10713) = (RNH, FA, T, RNI).

Ra is now in the range [AHMZ, AHMW] = [18744, 18767], so Rc is between √(19021² − 23² − 18767²) ≈ 3098.003 and √(19021² − 1² − 18744²) ≈ 3234.3, putting it in the range [3099 3234]. That then narrows the range for Ra to [18745, 18766], which narrows Rc to [3105, 3228], and so on until they settle at Ra in [18748, 18757] = [AHMD, AHMM] and Rc in [3158, 3211] = [EKN, EMS].

Suppose Ra = 18748: then Rc is between √(19021² − 18748² − 23²) ≈ 3210.98 and √(19021² − 18748² − 1²) ≈ 3211.1, ie, Rc = 3211, but that would make Rb = √(19021² − 18748² − 3211²) ≈ 20.4, not a whole number. Suppose instead Ra = 18749: we find Rc has to be between (approximately) 3205.1 and 3205.2, but there are no whole numbers in that range. Similarly, values of Ra = 18750 to 18755 allow no whole numbers for Rc. Ra = 18757 would make Rc = 3158, but then Rc ≈ 20.7, not a whole number. That leaves only Ra = 18756, Rc = 3164, Rb = 3. So box R is (18756, 3, 3164, 19021) = (AHML, C, EKT, AIZM).

Now Wb = WL = 564, so Wa ≤ √(12941² − 564² − 24²) ≈ 12928.7, so its range is slightly narrowed to [12916, 12928]. Of the 13 possibilities for Wa, only two give integer results for Wc, so box W is either (12916, 564, 573, 12941) = (VJD, WL, WU, VKE) or (12923, 564, 384, 12941) = (VJK, WL, PZ, VKE).

From the grid, Ua = J = 10 and Ub = _C. Ub < Ud = QG, so Ub is in the range [AC, QC] = [27, 411] (in steps of 24). Of the 17 possibilities for Ub, only three give integer results for Uc; one of them is (10, 315, 270, 415), which can be ruled out because all four dimensions share a common factor of 5. That leaves box U as either (10, 27, 414, 415) = (J, AC, QF, QG) or (10, 171, 378, 415) = (J, GC, OR, QG).

From the grid, Va = AIWP = 19576 and Vb = _Z, in the range [AZ, WZ] = [24, 552] (in steps of 24). Vb ≥ √(19585² − 19576² − 575²) ≈ 147.7, which reduces Vb’s range to [GZ, WZ] = [168, 552] (in steps of 24). Of the 17 possibilities, only Vb = 240 makes Vc = √(19585² − 19576² − Vb²) an integer, so box V is (19576, 240, 543, 19585) = (AIWP, JZ, VO, AJZA).

Now Sa = BJ = 58 and Sb is I_, in the range [216, 239]. Sc is between √(603² − 58² − 239²) ≈ 550.6 and √(603² − 58² − 216²) ≈ 559.99, so it’s in the range [551, 559] = [VW, WG]. Of the 9 possibilities for Sc, only Sc = 551 makes Sb = √(603² − 58² − Sc²) an integer, so box S is (58, 238, 551, 603) = (BJ, IV, VW, AAC).

From the grid, we have Lb = A__ which is in the range [576, 1151]. La is between √(17051² − 1151² − 23²) ≈ 17012.1 and √(17051² − 576² − 1²) ≈ 17041.3, so it’s in the range [17013, 17041] = [AELU, AENA] and we can enter the E and the third digit is restricted to the range [L, N].

Ca then ends in one of {L, M, N} so the last digit of Ca² is one of {Z, A, D} respectively; Cb ends in P so the last digit of Cb² is P; the last digit of Cd² is I. Cd² − Cb² ends in AI − P = Q, so Cc² = Cd² − Cb² − Ca² must end in {Q − Z, Q − A, Q − D} respectively, ie, {Q, P, M}, of which only P is a possible final digit for a base-24 square, so Cc² ends in P, with Ca = ABTM = 15469.

This makes Cb² + Cc² = Cd² − Ca² = 247568. As Ca is odd, both Cb and Cc must be even, so (in base 10) their squares can only end in 0, 4 or 6. To have a sum ending in 8, both squares must end in 4, so Cb ends in 2 or 8; this restricts Cb = 24n + 16 (for some integer n in [1, 23]) to {88, 112, 208, 232, 328, 352, 448, 472} (not 568 because 568² > 247568). Of these, only Cb = 208 makes Cc = √(247568 − Cb²) an integer, so box C is (15469, 208, 452, 15477) = (ABTM, HP, RT, ABTU).

The grid now gives Fa = BH = 56 and Fb = B_, in the range [48, 71]. Fc is between √(261² − 56² − 71²) ≈ 244.8 and √(261² − 56² − 48²) ≈ 250.4, so Fc is in the range [245, 250]. Of the 6 possibilities, only 248 gives an integer solution for Fb = √(261² − 56² − Fc²), so box F is (56, 59, 248, 261) = (BH, BK, JH, JU).

