How should I approach using two 8s and two 3s to make the number 24?Coppers - Make 24$tau$ is greatest! $tau$...
Why did the villain in the first Men in Black movie care about Earth's Cockroaches?
Hybrid flat to drop conversion advice
Using only 1s, make 29 with the minimum number of digits
I will be going to Sweden on business purpose .Can I visit London from Sweden and how much UK visa will cost?
What is a good reason for every spaceship to carry a weapon on board?
Looking for access to original paper for Category O
In Linux what happens if 1000 files in a directory are moved to another location while another 300 files were added to the source directory?
How to deal with an incendiary email that was recalled
Does theoretical physics suggest that gravity is the exchange of gravitons or deformation/bending of spacetime?
What is causing the chain skip after chain replacement?
Early credit roll before the end of the film
Why are these T-SQL jobs from different SQL Server instances executed on the same instance (Always On Availability Groups)
How can I get my players to come to the game session after agreeing to a date?
Cookies - Should the toggles be on?
How do you funnel food off a cutting board?
kill -0 <PID> は何をするのでしょうか?
Quickly creating a sparse array
Can a Pact of the Blade warlock use the correct existing pact magic weapon so it functions as a "Returning" weapon?
How should I handle players who ignore the session zero agreement?
Can I write a book of my D&D game?
Why avoid shared user accounts?
Do authors have to be politically correct in article-writing?
Is it possible to "destroy" a wallet?
Citing paywalled articles accessed via illegal web sharing
How should I approach using two 8s and two 3s to make the number 24?
Coppers - Make 24$tau$ is greatest! $tau$ is all (from 1 to 20)Express the number $2015$ using only the digit $2$ twiceLargest number with five 1's and five numeric operationsHow many consecutive integers can you make using only four digits?Use four 8's to make the number 2016Fix the perfectly ordered equationPalindromic number puzzle - make 505 from 20202Use 0, 5, 7 and 1 to make 89Use 6, 5 and 3 to make 57Create all numbers from 1-100 using 1,3,3,6
$begingroup$
Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, times , div$, parentheses) to make the number $24$.
(You may not join numbers together to form new numbers, like $ 8, 3rightarrow 83$)
I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?
mathematics formation-of-numbers
New contributor
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, times , div$, parentheses) to make the number $24$.
(You may not join numbers together to form new numbers, like $ 8, 3rightarrow 83$)
I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?
mathematics formation-of-numbers
New contributor
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
$endgroup$
– JonMark Perry
17 hours ago
1
$begingroup$
Possible duplicate of Coppers - Make 24
$endgroup$
– North
13 hours ago
8
$begingroup$
If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
$endgroup$
– Lunin
13 hours ago
$begingroup$
One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with ifno, and opens up more options ifyes.
$endgroup$
– Justin Time
13 hours ago
add a comment |
$begingroup$
Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, times , div$, parentheses) to make the number $24$.
(You may not join numbers together to form new numbers, like $ 8, 3rightarrow 83$)
I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?
mathematics formation-of-numbers
New contributor
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, times , div$, parentheses) to make the number $24$.
(You may not join numbers together to form new numbers, like $ 8, 3rightarrow 83$)
I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?
mathematics formation-of-numbers
mathematics formation-of-numbers
New contributor
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 mins ago
Omega Krypton
4,5551440
4,5551440
New contributor
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 19 hours ago
user9888273user9888273
563
563
New contributor
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
$endgroup$
– JonMark Perry
17 hours ago
1
$begingroup$
Possible duplicate of Coppers - Make 24
$endgroup$
– North
13 hours ago
8
$begingroup$
If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
$endgroup$
– Lunin
13 hours ago
$begingroup$
One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with ifno, and opens up more options ifyes.
$endgroup$
– Justin Time
13 hours ago
add a comment |
$begingroup$
see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
$endgroup$
– JonMark Perry
17 hours ago
1
$begingroup$
Possible duplicate of Coppers - Make 24
$endgroup$
– North
13 hours ago
8
$begingroup$
If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
$endgroup$
– Lunin
13 hours ago
$begingroup$
One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with ifno, and opens up more options ifyes.
