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I'm designing a puzzle game, and am trying to play-test some concepts. Can you solve these?

108153767833 = A + B + C. What are the components and why?

A = ?

B = ?

C = ?

Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.

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    $\begingroup$ Yep. They are related. $\endgroup$ – NigelMNZ Dec 7 '18 at 20:41
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    $\begingroup$ yeah never mind, I should have looked closer:-) $\endgroup$ – deep thought Dec 7 '18 at 20:42
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Partial:

C

Mazes with the following rules:
1. Take the shortest path possible starting at green and ending at red.
2. If yellow dots, all must be "retrieved" before reaching red.
3. Red cannot be crossed, thus only reached once.
4. Colored squares are "locked doors" that are only passable once you have "retrieved" the associated colored triangle.
Based on these rules and clarification provided by a hint and a comment, we can now solve the last maze, which gives us a solution of
$99453888111$


A

Found by @deep_thought. Please review their answer.


B

After working through this and making a mess of a wall of text, I found that @JonMark_Perry was better able to succinctly describe the process.
Let's look at #1 for an example though:
Example 1. a_1_z -> pp=11
p falls in the range of a to z, thus p is 1. As p is printed twice, the result is 11.

Example 2. a_1_ij_2_op_3_z -> pi = 31 ty =33
set p-z includes p, t, and y, thus p, t, and y are all 3.
set a-i includes i, thus i is 1

Example 3 gets a little interesting. Here, q_1_k translates to ^k-q = 1, or letters not in the set of k-q get +1.
3 also introduces the idea that set values can stack.
As such, p is in set a-p and not in set k-q, thus is 1 + 1 or 2.
l is in set a-p, k-q, l-z, thus is 1 + -1 or 2. (Doesn't get the 1 from ^k-q as l is not not in k-q.

Thus brings us to Example 5, which we must solve to find B. The rule sets are as follows:
a-z = 1
a-m = 1 n-z = 2
a-p = 2 q-z = 1
a-e = 1 f-z = 3
a-j = 1 k-s = 2 t-z = 3
a-p = -2 q-z = -1
These thus result in the following letter values:
p = 1 2 2 3 2 -2 = 8
i = 3 1 = 4
t = 1 2 3 3 = 9
y = 1 2 3 3 = 9
l = 1 1 3 2 = 7
e = 1 1 1 1 = 4
a = 1 1 1 1 = 4
s = 1 2 3 2 = 8
And when the word pityplease is decoded, we get
B = $8499874484$
However, this value appears to be incorrect.

Actually, what if a_-2_pq_-1_z were instead just a_-1_z:
p = 1 2 2 3 2 -1 = 9
i = 1 1 2 3 1 -1 = 7
t = 1 2 1 3 3 -1 = 9
y = 1 2 1 3 3 -1 = 9
l = 1 1 2 3 2 -1 = 8
e = 1 1 2 1 1 -1 = 5
a = 1 1 2 1 1 -1 = 5
s = 1 2 1 3 2 -1 = 8
$9799985585$
Perfect! Just what we needed!

Answer:

If T is $109253878934$
And A is confirmed $5238$
And C is confirmed $99453888111$
Then B must be $109253878934$ - $5238$ - $99453888111$ = $9799985585$
However, I can't get my B formula to match.
Actually, if we use the slightly modified rule for B5, then we get the desired result!
Technically not correct, but I'm happy with it for now.

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  • $\begingroup$ Looks like you are correct. Totally an oversight on my part. Nice job finding that. My intended solution was none of those three, interestingly enough. $\endgroup$ – NigelMNZ Dec 7 '18 at 21:30
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    $\begingroup$ you can have 94492888111 as well. $\endgroup$ – JMP Dec 7 '18 at 21:32
  • $\begingroup$ For clarity, my previous comment was in reference to your solution for C. $\endgroup$ – NigelMNZ Dec 7 '18 at 21:58
  • $\begingroup$ Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C. $\endgroup$ – NigelMNZ Dec 7 '18 at 22:50
  • $\begingroup$ Yes but my current C value actually works for that hint @NigelMNZ. $\endgroup$ – Dorrulf Dec 7 '18 at 22:51
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Partial answer

For A,

if we ignore tilt and position, then I can fit the four numbers at the top with: colour is power of ten, shape is digit (star is one), filled is positive, empty is negative; sum all shapes.
- filled red star = 1
- filled red triangle and filled red square = 3 + 4 = 7
- filled blue pentagon and empty red heptagon = 50 - 7 = 43
- filled green triangle and empty blue square and empty red enneagon = 300 - 40 - 9 = 251
This gives P^1 = 157, P^2 = 857, P^3 = 1201. These are prime factors of the number at the bottom. P^4 is the missing factor is 3023. Their sum is A = 5238.

Part B: no clue

Part C: already solved by @Dorrulf

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  • 1
    $\begingroup$ Nicely done for A! $\endgroup$ – Dorrulf Dec 7 '18 at 22:05
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    $\begingroup$ Indeed, very nice. $\endgroup$ – NigelMNZ Dec 7 '18 at 22:16
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B.

Every a_b_c triple forms a range in the alphabet that scores b points, for example a_1_m means that every letter in [a,m] scores 1 point, with wrapround if the range is backwards. I get $B=8699874484$.

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