Bb = T_ is in the range [480, 503], so Ba is between √(4043² − 503² − 575²) ≈ 3970.2 and √(4043² − 480² − 24²) ≈ 4014.3, so it’s in the range [3971, 4014] = [FUK, FWF]. Entering the initial F completes La = AEMF = 17022. Then Lb is between √(17051² − 17022² − 23²) ≈ 993.8 and √(17051² − 17022² − 1²) ≈ 994.04, so Lb = 994 and box L is (17022, 994, 9, 17051) = (AEMF, AQJ, I, AENK).

Jb is now Q_A, at least QAA (not QZA, which would make Nb start with Z). The maximum for Jb is √(10233² − 1² − 576²) ≈ 10216.8, so Jb ≤ 10216 = QQP, but it must end in A, so Jb is in the range [QAA, QQA] = [9817, 10201] (in steps of 24). As Jb is odd, both Ja and Jc must be even, so Ja is in the range [2, 22] (in steps of 2). Jc is between √(10233² − Jb² − 22²) and √(10233² − Jb² − 2²). Of the 17 possibilities, only Jb = 9817 makes Jc an even integer, so box J is (16, 9817, 2888, 10233) = (P, QAA, EZH, QRI).

Nb is now A_, in the range [24, 47], which means Na is between √(687² − 47² − 23²) ≈ 685.004 and √(687² − 24² − 1²) ≈ 686.6, so Na = 686 = ADN. Then Nb is between √(687² − 686² − 23²) ≈ 29.05 and √(687² − 686² − 1²) ≈ 37.04, so Nb is in the range [30, 37]. Nd² − Na² = 1373, so either Nb² ends in 4 and Nc² ends in 9 or vice versa, which restricts Nb to {32, 33, 37}. Of these, only Nb = 37 makes Nc = √(1373 − Nb²) an integer, so box N is (686, 37, 2, 687) = (ADN, AM, B, ADO).

Ga is now ANE_, in the range [22008, 22031]. Gc is between √(22215² − Ga² − 23²) and √(22215² − Ga² − 1²); of the 24 possibilities for Ga, only 22010 and 22030 make Gc an integer as well as Gb, which leaves two possibilities for box G: (22010, 2, 3011, 22215) = (ANEB, B, EEK, ANMO) and (22030, 2, 2861, 22215) = (ANEV, B, DWE, ANMO). We can enter Gb = B.

From Ba, we know Da is in the range [U, W] = [21, 23]. The minimum for Db is √(651² − 23² − 575²) ≈ 304.4, so Db ≥ 305 = LQ, but its last digit is B, so Db is in the range [MB, WB] = [314, 554] (in steps of 24). Dc is between √(651² − 23² − Db²) and √(651² − 21² − Db²); of the 11 possibilities for Db, only Db = 506 gives integer results for both Dc and Da. So box D is (22, 506, 409, 651) = (V, UB, QA, ACC).

Bd = GZK so Bd² ends in A. Bb is one of {TB, TV} = {482, 502} (from Ga); either way, Bb² ends in D (B×B = D or V×V = TD), which means either Ba² ends in I and Bc² ends in L or vice versa, so either Ba ends in one of {C, I, O, U} and Bc ends in one of {F, R}, or vice versa; so Ba is one of {FVC, FVF, FVI, FVO, FVR, FVU} = {3987, 3990, 3993, 3999, 4002, 4005}. Of the 12 possibilities for Ba and Bb, only Ba = 4002 and Bb = 502 make Bc = √(4043² − Ba² − Bb²) an integer, so box B is (4002, 502, 279, 4043) = (FVR, TV, KO, GZK). This fixes Ga = ANEV, so box G is (22030, 2, 2861, 22215) = (ANEV, B, DWE, ANMO).

Id = KRI so Id² ends in I; Ia = AD so Ia² ends in P. So either Ib² ends in A and Ic² ends in P or vice versa, which means either Ib ends in one of {A, E, G, K, M, Q, S, W} and Ic ends in one of {D, H, P, T} or vice versa. From the preamble, we know that the first two digits of Ib are two of {A, H, P, W}, and Ib < Id = KRI, so Ib must be one of {AH_, AP_, AW_, HA_, HP_, HW_}.

