ALL YOU HAVE TO DO IS TAKE A CUP OF FLOUR, ADD IT TO THE MIX
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add me on steam, it's not comfy to chat in comments
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I'm still not ok with the last part of the Q1: there is no limitation on how much gifts Tommy can send, except his wallet. Imagine Tommy is in fact Bill Gates in disguise, he can buy 3,000,000 copies (300,000 10copies bundles for example) even if he only has 2 friends. He'll waste 2,999,998 copies, which is AT LEAST 1 copie. It think you want to say AT MOST 1 copie, but there's still informations we're missing:
-If there is a restriction on how much bundles he can buy of each time.
-If he sent a gift to every of his friends the first time.
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If he wants to give away 2,999,998 copies, he can buy exactly that many.
If he want to give away, say, 7 copies, he can't buy exactly that many. The question is asking what is the highest number he can want to give away where he'll still be forced to buy extra copies.
I've figured that one out, but it's not really helping me at all anyway. Still, if that helps you out, that's great.
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Tommy can try to buy any number of gifts (games, copies). But he is forced (thanks Award for the word) to buy in bundles. And thus there are certain numbers of gifts (not bundles) that he can't get by any combinations of, say, 6,9 or 29 copies bundles, namely, 10. But there is a maximum number after which any number of gifts can be purchased by some combinations of these bundles.
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I like cookies. Do you?
Sugar cookie recipe!
Also: to cook you need a Timer
Also: don't hint, don't cheat, be nice. Thank you.
General hints:
Levels of difficulty for cookies:
Free Cookie - effortless
The Cookie of Wise Choice - very easy
The Inquisitive Cookie - easy
The Cookie of Understanding - normal
The Cookie of Perseverance - hard
The Greek Cookie - hard/very hard
The Obvious Cookie - normal?
Number of cookie-eaters, per cookie, at least:
Free Cookie (infinite)
The Cookie of Wise Choice (53)
The Inquisitive Cookie (9)
The Cookie of Understanding (6)
The Cookie of Perseverance (9)
The Greek Cookie (3)
The Obvious Cookie (13)
Anthrobomination, Award (7/7!)
froost (6/7+) first to get to original cookies
Sorry! Wrong alarm for Q3!
I should sound like Wheatley from Portal 2. "False alarm! False alarm!"
Clarifications/Corrections for math questions.
By the way i have solvers for all these questions. So they are solvable! :)
Q1. Concerning the following:
"What is the maximum number of copies Tommy can purchase but STILL waste at least 1 copy?"
Q3. Puzzle version is correct.
Q4. Consider that every game in those 15 similar chance giveaways is different, same for 6.
Also relevant:
In the Hall of Perseverance you can find Tom (not Tommy). It is Jimmy in disguise!
Also there (silly me!) for Stephanie it's 'hours' not 'minutes'!
HINTS.
Hint #1. Basic help.
If this very thread were "a cookie" its recipe (ingredients and order) would be:
Basic help comprehension test
Hint #2. Advanced help.
Additional hints:
Hint #3. The Obvious Cookie.
Get it here
Small mistake in Q3. Correct picture here
Solutions.
First, you had to get to the quiz.
The link was hidden in the Sugar Cookie Recipe.
The actual quiz was easy. You had multiple choice answers and could just guess the right answer.
By solving the quiz you are given certain pieces of information:
Link to the Cookie of Wise Choice.
The information that there are four other cookies hidden somewhere.
The link:
http://www.steamgifts.com/giveaway/xfxvX/cookie-of-wise-choice
That was invalid.
When closely inspected the link itself reveals the code 'lxbxl' while the clickable name shows 'xfxvX'.
It's a mixture of giveaway code itself and its format (x for lower case, X for upper or digits) usually given here on steamgifts.
When combined, it gives 'lfbvl' as a code and 'xxxxX' as format --> 'lfbvL'.
When getting to the giveaway you were hinted to look into the HINTS to the quiz.
“I hope you did not use Hints, because there are none! Ha-ha!”. Hints section was critical to get to the Cookie of Perseverance.
The Cookie of Perseverance.
This was on a straight path, but required math.
The actual ingredients were hidden in the hints to the main quiz. You had to solve math problems and get answers. All answers were numbers which you had to translate into symbols using the corresponding ASCII code.
Full solution here
Also you need drawings for Q2. Here
In the end you have the following numbers: 113, 53, 84, 75 ,72, which in ASCII is 'q5TKH'. This directs you here , the long forgotten giveaway. There you should find my comment and get to The Hall of Perseverance. Which is another quiz.