$endgroup$
– Justin Time
13 hours ago
$begingroup$
see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
$endgroup$
– JonMark Perry
17 hours ago
$begingroup$
see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
$endgroup$
– JonMark Perry
17 hours ago
1
1
$begingroup$
Possible duplicate of Coppers - Make 24
$endgroup$
– North
13 hours ago
$begingroup$
Possible duplicate of Coppers - Make 24
$endgroup$
– North
13 hours ago
8
8
$begingroup$
If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
$endgroup$
– Lunin
13 hours ago
$begingroup$
If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
$endgroup$
– Lunin
13 hours ago
$begingroup$
One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if
no, and opens up more options if yes.$endgroup$
– Justin Time
13 hours ago
$begingroup$
One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if
no, and opens up more options if yes.$endgroup$
– Justin Time
13 hours ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)
Here's one method of thinking to get to the answer:
1) Is this a trick question?
It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.
2) What do we need to do?
What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.
3) Can we simplify the problem down at all?
Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.
4) What components get us closer to the solution?
So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?
$1 = 8 div 8$ with $3,3$ leftover
$16 = 8 + 8$ with $3,3$ leftover- $64 = 8 times 8$
- $0 = 8 - 8$
$24 = 8 times 3$ with $8,3$ leftover- $11 = 8 + 3$
- $5 = 8 - 3$
- $2 frac{2}{3} = 8 div 3$
- ...
5) Work from the other end - what do the components of the solution look like?
Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:
$sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$
I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?
- $8 * 3 = 24$
- $8 / frac{1}{3} = 24$
- $27 - 3 = 24$
- $21 + 3 = 24$
- $32 - 8 = 24$
- ...
6) Connect the dots.
We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?
From the above, here's the link I've come up with:
$ 3 - 2 frac{2}{3} = frac{1}{3}$
That will lead us to a solution by putting it all together:
$8 div (3 - frac{8}{3})) = 24$
Fin
That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.
$endgroup$
13
$begingroup$
Hey look, someone actually answered the question asked! :)
$endgroup$
– Rubio♦
10 hours ago
add a comment |
$begingroup$
Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).
$8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
$= 8 div frac{1}{3}$
$= 24$
If we allow square roots, a simpler solution is possible.
$sqrt{8 times 8 times 3 times 3}$
$= 8 times 3$
$= 24$
In fact, there are many more solutions if you allow more operations.
$endgroup$
add a comment |
$begingroup$
Here's a fairly simple one:
$3!times(frac88+3)$
$endgroup$
4
$begingroup$
Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
$endgroup$
– eedrah
10 hours ago
add a comment |
$begingroup$
Another simple solution :
$(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$
New contributor
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Factorials are not allowed, unfortunately....
$endgroup$
– Glorfindel
7 hours ago
add a comment |
$begingroup$
Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:
$(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$
$endgroup$
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
});
}
});
user9888273 is a new contributor. Be nice, and check out our Code of Conduct.
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%2f80039%2fhow-should-i-approach-using-two-8s-and-two-3s-to-make-the-number-24%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)
Here's one method of thinking to get to the answer:
1) Is this a trick question?
It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.
2) What do we need to do?
What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.
3) Can we simplify the problem down at all?
Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.
4) What components get us closer to the solution?
So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?
$1 = 8 div 8$ with $3,3$ leftover
$16 = 8 + 8$ with $3,3$ leftover- $64 = 8 times 8$
- $0 = 8 - 8$
$24 = 8 times 3$ with $8,3$ leftover- $11 = 8 + 3$
- $5 = 8 - 3$
- $2 frac{2}{3} = 8 div 3$
- ...
5) Work from the other end - what do the components of the solution look like?
Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:
$sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$
I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?
- $8 * 3 = 24$
- $8 / frac{1}{3} = 24$
- $27 - 3 = 24$
- $21 + 3 = 24$
- $32 - 8 = 24$
- ...
6) Connect the dots.
We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?
From the above, here's the link I've come up with:
$ 3 - 2 frac{2}{3} = frac{1}{3}$
That will lead us to a solution by putting it all together:
$8 div (3 - frac{8}{3})) = 24$
Fin
That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.
$endgroup$
13
$begingroup$
Hey look, someone actually answered the question asked! :)
$endgroup$
– Rubio♦
10 hours ago
add a comment |
$begingroup$
While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)
Here's one method of thinking to get to the answer:
1) Is this a trick question?
It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.
2) What do we need to do?
What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.
