Mixing
1984 - take the digits 1,9, 8 and 4 and make 369
Clash Royale CLAN TAG#URR8PPP
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Part II... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ...
Warm up
Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(sqrt4)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.
Main Event
If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only.
Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.
Note that in all the puzzles above
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles.
many thanks to the authors of the similar questions below for inspiring this question.
This is part II after the first in this series was solved
- Use 2 0 1 and 8 to make 67
- Make numbers 93 using the digits 2, 0, 1, 8
- Make numbers 1 - 30 using the digits 2, 0, 1, 8
mathematics formation-of-numbers
asked 5 hours ago
tom
1,718324
add a comment |Â
up vote
1
down vote
favorite
Part II... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ...
Warm up
Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(sqrt4)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.
Main Event
If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only.
Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.
Note that in all the puzzles above
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles.
many thanks to the authors of the similar questions below for inspiring this question.
This is part II after the first in this series was solved
- Use 2 0 1 and 8 to make 67
- Make numbers 93 using the digits 2, 0, 1, 8
- Make numbers 1 - 30 using the digits 2, 0, 1, 8
mathematics formation-of-numbers
asked 5 hours ago
tom
1,718324
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Part II... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ...
Warm up
Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(sqrt4)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.
Main Event
If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only.
Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.
Note that in all the puzzles above
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles.
many thanks to the authors of the similar questions below for inspiring this question.
This is part II after the first in this series was solved
- Use 2 0 1 and 8 to make 67
- Make numbers 93 using the digits 2, 0, 1, 8
- Make numbers 1 - 30 using the digits 2, 0, 1, 8
mathematics formation-of-numbers
asked 5 hours ago
tom
1,718324
Part II... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ...
Warm up
Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(sqrt4)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.
Main Event
If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only.
Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.
Note that in all the puzzles above
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles.
many thanks to the authors of the similar questions below for inspiring this question.
This is part II after the first in this series was solved
- Use 2 0 1 and 8 to make 67
- Make numbers 93 using the digits 2, 0, 1, 8
- Make numbers 1 - 30 using the digits 2, 0, 1, 8
mathematics formation-of-numbers
mathematics formation-of-numbers
asked 5 hours ago
tom
1,718324
asked 5 hours ago
tom
1,718324
asked 5 hours ago
tom
1,718324
asked 5 hours ago
tom
1,718324
asked 5 hours ago
tom
1,718324
1,718324
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$
Or, with an infinite number of functions:
$frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above
answered 4 hours ago
elias
7,71232051
1
brilliant good job, plus one - this looks like the first answer to the main event :-)
– tom
4 hours ago
If only I didn't use the spare $0$... Eh well, you deserved it :P
– user477343
4 hours ago
add a comment |Â
up vote
3
down vote
Well, for the warm-up
$$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$
Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...
$$big((8-1-0!)!div sqrt4big)+9=boxed369$$
:D
answered 4 hours ago
user477343
3,4471745
Good job - looks like this is the first correct answer to the warm up - great job
– tom
4 hours ago
@tom Now to not use any concatenation... :P
– user477343
4 hours ago
add a comment |Â
up vote
2
down vote
Perhaps, for the warm-up
$41times 9times sqrtsqrtsqrt...8$, with infinite roots
$=369times sqrtsqrtsqrt...8$.
edited 4 hours ago
user477343
3,4471745
answered 4 hours ago
binu kurian
312
New contributor
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
– tom
4 hours ago
1
I added roll over spoiler hider - so that people have to roll over to see your answer
– tom
4 hours ago
Thanks @tom. As a newbie I am still learning how to format my answers
– binu kurian
3 hours ago
no problem, but I think @user477343 sorted it out.
