THIS IS A READONLY ARCHIVE FROM THE SORABJI.COM MESSAGE BOARDS (19952016). 

Each letter has an associated numerical value and the total of all the letters is the bird's weight. For example, if the letter A, B, C, D, I, K, L and R had the values 15, 3, 22, 7, 20, 12, 19 and 1 respectively, the weight of a BLACKBIRD would be 102 (3+19+15+22+12+3+20+1+7). Unfortunately, the scales broke just before Codey weighed the last bird. Your puzzle solving mission is to figure out what that last birdthe hummingbirdshould have weighed. Your puzzle solving mission is to figure out what that last birdthe hummingbirdshould have weighed. barn owl 56 oriole 49 condor 32 osprey 56 egret 67 pelican 63 eider 53 pintail 64 falcon 54 quail 49 finch 58 robin 27 grebe 59 snipe 43 grouse 62 sparrow 41 gull 72 stork 50 hawk 59 tanager 64 heron 42 tern 38 ibis 34 vireo 44 junco 48 warbler 66 loon 28 wren 43 hummingbird ? 
This puzzle was created by one of NSA's master puzzle makers. It requires you to solve linear equations and do some cryptanalysis. The puzzle is more difficult than it initially appears. Just remember that it can be solved using the information provided without the benefit of a computer. A pencil and paper and some persistence are all you need. Of course, if you want to write a program to help solve the puzzle, that's okay, too. 
hint: it's a mouthful 
Do you want the proof? 
BARN OWLLOONWARBLER+ORIOLE reduces to O+I=11 WRENTERN reduces to WT=5 HERONTERN reduces to H+OT=4 TANAGERTERN reduces to 2A+G=26 EGRETGREBE reduces to TB=8 ETC. Using O+I=11, (GREBE+TERN+(O+I)EGRETROBIN reduces to E=14 Using E=14 and 2A+G=26, 2(TANAGER)EGRETTERN reduces to N+2A=11 Make assumptions and produce putative partial solutions. If N+2A=11 and N>0 and A>0 then A=1,2,3,4 or 5. Create a table in which each row represents a putative partial solution derived using the assumed value of A and E=14. For example, from step 1 we know that 2A+G=26. If a putative solution for A is found then a solution for G can be found also. Using WREN and assuming values for N results in putative values for W+R. Likewise GREBE produces putative values for R+B and WARBLER produces putative values for L. LOON produces putative values for O. Start eliminating putative solutions. Putative Values for Variables Solution A N G W+R R+B L O 1 1 9 24 20 7 24 2 2 7 22 22 9 19 1 3 3 5 20 24 11 14 4 4 3 18 26 13 9 8 5 5 1 15 28 15 4 LOON has a weight of 28 and therefore L+N must be divisible by two. Eliminate solutions 1, 3 and 5. Eliminate solution 4. Calculate putative values for I and R using O+I=11 and ORIOLE. Use EIDER to calculate putative values for D. Substituting D=15 into CONDOR yields C=9. Therefore eliminate solution 4. You are then left with this: Solution A N G W+R R+B L O I R D C 2 2 7 22 22 9 19 1 10 4 11 8 4 4 3 18 26 13 9 8 3 7 15 9 By a simple process of elimination, A=2, N=7, G=22, L=19, O=1, etc. is the solution. Solve for remaining letters. Complete the mixed alphabet = OAPRBQNCSIDUTEVHFWLJ_GK_Y_ except for M, X, and Z Recover key matrix and determine M=26 O R N I T H L G Y A B C D E F J K M P Q S U V W X Z Calculate HUMMINGBIRD = 149. Very good, Christopher. May I ask, where are you from? 
http://www.nsa.gov/programs/kids/standard/kitchen/master/solution.shtml 
The Abominable Dr. Phibes = A Hibernated Bimbo Helps. Pretty cool, ain't it? 
Actually, I work for the CPSG (Cryptology Program Systems Group) division of AIA (Air Intelligence Agency). It is part of DISA (Defense Information Systems Agency). Part of my job IS electronic encryption. 


and then he drank it. heehee. 

LS 