Counting all the hearts
$begingroup$
The Arthur family ( Henrik, Olga, Heather and Kristophe) are playing Bridge at the dining table with a standard deck of cards.
Taking into account every possibility
How many hearts are at that table?
lateral-thinking
asked 10 hours ago
DEEMDEEM
6,349120113
$endgroup$
add a comment |
$begingroup$
The Arthur family ( Henrik, Olga, Heather and Kristophe) are playing Bridge at the dining table with a standard deck of cards.
Taking into account every possibility
How many hearts are at that table?
lateral-thinking
asked 10 hours ago
DEEMDEEM
6,349120113
$endgroup$
add a comment |
$begingroup$
The Arthur family ( Henrik, Olga, Heather and Kristophe) are playing Bridge at the dining table with a standard deck of cards.
Taking into account every possibility
How many hearts are at that table?
lateral-thinking
asked 10 hours ago
DEEMDEEM
6,349120113
$endgroup$
The Arthur family ( Henrik, Olga, Heather and Kristophe) are playing Bridge at the dining table with a standard deck of cards.
Taking into account every possibility
How many hearts are at that table?
lateral-thinking
lateral-thinking
asked 10 hours ago
DEEMDEEM
6,349120113
asked 10 hours ago
DEEMDEEM
6,349120113
asked 10 hours ago
DEEMDEEM
6,349120113
asked 10 hours ago
DEEMDEEM
6,349120113
asked 10 hours ago
DEEMDEEM
6,349120113
6,349120113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First,
there are $4$ human hearts.
Then,
Considering this image of standard playing cards:
The cards have $2$ hearts on each card next to the name of the card, each face card has $2$ additional hearts in the art, and every other card has its number of hearts, totaling $$2cdot13+3cdot2+(1+2+dots+10)=26+6+55=87.$$
Finally,
The family's names contain letters that form the word heart. We will use Dr Xorile's suggestion to count the total number of ways to form heart using different instances of the letters in their names. If we take their full names: Henrik Arthur, Olga Arthur, Heather Arthur, and Kristophe Arthur, we count $8$ H's, $4$ E's, $6$ A's, $11$ R's, and $6$ T's, giving $$8cdot4cdot6cdot11cdot6=12672.$$
which adds up to
$4+87+12672=12763$.
answered 10 hours ago
noednenoedne
6,60711956
$endgroup$
1
$begingroup$
You have to take into account every possibility: what if one of them is a timelord?
$endgroup$
– Arnaud Mortier
10 hours ago
1
$begingroup$
Do the people portrayed in face cards have hearts? (I say no, they are mere abstract sketches of people. Perhaps DEEM thinks otherwise.)
$endgroup$
– Gareth McCaughan♦
9 hours ago
1
$begingroup$
@GarethMcCaughan I was thinking the same. After all, the King cannot still have a heart with that sword stuck in his head for so long.
$endgroup$
– noedne
9 hours ago
2
$begingroup$
At some point we might begin to suspect that the Arthurs are octupuses.
$endgroup$
– noedne
9 hours ago
1
$begingroup$
Every possibility? Well, for all we know there's another part of the table piled high with dozens of other packs of cards...
$endgroup$
– Gareth McCaughan♦
9 hours ago
|
show 5 more comments
$begingroup$
Starting with noedne's analysis of
87 hearts
from the card deck alone. We also have:
four (I assume) humans, that have one heart each... except that the two women could be pregnant (take into account every possibility!), and with humans twinning is reasonable, but triplets are pretty rare, so I would say up to 8 human hearts.
Oops, almost forgot to look at:
The text itself! The Arthur family has a cleverly hidden heart, and there are enough letters in the other names for 2 more, adding these to the human and card hearts are 87+8+3 for a grand total of 98 hearts. I think this is a stretch, but the fact that there are 52 cards in a deck also looks like a heart(
), so that would make 99.
Or if you want to be absolutely ridiculous:
Octuplets have been born a few times, so you could have 4+8+8=20 human hearts for a grand total of 110, but at that rate do you count multiple births higher than eight as long as they can have a beating heart in the womb? That's why I consider this a ridiculous option.
answered 5 hours ago
NH.NH.
32119
$endgroup$
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%2f80562%2fcounting-all-the-hearts%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First,
there are $4$ human hearts.