There you have to answer 5 questions correctly and use the correct sum as an answer.
Answers for the questions:
Q1.
Any number, but 3, can be represented by combination of 6 and 9.
29 mod 3 = 2
29*2 mod 3 = 58 mod 3 = 1.
So the maximum number is 58+3=61.
Q2. You have it already. 22.
Q3. So it should be 'hours' as pointed out. It is 7+33+1 = 41
Q4. To win at least 1 is not to lose at all. To lose all giveaways: (1/2) (3/4) (9/10) = 27/80
To win at least 1 is 1 – 27/80 = 53/80 = 0,6625
Q5. About a Mite, Fishing Floor and Super Fleet Buoy will sum up to 35. Fees for submission: 1+2+4+8+16+32+64=127. Total 35+127=162
The sum of answers is 162+0,6625+41+22+61 = 286,6625
The Inquisitive Cookie.
If you were inquisitive enough you'd check the main quiz for all multiple answers. And find out, that in every question exactly one was wrong. You could do it even without math just by checking 'letters'.
The letters corresponding to wrong answers are: c,C(!),f,d,a. What is this? A giveaway code: cCfda. It goes here. There you need to find my comment and get to The Hall of Inquisitives .
A simple name the game quiz! Do it yourself if you want to get to the giveaway.
The Cookie of Understanding.
And it was hiding in the same place as the Inquisitive Cookie! In the wrong answers (“check answers twice!” hint). Wrong numbers are: 55,110, 73 (this one is tricky), 65 and 48. Or in ASCII it is '7nIA0'.
'73' is tricky because it's the only wrong number in sequence.
Now that you have the code you should go here. Find my comment and proceed to the Hall of Understanding. There you need to answer 3 easy questions.
Answers:
Q1. 191
Q2. 201
Q3. 75
Answer to the quiz is 750
The Greek Cookie.
That's my favorite. The only explicit hint you had is word 'Greek' written in italics. So you had to find something written in italics in the main quiz. And that what you should see:
Leopard Hot lines
Inhuman Car Loop
Too Intimate Roach
A Mad Magpie Battle
About a mite
Again you could be wondering what it is. The answer is simple: it's Greek, it's math, you have 2 examples in Q2 (alpha, beta). So you need to use the Greek alphabet. And those strange names are just anagrams. Of Greek letters. Besides the number of words means how many letters to look for.
So,
Leopard Hot lines = rho + delta + epsilon
Inhuman Car Loop = alpha + omicron + nu
Too Intimate Roach = omicron + iota + theta
A Mad Magpie Battle = gamma + beta + delta + pi
About a mite = mu + iota + beta
But then, again. What should you do with them? The answer is Greek numerals.
Leopard Hot lines = rho + delta + epsilon = 100 + 4 + 5 = 109
Inhuman Car Loop = alpha + omicron + nu = 1 + 70 + 50 = 121
Too Intimate Roach = omicron + iota + theta = 70 + 10 + 9 = 89
A Mad Magpie Battle = gamma + beta + delta + pi = 3 + 2 + 4 + 80 = 89
About a mite = mu + iota + beta = 40 + 10 + 2 = 52
Turning to ASCII (myYY4) leads you here. There you should find my comment and proceed to the Greek Hall
In the Hall again you are asked 3 questions. Answers are:
Q1. Aeronautic Tomb = beta + omicron + tau = 2 + 70 + 300 = 372
Q2. Adult Enigmas = sigma + delta + nu = 254
Q3. Immolated Organics = omicron + sigma + delta = 274
Q1 + Q2 + Q3 = 900.
But that was not the answer. Because you should have used the Greek letter instead of the numbers. And this letter is 'sampi'.
The Obvious Cookie.
Answers to the quiz:
Q1. Easy to calculate: (1626 - 4)/2 = 811, (811 - 3)/2 = 404, (404 - 2)/2 = 201, (201 - 1)/2 = 100, (100-0)/2 =50.
Answer is 50.
Q2. The probability can be calculated like this: the chance to pick out letter 'c' as first is 1/6, letter 'o' as second = 2/5, 'o' as third = 1/4, 'k' as fourth = 1/3, 'i' as fifth = 1/2, 'e' as sixth = 1. The total probability is the product of these, because they need to happen simultaneously. So P = (1/6) (2/5) (1/4) (1/3) (1/2) * (1) = 2/720 = 1/360.
Answer is 1/360.
Q3. The answer is 'cake'. Full explanation for that on the giveaway page, in one of the comments.
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