3) Can we simplify the problem down at all?
Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.
4) What components get us closer to the solution?
So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?
$1 = 8 div 8$ with $3,3$ leftover
$16 = 8 + 8$ with $3,3$ leftover- $64 = 8 times 8$
- $0 = 8 - 8$
$24 = 8 times 3$ with $8,3$ leftover- $11 = 8 + 3$
- $5 = 8 - 3$
- $2 frac{2}{3} = 8 div 3$
- ...
5) Work from the other end - what do the components of the solution look like?
Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:
$sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$
I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?
- $8 * 3 = 24$
- $8 / frac{1}{3} = 24$
- $27 - 3 = 24$
- $21 + 3 = 24$
- $32 - 8 = 24$
- ...
6) Connect the dots.
We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?
From the above, here's the link I've come up with:
$ 3 - 2 frac{2}{3} = frac{1}{3}$
That will lead us to a solution by putting it all together:
$8 div (3 - frac{8}{3})) = 24$
Fin
That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.
$endgroup$
13
$begingroup$
Hey look, someone actually answered the question asked! :)
$endgroup$
– Rubio♦
10 hours ago
add a comment |
$begingroup$
While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)
Here's one method of thinking to get to the answer:
1) Is this a trick question?
It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.
2) What do we need to do?
What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.
3) Can we simplify the problem down at all?
Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.
4) What components get us closer to the solution?
So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?
$1 = 8 div 8$ with $3,3$ leftover
$16 = 8 + 8$ with $3,3$ leftover- $64 = 8 times 8$
- $0 = 8 - 8$
$24 = 8 times 3$ with $8,3$ leftover- $11 = 8 + 3$
- $5 = 8 - 3$
- $2 frac{2}{3} = 8 div 3$
- ...
5) Work from the other end - what do the components of the solution look like?
Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:
$sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$
I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?
- $8 * 3 = 24$
- $8 / frac{1}{3} = 24$
- $27 - 3 = 24$
- $21 + 3 = 24$
- $32 - 8 = 24$
- ...
6) Connect the dots.
We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?
From the above, here's the link I've come up with:
$ 3 - 2 frac{2}{3} = frac{1}{3}$
That will lead us to a solution by putting it all together:
$8 div (3 - frac{8}{3})) = 24$
Fin
That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.
$endgroup$
While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)
Here's one method of thinking to get to the answer:
1) Is this a trick question?
It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.
2) What do we need to do?
What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.
3) Can we simplify the problem down at all?
Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.
4) What components get us closer to the solution?
So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?
$1 = 8 div 8$ with $3,3$ leftover
$16 = 8 + 8$ with $3,3$ leftover- $64 = 8 times 8$
- $0 = 8 - 8$
$24 = 8 times 3$ with $8,3$ leftover- $11 = 8 + 3$
- $5 = 8 - 3$
- $2 frac{2}{3} = 8 div 3$
- ...
5) Work from the other end - what do the components of the solution look like?
Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:
$sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$
I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?
- $8 * 3 = 24$
- $8 / frac{1}{3} = 24$
- $27 - 3 = 24$
- $21 + 3 = 24$
- $32 - 8 = 24$
- ...
6) Connect the dots.
We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?
From the above, here's the link I've come up with:
$ 3 - 2 frac{2}{3} = frac{1}{3}$
That will lead us to a solution by putting it all together:
$8 div (3 - frac{8}{3})) = 24$
Fin
That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.
edited 8 hours ago
boboquack
15.5k149118
15.5k149118
answered 10 hours ago
eedraheedrah
875118
875118
13
$begingroup$
Hey look, someone actually answered the question asked! :)
$endgroup$
– Rubio♦
10 hours ago
add a comment |
13
$begingroup$
Hey look, someone actually answered the question asked! :)
$endgroup$
– Rubio♦
10 hours ago
13
13
$begingroup$
Hey look, someone actually answered the question asked! :)
$endgroup$
– Rubio♦
10 hours ago
$begingroup$
Hey look, someone actually answered the question asked! :)
$endgroup$
– Rubio♦
10 hours ago
add a comment |
$begingroup$
Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).
$8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
$= 8 div frac{1}{3}$
$= 24$
If we allow square roots, a simpler solution is possible.
$sqrt{8 times 8 times 3 times 3}$
$= 8 times 3$
$= 24$
In fact, there are many more solutions if you allow more operations.