Taking these in order, if Ib is in the range [AHA, AHW] (not AHZ, because it can’t end in Z), then Ic is between √(6777² − 28² − 791²) ≈ 6730.6 and √(6777² − 28² − 769²) ≈ 6733.2, giving the range [6731, 6733]; none of those 3 possibilities gives an integer value for Ib = √(6777² − 28² − Ic²). If Ib is in the range [APA, APW], then Ic is between √(6777² − 28² − 983²) ≈ 6705.3 and √(6777² − 28² − 961²) ≈ 6708.5, giving the range [6706, 6708]; none of those 3 possibilities gives an integer value for Ib. If Ib is in the range [AWA, AWW], then Ic is between √(6777² − 28² − 1151²) ≈ 6678.5 and √(6777² − 28² − 1129²) ≈ 6682.2, giving the range [6579, 6682]; of those 4 possibilities, only Ic = 6679 gives an integer value, Ib = 1148 = AWT.

For Ib in the range [HAA, HAW], Ic is in [4926, 4945] = [HMF, HNA], which with last-digit restrictions means it’s one of {HMG, HMH, HMK, HMM, HMP, HMQ, HMS, HMT, HMW, HNA} but none of those 10 possibilities gives an integer value for Ib. For Ib in [HPA, HPW], Ic is in [4559, 4582] = [GUW, GVV], giving {GUW, GVA, GVD, GVE, GVG, GVH, GVK, GVM, GVP, GVQ, GVS, GVT} but none of those 12 possibilities gives an integer value for Ib. For Ib in [HWA, HWW], Ic is in [4367, 4392] = [GMW, GOZ], giving {GMW, GNA, GND, GNE, GNG, GNH, GNK, GNM, GNP, GNQ, GNS, GNT, GNW} but none of those 13 possibilities gives an integer value for Ib. So the only solution for box I is (28, 1148, 6679, 6777) = (AD, AWT, KNG, KRI).

Pd = AZCO so Pd² ends in I; Pa = AN so Pa² ends in D. So either Pb² ends in A and Pc² ends in D or vice versa, which means either Pb ends in one of {A, E, G, K, M, Q, S, W} and Pc ends in one of {B, J, N, V} or vice versa. From the grid, we know Pb’s last digit is in the range [K, R], so it must be one of {M, N, Q}. From the preamble, we know the first two digits are HP or PH (as the A and W are now used in Ib), so Pb is one of {HPM, HPN, HPQ, PHM, PHN, PHQ} = {5005, 5006, 5009, 9421, 9422, 9425}. Of those 6 possibilities, only Pb = 5006 makes Pc = √(13911² − 38² − Pb²) an integer, so box P is (38, 5006, 12979, 13911) = (AN, HPN, VLS, AZCO).

Hd² ends in A and Hb is now BPSN, so Hb² ends in D, so either Ha² ends in I and Hc² ends in L or vice versa, which means either Ha ends in one of {C, I, O, U} and Hc ends in one of {F, R} or vice versa.

The ambiguities for boxes U and W (see earlier) clearly cannot be resolved by the remaining clues, so there must be a thematic element involved.

The preamble says that the heights (not entered in the grid) “should finally reveal” something. The final digits of the ones we have so far spell _OTA_HE[C/F/I/O/R/U]GHTINBAS_TW_[R/F]O[Z/U]_, which, given the context, strongly points to TOTAL HEIGHT IN BASE TWO-FOUR. This fixes box U as (10, 27, 414, 415) = (J, AC, QF, QG) and box W as (12916, 564, 573, 12941) = (VJD, WL, WU, VKE). It also provides final digits for all other c elements, helping to solve the remaining clues.

We have Hc = _I, in the range [AI, WI] = [33, 561] (in steps of 24). Of those 23 possibilities, the only ones that give integer values for Ha = √(37339² − 37334² − Hc²) are Hc = 249, Ha = 558 and Hc = 297, Ha = 534; so Ha is one of {534, 558} = {VF, WF}.

Ed² ends in A, and Ec = _L so Ec² ends in Z, which means the last digits of Ea² and Eb² are {A, Z} or {I, P}, in either order. Ha tells us that Ea is one of {V, W}, but it can’t be V because V² ends in D. So box E is (23, 456, 516, 689) = (W, SZ, UL, ADQ), which fixes box H as (558, 37334, 249, 37339) = (WF, BPSN, JI, BPSS).

Ad² ends in A, Ab² = _Z² which ends in Z, and Ac² = _T² which ends in P, so Aa² ends in I, which means Aa ends in one of {C, I, O, U}. We already have Aa in the range [ABSH, ABSW], so Aa is one of {ABSI, ABSO, ABSU} = {15441, 15447, 15453}.

Taking these in turn, if Aa = 15441, Ac ≥ √(15461² − 15441² − 552²) ≈ 559.8, giving the lower bound WH, so Ac can only be WT. But then Ab = √(15461² − 15441² − 572²) ≈ 539.3 is not an integer.

If Aa = 15447, Ac ≥ √(15461² − 15447² − 552²) ≈ 357.8, giving the lower bound NV, so Ac is in the range [OT, WT] (in steps of 24). Of the 9 possibilities, none makes Ab = √(15461² − 15447² − Ac²) an integer.