Then,
Considering this image of standard playing cards:
The cards have $2$ hearts on each card next to the name of the card, each face card has $2$ additional hearts in the art, and every other card has its number of hearts, totaling $$2cdot13+3cdot2+(1+2+dots+10)=26+6+55=87.$$
Finally,
The family's names contain letters that form the word heart. We will use Dr Xorile's suggestion to count the total number of ways to form heart using different instances of the letters in their names. If we take their full names: Henrik Arthur, Olga Arthur, Heather Arthur, and Kristophe Arthur, we count $8$ H's, $4$ E's, $6$ A's, $11$ R's, and $6$ T's, giving $$8cdot4cdot6cdot11cdot6=12672.$$
which adds up to
$4+87+12672=12763$.
answered 10 hours ago
noednenoedne
6,60711956
$endgroup$
1
$begingroup$
You have to take into account every possibility: what if one of them is a timelord?
$endgroup$
– Arnaud Mortier
10 hours ago
1
$begingroup$
Do the people portrayed in face cards have hearts? (I say no, they are mere abstract sketches of people. Perhaps DEEM thinks otherwise.)
$endgroup$
– Gareth McCaughan♦
9 hours ago
1
$begingroup$
@GarethMcCaughan I was thinking the same. After all, the King cannot still have a heart with that sword stuck in his head for so long.
$endgroup$
– noedne
9 hours ago
2
$begingroup$
At some point we might begin to suspect that the Arthurs are octupuses.
$endgroup$
– noedne
9 hours ago
1
$begingroup$
Every possibility? Well, for all we know there's another part of the table piled high with dozens of other packs of cards...
$endgroup$
– Gareth McCaughan♦
9 hours ago
|
show 5 more comments
$begingroup$
First,
there are $4$ human hearts.
Then,
Considering this image of standard playing cards:
The cards have $2$ hearts on each card next to the name of the card, each face card has $2$ additional hearts in the art, and every other card has its number of hearts, totaling $$2cdot13+3cdot2+(1+2+dots+10)=26+6+55=87.$$
Finally,
The family's names contain letters that form the word heart. We will use Dr Xorile's suggestion to count the total number of ways to form heart using different instances of the letters in their names. If we take their full names: Henrik Arthur, Olga Arthur, Heather Arthur, and Kristophe Arthur, we count $8$ H's, $4$ E's, $6$ A's, $11$ R's, and $6$ T's, giving $$8cdot4cdot6cdot11cdot6=12672.$$
which adds up to
$4+87+12672=12763$.
answered 10 hours ago
noednenoedne
6,60711956
$endgroup$
1
$begingroup$
You have to take into account every possibility: what if one of them is a timelord?
$endgroup$
– Arnaud Mortier
10 hours ago
1
$begingroup$
Do the people portrayed in face cards have hearts? (I say no, they are mere abstract sketches of people. Perhaps DEEM thinks otherwise.)
$endgroup$
– Gareth McCaughan♦
9 hours ago
1
$begingroup$
@GarethMcCaughan I was thinking the same. After all, the King cannot still have a heart with that sword stuck in his head for so long.
$endgroup$
– noedne
9 hours ago
2
$begingroup$
At some point we might begin to suspect that the Arthurs are octupuses.
$endgroup$
– noedne
9 hours ago
1
$begingroup$
Every possibility? Well, for all we know there's another part of the table piled high with dozens of other packs of cards...
$endgroup$
– Gareth McCaughan♦
9 hours ago
|
show 5 more comments
$begingroup$
First,
there are $4$ human hearts.
Then,
Considering this image of standard playing cards:
The cards have $2$ hearts on each card next to the name of the card, each face card has $2$ additional hearts in the art, and every other card has its number of hearts, totaling $$2cdot13+3cdot2+(1+2+dots+10)=26+6+55=87.$$
Finally,
The family's names contain letters that form the word heart. We will use Dr Xorile's suggestion to count the total number of ways to form heart using different instances of the letters in their names. If we take their full names: Henrik Arthur, Olga Arthur, Heather Arthur, and Kristophe Arthur, we count $8$ H's, $4$ E's, $6$ A's, $11$ R's, and $6$ T's, giving $$8cdot4cdot6cdot11cdot6=12672.$$
which adds up to
$4+87+12672=12763$.
answered 10 hours ago
noednenoedne
6,60711956
$endgroup$
First,
there are $4$ human hearts.
