Solution: RC1005 Water Traversal

Answer: CHANNEL SURFING

This puzzle comprises 700 seemingly random characters arranged in a rectangle, but starting from the first character and reading along spells out an instruction: TAKE OUT ALL D'S, RESPACE TO SQUARE, CHECK DIAGONAL. There are 24 instances of the letter D in the rectangle, so removing all of them would leave 676 = 26² characters behind. This lets us rearrange the leftover characters into a 26 by 26 square:


			TAKEOUTALLSRESPACETOSQUARE

			CHECKIAGONALHSKT6REJJZNFOZ

			Y0EUPHT5Y50VBPZ09ALB1CLQ1H

			QJJN7705Y5U61EPLX1A9L1IPN1

			0Z97H1F725O1Z60WI80VXECFTI

			0MM7MIM5M71E71M3M15A5MM5M1

			U8F5HGG515O50202517035L5ZV

			LKJX23UH6SO73247K55N7425ZS

			M0MM000MLMM5M515MM3M2MM2MM

			B90I89F11IN2E5G20Z502XL7NN

			0NICE1JK8TGC1LLNJ0Z9V6R8T0

			Z66C9OYXFB4HOBTHROK9R99QGU

			O1MMFMM0MMG0T57MM0MGXF0MPF

			1523X06C3Z0T3E6F1G50H3UGU9

			F70TCT1L7YF0G5A0465TN7122F

			P730JTU13CC41Q5CU05L1362G3

			Z2JPZ0JP5F0N571THX5Y650371

			J275M8MM533OOHMMLMI77CP7F5

			E09A89880X8VS641AZFC049503

			RO05331XQ3730FW0WXPOFW1UHS

			0LT20YI4ZQ7ANA1NH1XXR37591

			R8022SMM1M7M0U2MM7MM5M0M1H

			R1U2PJ0ZZG71L075U3Y07JOSEC

			PBT51GBN02770N2Y238F313RC1

			HNMAF9MPMMOMOT7MM7MM5MM5SX

			94ZWQJQFP9FX8Q5XK5RKL253YE

Reading along the main (top-left to bottom-right) diagonal of this square, as instructed, spells out the next step THEN HIGHLIGHT EACH M FOR MORSE. Finding and highlighting each instance of the letter M in the square results in the following:


			TAKEOUTALLSRESPACETOSQUARE

			CHECKIAGONALHSKT6REJJZNFOZ

			Y0EUPHT5Y50VBPZ09ALB1CLQ1H

			QJJN7705Y5U61EPLX1A9L1IPN1

			0Z97H1F725O1Z60WI80VXECFTI

			0██7█I█5█71E71█3█15A5██5█1

			U8F5HGG515O50202517035L5ZV

			LKJX23UH6SO73247K55N7425ZS

			█0██000█L██5█515██3█2██2██

			B90I89F11IN2E5G20Z502XL7NN

			0NICE1JK8TGC1LLNJ0Z9V6R8T0

			Z66C9OYXFB4HOBTHROK9R99QGU

			O1██F██0██G0T57██0█GXF0█PF

			1523X06C3Z0T3E6F1G50H3UGU9

			F70TCT1L7YF0G5A0465TN7122F

			P730JTU13CC41Q5CU05L1362G3

			Z2JPZ0JP5F0N571THX5Y650371

			J275█8██533OOH██L█I77CP7F5

			E09A89880X8VS641AZFC049503

			RO05331XQ3730FW0WXPOFW1UHS

			0LT20YI4ZQ7ANA1NH1XXR37591

			R8022S██1█7█0U2██7██5█0█1H

			R1U2PJ0ZZG71L075U3Y07JOSEC

			PBT51GBN02770N2Y238F313RC1

			HN█AF9█P██O█OT7██7██5██5SX

			94ZWQJQFP9FX8Q5XK5RKL253YE

This can be read as Morse code, giving -... .. -. .- .-. -.-- --- -. . .- -. -.. --.. . .-. ---, which translates into English as BINARY ONE AND ZERO. This hints at a similar next step; some of the rows of text contain exactly 5 binary digits (0 or 1), which we can extract and convert via five-bit binary to letters:


			TAKEOUTALLSRESPACETOSQUARE             

			CHECKIAGONALHSKT6REJJZNFOZ             

			Y0EUPHT5Y50VBPZ09ALB1CLQ1H    00011 = C

			QJJN7705Y5U61EPLX1A9L1IPN1    01111 = O

			0Z97H1F725O1Z60WI80VXECFTI    01100 = L

			0MM7MIM5M71E71M3M15A5MM5M1    01111 = O

			U8F5HGG515O50202517035L5ZV    10010 = R

			LKJX23UH6SO73247K55N7425ZS             

			M0MM000MLMM5M515MM3M2MM2MM    00001 = A

			B90I89F11IN2E5G20Z502XL7NN    01100 = L

			0NICE1JK8TGC1LLNJ0Z9V6R8T0    01100 = L

			Z66C9OYXFB4HOBTHROK9R99QGU             

			O1MMFMM0MMG0T57MM0MGXF0MPF    10000 = P

			1523X06C3Z0T3E6F1G50H3UGU9    10010 = R

			F70TCT1L7YF0G5A0465TN7122F    01001 = I

			P730JTU13CC41Q5CU05L1362G3    01101 = M

			Z2JPZ0JP5F0N571THX5Y650371    00101 = E

			J275M8MM533OOHMMLMI77CP7F5             

			E09A89880X8VS641AZFC049503    00100 = D

			RO05331XQ3730FW0WXPOFW1UHS    01001 = I

			0LT20YI4ZQ7ANA1NH1XXR37591    00111 = G

			R8022SMM1M7M0U2MM7MM5M0M1H    01001 = I

			R1U2PJ0ZZG71L075U3Y07JOSEC    10100 = T

			PBT51GBN02770N2Y238F313RC1    10011 = S

			HNMAF9MPMMOMOT7MM7MM5MM5SX             

			94ZWQJQFP9FX8Q5XK5RKL253YE

The binary spells out the instruction COLOR ALL PRIME DIGITS. Similar to the Morse code step, we can shade in all the instances of 2, 3, 5, and 7 in the square, giving the following:


			TAKEOUTALLSRESPACETOSQUARE

			CHECKIAGONALHSKT6REJJZNFOZ

			Y0EUPHT█Y█0VBPZ09ALB1CLQ1H

			QJJN██0█Y█U61EPLX1A9L1IPN1

			0Z9█H1F███O1Z60WI80VXECFTI

			0MM█MIM█M█1E█1M█M1█A█MM█M1

			U8F█HGG█1█O█0█0██1█0██L█ZV

			LKJX██UH6SO███4█K██N█4██ZS

			M0MM000MLMM█M█1█MM█M█MM█MM

			B90I89F11IN█E█G█0Z█0█XL█NN

			0NICE1JK8TGC1LLNJ0Z9V6R8T0

			Z66C9OYXFB4HOBTHROK9R99QGU

			O1MMFMM0MMG0T██MM0MGXF0MPF

			1███X06C█Z0T█E6F1G█0H█UGU9

			F█0TCT1L█YF0G█A046█TN█1██F

			P██0JTU1█CC41Q█CU0█L1█6█G█

			Z█JPZ0JP█F0N██1THX█Y6█0██1

			J███M8MM███OOHMMLMI██CP█F█

			E09A89880X8VS641AZFC049█0█

			RO0███1XQ███0FW0WXPOFW1UHS

			0LT█0YI4ZQ█ANA1NH1XXR███91

			R80██SMM1M█M0U█MM█MM█M0M1H

			R1U█PJ0ZZG█1L0██U█Y0█JOSEC

			PBT█1GBN0███0N█Y██8F█1█RC1

			HNMAF9MPMMOMOT█MM█MM█MM█SX

			94ZWQJQFP9FX8Q█XK█RKL███YE

The shapes of the shaded prime digits spell out (roughly) the final answer, CHANNEL SURFING.