If Aa = 15453, Ac ≤ √(15461² − 15453² − 24²) ≈ 496.7, giving the upper bound TP, so Ac is in the range [AT, ST] (in steps of 24). Then Ab is between √(15461² − 15453² − 476²) = 144 and √(15461² − 15453² − 44²) ≈ 495.4, giving the range [FZ, TO], putting Ab in [FZ, SZ] (in steps of 24). Of the 14 possibilities, only Ac = 144 and Ac = 384 make Ab = √(15461² − 15447² − Ac²) an integer; the latter gives Ab = 316 = MD, which doesn’t end in T, so box A is (15453, 144, 476, 15461) = (ABSU, FZ, ST, ABTE).

From the grid, Qb is one of {AZZT, AZAT}. If Qb = AZZT = 13844, Qc ≥ √(13869² − 13844² − 575²) ≈ 601.8, giving the lower bound AAB, but that’s ruled out because Qc has only two digits, so Qc = AZAT = 13868. Then Qc ≤ √(13869² − 13868² − 24²) ≈ 164.8, giving the upper bound FT; we know it ends in E, so Qc is in the range [AE, FE] (in steps of 24). Of the 6 possibilities, only Qc = 29 makes Qa = √(13869² − 13868² − Qc²) an integer, so box Q is (164, 13868, 29, 13869) = (FT, AZAT, AE, AZAU).

Td = QS, Tc = _O and from the grid Ta = F; their squares end in A, I and L, respectively, so Tb² ends in D and Tb ends in one of {B, J, N, V}. As it has to be less than Td = QS, Tb is in the range [AB, QN] = [26, 422].

Tc is between √(427² − 6² − 422²) ≈ 64.9 and √(427² − 6² − 26²) ≈ 426.2, giving the range [BQ, QR], putting Tc in [CO, QO] = [87, 423] (in steps of 24). Of the 15 possibilities, only Tc = 87 and Tc = 423 make Tb = √(427² − 6² − Tc²) an integer, so box T is either (6, 418, 87, 427) = (F, QJ, CO, QS) or (6, 58, 423, 427) = (F, BJ, QO, QS). We can enter J as the last digit of Tb, but the unchecked digit remains ambiguous.

Xd = ABAS, Xc is _R and from the grid Xb is now _J; their squares end in A, L and D, respectively, so Xa² ends in I and Xa ends in one of {C, I, O, U}. We already know that Xa is ABA_ and it’s < Xd, so Xa is one of {ABAC, ABAI, ABAO} = {15003, 15009, 15015}.

In base 10, Xb = 24m + 10 and Xc = 24n + 18, for some values of m and n in the range [1, 23]). They’re both even, so their squares can only end in 0, 4 or 6.

If Xa = 15003, the sum Xb² + Xc² = Xd² − Xa² = 480352. For the sum of the two squares to end in 2, both Xb² and Xc² must end in 6, so each of Xb and Xc ends in either 4 or 6. For Xc, the values of 24n + 18 that end in 4 or 6 are {66, 114, 186, 234, 306, 354, 426, 474, 546}. Of these 9 possibilities, none makes Xb = √(15019² − 15003² − Xc²) an integer.

If Xa = 15009, the sum Xb² + Xc² = Xd² − Xa² = 300280. For the sum to end in 0, either both squares end in 0, or one ends in 4 and the other in 6. Any square ending in 0 must be the square of a multiple of 10, so it ends in 00; but two squares ending in 00 can’t give a sum ending in 80. So either Xb ends in one of {2, 8} and Xc ends in one of {4, 6}, or vice versa. Of the 18 possibilities for Xc ending in one of {2, 4, 6, 8}, none makes Xb = √(15019² − 15009² − Xc²) an integer.

If Xa = 15015, the sum Xb² + Xc² = Xd² − Xa² = 120136. For the sum to end in 6, one square must end in 0 and the other in 6; furthermore, one square ends in 00 and the other in 36 (so that the sum ends in 36). The possible values for Xc are {90, 210, 306, 330, 450, 570}, of which only Xc = 330 makes Xb = √(15019² − 15015² − Xc²) an integer. So box X is (15015, 106, 330, 15019) = (ABAO, DJ, MR, ABAS).

The grid is now complete except for Tb = BJ or QJ. The message from last digits of the c values is “total height in base two-four” and the preamble tells us that it reveals an appropriate word. There are two possibilities for the total of all the c values, depending on whether Tc = 87 or 423: the former gives 42173 = CAEE and the latter 42509 = CASE. Only the latter is a word, and it’s appropriate because a case is a box of sorts, so we have box T as (6, 58, 423, 427) = (F, BJ, QO, QS).

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Listener Crossword 4256: Solution Notes

Boxes by Radix

Puzzle explanation

Puzzle solution process