$endgroup$
add a comment |
$begingroup$
Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).
$8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
$= 8 div frac{1}{3}$
$= 24$
If we allow square roots, a simpler solution is possible.
$sqrt{8 times 8 times 3 times 3}$
$= 8 times 3$
$= 24$
In fact, there are many more solutions if you allow more operations.
$endgroup$
add a comment |
$begingroup$
Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).
$8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
$= 8 div frac{1}{3}$
$= 24$
If we allow square roots, a simpler solution is possible.
$sqrt{8 times 8 times 3 times 3}$
$= 8 times 3$
$= 24$
In fact, there are many more solutions if you allow more operations.
$endgroup$
Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).
$8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
$= 8 div frac{1}{3}$
$= 24$
If we allow square roots, a simpler solution is possible.
$sqrt{8 times 8 times 3 times 3}$
$= 8 times 3$
$= 24$
In fact, there are many more solutions if you allow more operations.
edited 19 hours ago
answered 19 hours ago
HughHugh
2,2651926
2,2651926
add a comment |
add a comment |
$begingroup$
Here's a fairly simple one:
$3!times(frac88+3)$
$endgroup$
4
$begingroup$
Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
$endgroup$
– eedrah
10 hours ago
add a comment |
$begingroup$
Here's a fairly simple one:
$3!times(frac88+3)$
$endgroup$
4
$begingroup$
Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
$endgroup$
– eedrah
10 hours ago
add a comment |
$begingroup$
Here's a fairly simple one:
$3!times(frac88+3)$
$endgroup$
Here's a fairly simple one:
$3!times(frac88+3)$
answered 18 hours ago
JonMark PerryJonMark Perry
19.5k63991
19.5k63991
4
$begingroup$
Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
$endgroup$
– eedrah
10 hours ago
add a comment |
4
$begingroup$
Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
$endgroup$
– eedrah
10 hours ago
4
4
$begingroup$
Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
$endgroup$
– eedrah
10 hours ago
$begingroup$
Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
$endgroup$
– eedrah
10 hours ago
add a comment |
$begingroup$
Another simple solution :
$(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$
New contributor
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Factorials are not allowed, unfortunately....
$endgroup$
– Glorfindel
7 hours ago
add a comment |
$begingroup$
Another simple solution :
$(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$
New contributor
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Factorials are not allowed, unfortunately....
$endgroup$
– Glorfindel
7 hours ago
add a comment |
$begingroup$
Another simple solution :
$(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$
New contributor
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Another simple solution :
$(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$
New contributor
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
Omega Krypton
4,5551440
4,5551440
New contributor
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
Srinivas TamtamSrinivas Tamtam
9
9
New contributor
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Factorials are not allowed, unfortunately....
$endgroup$
– Glorfindel
7 hours ago
add a comment |
1
$begingroup$
Factorials are not allowed, unfortunately....
$endgroup$
– Glorfindel
7 hours ago
1
1
$begingroup$
Factorials are not allowed, unfortunately....
$endgroup$
– Glorfindel
7 hours ago
$begingroup$
Factorials are not allowed, unfortunately....
$endgroup$
– Glorfindel
7 hours ago
add a comment |
$begingroup$
Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:
$(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$
$endgroup$
add a comment |
$begingroup$
Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:
$(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$
$endgroup$
add a comment |
$begingroup$
Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:
$(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$
$endgroup$
Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:
$(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$
answered 10 hours ago
eedraheedrah
875118
875118
add a comment |
add a comment |
user9888273 is a new contributor. Be nice, and check out our Code of Conduct.
user9888273 is a new contributor. Be nice, and check out our Code of Conduct.
user9888273 is a new contributor. Be nice, and check out our Code of Conduct.
user9888273 is a new contributor. Be nice, and check out our Code of Conduct.
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%2f80039%2fhow-should-i-approach-using-two-8s-and-two-3s-to-make-the-number-24%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
$begingroup$
see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
$endgroup$
– JonMark Perry
17 hours ago
1
$begingroup$
Possible duplicate of Coppers - Make 24
$endgroup$
– North
13 hours ago
8
$begingroup$
If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
$endgroup$
– Lunin
13 hours ago
$begingroup$
One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if
no, and opens up more options ifyes.$endgroup$
– Justin Time
13 hours ago