Three Visual Puzzles
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?
Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.
mathematics visual mazes alphametic
add a comment |
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?
Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.
mathematics visual mazes alphametic
2
Yep. They are related.
– NigelMNZ
Dec 7 '18 at 20:41
1
yeah never mind, I should have looked closer:-)
– deep thought
Dec 7 '18 at 20:42
add a comment |
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?
Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.
mathematics visual mazes alphametic
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?
Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.
mathematics visual mazes alphametic
mathematics visual mazes alphametic
edited Dec 8 '18 at 6:19
asked Dec 7 '18 at 20:38
NigelMNZ
1406
1406
2
Yep. They are related.
– NigelMNZ
Dec 7 '18 at 20:41
1
yeah never mind, I should have looked closer:-)
– deep thought
Dec 7 '18 at 20:42
add a comment |
2
Yep. They are related.
– NigelMNZ
Dec 7 '18 at 20:41
1
yeah never mind, I should have looked closer:-)
– deep thought
Dec 7 '18 at 20:42
2
2
Yep. They are related.
– NigelMNZ
Dec 7 '18 at 20:41
Yep. They are related.
– NigelMNZ
Dec 7 '18 at 20:41
1
1
yeah never mind, I should have looked closer:-)
– deep thought
Dec 7 '18 at 20:42
yeah never mind, I should have looked closer:-)
– deep thought
Dec 7 '18 at 20:42
add a comment |
3 Answers
3
active
oldest
votes
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.
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.
– NigelMNZ
Dec 7 '18 at 21:30
1
you can have 94492888111 as well.
– JonMark Perry
Dec 7 '18 at 21:32
For clarity, my previous comment was in reference to your solution for C.
– NigelMNZ
Dec 7 '18 at 21:58
Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
– NigelMNZ
Dec 7 '18 at 22:50
Yes but my current C value actually works for that hint @NigelMNZ.
– Dorrulf
Dec 7 '18 at 22:51
|
show 2 more comments
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
1
Nicely done for A!
– Dorrulf
Dec 7 '18 at 22:05
1
Indeed, very nice.
– NigelMNZ
Dec 7 '18 at 22:16
add a comment |
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 $B=8699874484$.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "559"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f76180%2fthree-visual-puzzles%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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.
– NigelMNZ
Dec 7 '18 at 21:30
1
you can have 94492888111 as well.
– JonMark Perry
Dec 7 '18 at 21:32
For clarity, my previous comment was in reference to your solution for C.
– NigelMNZ
Dec 7 '18 at 21:58
Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
– NigelMNZ
Dec 7 '18 at 22:50
Yes but my current C value actually works for that hint @NigelMNZ.
– Dorrulf
Dec 7 '18 at 22:51
|
show 2 more comments
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.
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.
– NigelMNZ
Dec 7 '18 at 21:30
1
you can have 94492888111 as well.
– JonMark Perry
Dec 7 '18 at 21:32
For clarity, my previous comment was in reference to your solution for C.
– NigelMNZ
Dec 7 '18 at 21:58
Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
– NigelMNZ
Dec 7 '18 at 22:50
Yes but my current C value actually works for that hint @NigelMNZ.
– Dorrulf
Dec 7 '18 at 22:51
|
show 2 more comments
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.
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.
edited Dec 7 '18 at 23:56
answered Dec 7 '18 at 21:12
Dorrulf
2,26219
2,26219
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.
– NigelMNZ
Dec 7 '18 at 21:30
1
you can have 94492888111 as well.
– JonMark Perry
Dec 7 '18 at 21:32
For clarity, my previous comment was in reference to your solution for C.
– NigelMNZ
Dec 7 '18 at 21:58
Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
– NigelMNZ
Dec 7 '18 at 22:50
Yes but my current C value actually works for that hint @NigelMNZ.
– Dorrulf
Dec 7 '18 at 22:51
|
show 2 more comments
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.
– NigelMNZ
Dec 7 '18 at 21:30
1
you can have 94492888111 as well.
– JonMark Perry
Dec 7 '18 at 21:32
For clarity, my previous comment was in reference to your solution for C.
– NigelMNZ
Dec 7 '18 at 21:58
Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
– NigelMNZ
Dec 7 '18 at 22:50
Yes but my current C value actually works for that hint @NigelMNZ.
– Dorrulf
Dec 7 '18 at 22:51
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.
– NigelMNZ
Dec 7 '18 at 21:30
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.
– NigelMNZ
Dec 7 '18 at 21:30
1
1
you can have 94492888111 as well.
– JonMark Perry
Dec 7 '18 at 21:32
you can have 94492888111 as well.
– JonMark Perry
Dec 7 '18 at 21:32
For clarity, my previous comment was in reference to your solution for C.
– NigelMNZ
Dec 7 '18 at 21:58
For clarity, my previous comment was in reference to your solution for C.
– NigelMNZ
Dec 7 '18 at 21:58
Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
– NigelMNZ
Dec 7 '18 at 22:50
Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
– NigelMNZ
Dec 7 '18 at 22:50
Yes but my current C value actually works for that hint @NigelMNZ.
– Dorrulf
Dec 7 '18 at 22:51
Yes but my current C value actually works for that hint @NigelMNZ.
– Dorrulf
Dec 7 '18 at 22:51
|
show 2 more comments
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
1
Nicely done for A!
– Dorrulf
Dec 7 '18 at 22:05
1
Indeed, very nice.
– NigelMNZ
Dec 7 '18 at 22:16
add a comment |
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
1
Nicely done for A!
– Dorrulf
Dec 7 '18 at 22:05
1
Indeed, very nice.
– NigelMNZ
Dec 7 '18 at 22:16
add a comment |
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
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
edited Dec 7 '18 at 22:02
answered Dec 7 '18 at 21:52
deep thought
2,6921734
2,6921734
1
Nicely done for A!
– Dorrulf
Dec 7 '18 at 22:05
1
Indeed, very nice.
– NigelMNZ
Dec 7 '18 at 22:16
add a comment |
1
Nicely done for A!
– Dorrulf
Dec 7 '18 at 22:05
1
Indeed, very nice.
– NigelMNZ
Dec 7 '18 at 22:16
1
1
Nicely done for A!
– Dorrulf
Dec 7 '18 at 22:05
Nicely done for A!
– Dorrulf
Dec 7 '18 at 22:05
1
1
Indeed, very nice.
– NigelMNZ
Dec 7 '18 at 22:16
Indeed, very nice.
– NigelMNZ
Dec 7 '18 at 22:16
add a comment |
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 $B=8699874484$.
add a comment |
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 $B=8699874484$.
add a comment |
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 $B=8699874484$.
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 $B=8699874484$.
answered Dec 7 '18 at 22:34
JonMark Perry
17.5k63584
17.5k63584
add a comment |
add a comment |
Thanks for contributing an answer to Puzzling Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f76180%2fthree-visual-puzzles%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
Yep. They are related.
– NigelMNZ
Dec 7 '18 at 20:41
1
yeah never mind, I should have looked closer:-)
– deep thought
Dec 7 '18 at 20:42