Then,
Considering this image of standard playing cards:
The cards have $2$ hearts on each card next to the name of the card, each face card has $2$ additional hearts in the art, and every other card has its number of hearts, totaling $$2cdot13+3cdot2+(1+2+dots+10)=26+6+55=87.$$
Finally,
The family's names contain letters that form the word heart. We will use Dr Xorile's suggestion to count the total number of ways to form heart using different instances of the letters in their names. If we take their full names: Henrik Arthur, Olga Arthur, Heather Arthur, and Kristophe Arthur, we count $8$ H's, $4$ E's, $6$ A's, $11$ R's, and $6$ T's, giving $$8cdot4cdot6cdot11cdot6=12672.$$
which adds up to
$4+87+12672=12763$.
answered 10 hours ago
noednenoedne
6,60711956
edited 9 hours ago
answered 10 hours ago
noednenoedne
6,60711956
answered 10 hours ago
noednenoedne
6,60711956
answered 10 hours ago
noednenoedne
6,60711956
6,60711956
1
$begingroup$
You have to take into account every possibility: what if one of them is a timelord?
$endgroup$
– Arnaud Mortier
10 hours ago
1
$begingroup$
Do the people portrayed in face cards have hearts? (I say no, they are mere abstract sketches of people. Perhaps DEEM thinks otherwise.)
$endgroup$
– Gareth McCaughan♦
9 hours ago
1
$begingroup$
@GarethMcCaughan I was thinking the same. After all, the King cannot still have a heart with that sword stuck in his head for so long.
$endgroup$
– noedne
9 hours ago
2
$begingroup$
At some point we might begin to suspect that the Arthurs are octupuses.
$endgroup$
– noedne
9 hours ago
1
$begingroup$
Every possibility? Well, for all we know there's another part of the table piled high with dozens of other packs of cards...
$endgroup$
– Gareth McCaughan♦
9 hours ago
|
show 5 more comments
1
$begingroup$
You have to take into account every possibility: what if one of them is a timelord?
$endgroup$
– Arnaud Mortier
10 hours ago
1
$begingroup$
Do the people portrayed in face cards have hearts? (I say no, they are mere abstract sketches of people. Perhaps DEEM thinks otherwise.)
$endgroup$
– Gareth McCaughan♦
9 hours ago
1
$begingroup$
@GarethMcCaughan I was thinking the same. After all, the King cannot still have a heart with that sword stuck in his head for so long.
$endgroup$
– noedne
9 hours ago
2
$begingroup$
At some point we might begin to suspect that the Arthurs are octupuses.
$endgroup$
– noedne
9 hours ago
1
$begingroup$
Every possibility? Well, for all we know there's another part of the table piled high with dozens of other packs of cards...
$endgroup$
– Gareth McCaughan♦
9 hours ago
1
1
$begingroup$
You have to take into account every possibility: what if one of them is a timelord?
$endgroup$
– Arnaud Mortier
10 hours ago
$begingroup$
You have to take into account every possibility: what if one of them is a timelord?
$endgroup$
– Arnaud Mortier
10 hours ago
1
1
$begingroup$
Do the people portrayed in face cards have hearts? (I say no, they are mere abstract sketches of people. Perhaps DEEM thinks otherwise.)
$endgroup$
– Gareth McCaughan♦
9 hours ago
$begingroup$
Do the people portrayed in face cards have hearts? (I say no, they are mere abstract sketches of people. Perhaps DEEM thinks otherwise.)
$endgroup$
– Gareth McCaughan♦
9 hours ago
1
1
$begingroup$
@GarethMcCaughan I was thinking the same. After all, the King cannot still have a heart with that sword stuck in his head for so long.
$endgroup$
– noedne
9 hours ago
$begingroup$
@GarethMcCaughan I was thinking the same. After all, the King cannot still have a heart with that sword stuck in his head for so long.
$endgroup$
– noedne
9 hours ago
2
2
$begingroup$
At some point we might begin to suspect that the Arthurs are octupuses.
$endgroup$
– noedne
9 hours ago
$begingroup$
At some point we might begin to suspect that the Arthurs are octupuses.
