How about
for proper punctuation? I’ve been playing around with this thing for a while in Word with the fonts, etc., haven’t seen anything.
Clint, “not the first letters I would choose” kind of strikes me because I was thinking it could be something like picking certain letters out of the message.
PS I RECOMMEND USING TIMES NEW ROMAN AND PROPER PUNCTUATION
Maybe we should be looking for the anagram in “times new roman” and “proper punctuation”
I’ve been lurking on this thread for a while but didn’t bother trying to solve any of the puzzles because I figured, if the others on the board haven’t had success, what chance did I have. Leason learned: don’t underestimate yourself.
In Microsoft Word, I pasted the properly punctuated text in Times New Roman and shrunk the margins until the message filled 11 lines. (One for each letter of “it’s an egnigma.”)
Taking the first letter of each line produced the anagram “inane stigma.” Go ahead and send an e-mail to inanestigma@gmail.com.
Wow. Nice work; that has to be correct.
And the canned response is a picture I can’t see which links to this:
Which is an integral which will lead to a zip file
Wow ! Nice work !
Great job! I knew it had to be the “first letter” of something, and I tried splitting it up after each punctuation, first letters of various phrases, but I didn’t think of that. Makes sense now, of course. Cool.
So, I’m assuming that solving the equation gives us the password to the .zip file. Any thoughts? I’m looking at it now and still retain maybe some lingering calculus skills, but the pictures … weird. The number in each photo name doesn’t match the number of the file on Flickr. As far as I know, the flickr file number is assigned, but if the number in the name is different then it must be important. And of course, the number in the title bar of the simianagent page is another 8-digit number.
The log is base five. The formulas for the summation are here. K(K+1) = K[sup]2[/sup] + K, so the summation should be (n[sup]2[/sup] + 1)/2 + (1/6)n(n+1)(2n+1). The integral is just x, so it’s the two numbers multiplied together.
Doing the math, I get either 890523362175408000000000000000000 or 20644753261387400000000000000000, depending on whether the order of operations is (log[sub]5[/sub]x)[sup]3[/sup] or log[sub]5/sub. Neither is the password to the zipfile. (Not guaranteeing I don’t have a typo in my math.)
And just to share the obvious - the photos in the equation were posted by Nina G Tamise.
My problem is precision, unfortunately. Shouldn’t have those trailing zeroes.
I’ll also point out that the file IDs for the 5 photos are:
58401196
83955266
29580037
00606004
96536514
Maybe nothing here - but if the log is base 5, and there’s 5 photos, well . . . . .
And on preview - the first number from each ID is 58209 which is an anagram for 89052, the first 5 digits from zut’s answer.
Do you mean (n[sup]2[/sup]+n)/2 instead of n[sup]2[/sup]+1?
Oops. Yes I do.
Don’t have time to look further into it, but there are a bunch more 8-digit combos on the page. The whole equation is made of multiple pic files with 8-digit numbers. For instance right clicking the top of the ‘d’ in ‘dx’ you get the filename 43480419.jpg with an alt text of 19709881. The easiest way to see them all is probably just looking at the page source.
OK, I downloaded an ultra-high-precision calculator from here, and ran the following through it:
13002966*((log(89387894)/log(5))^3)((51871595^2 + 51871595)/2 + (1/6)51871595(51871595+1)(251871595+1)) + 9490698171869188
Where log(89387894)/log(5) is the base-five log of 89387894, ((51871595^2 + 51871595)/2 + (1/6)51871595(51871595+1)(251871595+1)) is the summation of K(K+1) up to 51871595, and 94906981*71869188 is the integral. The calculation yields 890523362175408206188016635678243.567084228102494438350222308082666748475930299893. Neither 890523362175408206188016635678243 nor 890523362175408206188016635678244 is the password.
Alternatively, cubing before taking the log yields 20644753261387387011306613304332.9851279314732864376105917483430033636527413910648, which rounds off much more nicely. However, neither 20644753261387387011306613304332 nor 20644753261387387011306613304333 works.
89052 is the ZIP code for Henderson Nevada, not sure if that means anything?
whoops did not see the file download
These are the embedded codes:
25312720, 12836832, 62372751, 96536514, 77520022, 40943998, 71869188, 12438520, 46999099, 21355409, 84912151, 60624886, 51871595, 98867990, 61438018, 89555379, 58118313, 21355409, 41789453, 57690853, 64408695, 21355409, 30306914, 62586777, 75526364, 21355409, 13002966, 76764719, 35786040, 33301939, 89387894, 56426502, 89064070, 70352986, 56298251, 39947909, 94906981, 25317803, 19709881, 43480419, 12997724, 21355409, 81119936, 29352282, 71460262, 30956895, 96705443, 74500181, 96733413, 21355409
The page has a title with 89585544, and the 5 images are 54801196, 83955266, 29580037, 00606004 and 96536514.
So 56 8-digit sets by my count.
I am trying to figure the problem, but my calculus is quite weak. Can you just plug the image numbers into the equation where the images appear?