$endgroup$
– noedne
9 hours ago
1
1
$begingroup$
Every possibility? Well, for all we know there's another part of the table piled high with dozens of other packs of cards...
$endgroup$
– Gareth McCaughan♦
9 hours ago
$begingroup$
Every possibility? Well, for all we know there's another part of the table piled high with dozens of other packs of cards...
$endgroup$
– Gareth McCaughan♦
9 hours ago
|
show 5 more comments
$begingroup$
Starting with noedne's analysis of
87 hearts
from the card deck alone. We also have:
four (I assume) humans, that have one heart each... except that the two women could be pregnant (take into account every possibility!), and with humans twinning is reasonable, but triplets are pretty rare, so I would say up to 8 human hearts.
Oops, almost forgot to look at:
The text itself! The Arthur family has a cleverly hidden heart, and there are enough letters in the other names for 2 more, adding these to the human and card hearts are 87+8+3 for a grand total of 98 hearts. I think this is a stretch, but the fact that there are 52 cards in a deck also looks like a heart(
Or if you want to be absolutely ridiculous:
Octuplets have been born a few times, so you could have 4+8+8=20 human hearts for a grand total of 110, but at that rate do you count multiple births higher than eight as long as they can have a beating heart in the womb? That's why I consider this a ridiculous option.
answered 5 hours ago
NH.NH.
32119
$endgroup$
add a comment |
$begingroup$
Starting with noedne's analysis of
87 hearts
from the card deck alone. We also have:
four (I assume) humans, that have one heart each... except that the two women could be pregnant (take into account every possibility!), and with humans twinning is reasonable, but triplets are pretty rare, so I would say up to 8 human hearts.
Oops, almost forgot to look at:
The text itself! The Arthur family has a cleverly hidden heart, and there are enough letters in the other names for 2 more, adding these to the human and card hearts are 87+8+3 for a grand total of 98 hearts. I think this is a stretch, but the fact that there are 52 cards in a deck also looks like a heart(
Or if you want to be absolutely ridiculous:
Octuplets have been born a few times, so you could have 4+8+8=20 human hearts for a grand total of 110, but at that rate do you count multiple births higher than eight as long as they can have a beating heart in the womb? That's why I consider this a ridiculous option.
answered 5 hours ago
NH.NH.
32119
$endgroup$
add a comment |
$begingroup$
Starting with noedne's analysis of
87 hearts
from the card deck alone. We also have:
four (I assume) humans, that have one heart each... except that the two women could be pregnant (take into account every possibility!), and with humans twinning is reasonable, but triplets are pretty rare, so I would say up to 8 human hearts.
Oops, almost forgot to look at:
The text itself! The Arthur family has a cleverly hidden heart, and there are enough letters in the other names for 2 more, adding these to the human and card hearts are 87+8+3 for a grand total of 98 hearts. I think this is a stretch, but the fact that there are 52 cards in a deck also looks like a heart(
Or if you want to be absolutely ridiculous:
Octuplets have been born a few times, so you could have 4+8+8=20 human hearts for a grand total of 110, but at that rate do you count multiple births higher than eight as long as they can have a beating heart in the womb? That's why I consider this a ridiculous option.
answered 5 hours ago
NH.NH.
32119
$endgroup$
Starting with noedne's analysis of
87 hearts
from the card deck alone. We also have:
four (I assume) humans, that have one heart each... except that the two women could be pregnant (take into account every possibility!), and with humans twinning is reasonable, but triplets are pretty rare, so I would say up to 8 human hearts.
Oops, almost forgot to look at:
The text itself! The Arthur family has a cleverly hidden heart, and there are enough letters in the other names for 2 more, adding these to the human and card hearts are 87+8+3 for a grand total of 98 hearts. I think this is a stretch, but the fact that there are 52 cards in a deck also looks like a heart(
Or if you want to be absolutely ridiculous:
Octuplets have been born a few times, so you could have 4+8+8=20 human hearts for a grand total of 110, but at that rate do you count multiple births higher than eight as long as they can have a beating heart in the womb? That's why I consider this a ridiculous option.
answered 5 hours ago
NH.NH.
32119
edited 5 hours ago
answered 5 hours ago
NH.NH.
32119
answered 5 hours ago
NH.NH.
32119
answered 5 hours ago
NH.NH.
32119
32119
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.
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%2f80562%2fcounting-all-the-hearts%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