– tom
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$
Or, with an infinite number of functions:
$frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above
answered 4 hours ago
elias
7,71232051
1
brilliant good job, plus one - this looks like the first answer to the main event :-)
– tom
4 hours ago
If only I didn't use the spare $0$... Eh well, you deserved it :P
– user477343
4 hours ago
add a comment |Â
up vote
1
down vote
accepted
$frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$
Or, with an infinite number of functions:
$frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above
answered 4 hours ago
elias
7,71232051
1
brilliant good job, plus one - this looks like the first answer to the main event :-)
– tom
4 hours ago
If only I didn't use the spare $0$... Eh well, you deserved it :P
– user477343
4 hours ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$
Or, with an infinite number of functions:
$frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above
answered 4 hours ago
elias
7,71232051
$frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$
Or, with an infinite number of functions:
$frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above
answered 4 hours ago
elias
7,71232051
answered 4 hours ago
elias
7,71232051
answered 4 hours ago
elias
7,71232051
answered 4 hours ago
elias
7,71232051
7,71232051
1
brilliant good job, plus one - this looks like the first answer to the main event :-)
– tom
4 hours ago
If only I didn't use the spare $0$... Eh well, you deserved it :P
– user477343
4 hours ago
add a comment |Â
1
brilliant good job, plus one - this looks like the first answer to the main event :-)
– tom
4 hours ago
If only I didn't use the spare $0$... Eh well, you deserved it :P
– user477343
4 hours ago
1
1
brilliant good job, plus one - this looks like the first answer to the main event :-)
– tom
4 hours ago
brilliant good job, plus one - this looks like the first answer to the main event :-)
– tom
4 hours ago
If only I didn't use the spare $0$... Eh well, you deserved it :P
– user477343
4 hours ago
If only I didn't use the spare $0$... Eh well, you deserved it :P
– user477343
4 hours ago
add a comment |Â
up vote
3
down vote
Well, for the warm-up
$$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$
Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...
$$big((8-1-0!)!div sqrt4big)+9=boxed369$$
:D
answered 4 hours ago
user477343
3,4471745
Good job - looks like this is the first correct answer to the warm up - great job
– tom
4 hours ago
@tom Now to not use any concatenation... :P
– user477343
4 hours ago
add a comment |Â
up vote
3
down vote
Well, for the warm-up
$$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$
Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...
$$big((8-1-0!)!div sqrt4big)+9=boxed369$$
:D
answered 4 hours ago
user477343
3,4471745
Good job - looks like this is the first correct answer to the warm up - great job
– tom
4 hours ago
@tom Now to not use any concatenation... :P
– user477343
4 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Well, for the warm-up
$$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$
Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...
$$big((8-1-0!)!div sqrt4big)+9=boxed369$$
:D
answered 4 hours ago
user477343
3,4471745
Well, for the warm-up
$$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$
Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...
$$big((8-1-0!)!div sqrt4big)+9=boxed369$$
:D
answered 4 hours ago
user477343
3,4471745
edited 4 hours ago
answered 4 hours ago
user477343
3,4471745
answered 4 hours ago
user477343
3,4471745
answered 4 hours ago
user477343
3,4471745
3,4471745
Good job - looks like this is the first correct answer to the warm up - great job
– tom
4 hours ago
@tom Now to not use any concatenation... :P
– user477343
4 hours ago
add a comment |Â
Good job - looks like this is the first correct answer to the warm up - great job
– tom
4 hours ago
@tom Now to not use any concatenation... :P
– user477343
4 hours ago
Good job - looks like this is the first correct answer to the warm up - great job
– tom
4 hours ago
Good job - looks like this is the first correct answer to the warm up - great job
– tom
4 hours ago
@tom Now to not use any concatenation... :P
– user477343
4 hours ago
@tom Now to not use any concatenation... :P
– user477343
4 hours ago
add a comment |Â
up vote
2
down vote
Perhaps, for the warm-up
$41times 9times sqrtsqrtsqrt...8$, with infinite roots
$=369times sqrtsqrtsqrt...8$.
edited 4 hours ago
user477343
3,4471745
answered 4 hours ago
binu kurian
312
New contributor
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
– tom
4 hours ago
1
I added roll over spoiler hider - so that people have to roll over to see your answer
– tom
4 hours ago
Thanks @tom. As a newbie I am still learning how to format my answers
– binu kurian
3 hours ago
no problem, but I think @user477343 sorted it out.
– tom
3 hours ago
add a comment |Â
up vote
2
down vote
Perhaps, for the warm-up
$41times 9times sqrtsqrtsqrt...8$, with infinite roots
$=369times sqrtsqrtsqrt...8$.
edited 4 hours ago
user477343
3,4471745
answered 4 hours ago
binu kurian
312
New contributor
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
– tom
4 hours ago
1
I added roll over spoiler hider - so that people have to roll over to see your answer
– tom
4 hours ago
Thanks @tom. As a newbie I am still learning how to format my answers
– binu kurian
3 hours ago
no problem, but I think @user477343 sorted it out.
– tom
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Perhaps, for the warm-up
$41times 9times sqrtsqrtsqrt...8$, with infinite roots
$=369times sqrtsqrtsqrt...8$.
edited 4 hours ago
user477343
3,4471745
answered 4 hours ago
binu kurian
312
New contributor
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Perhaps, for the warm-up
$41times 9times sqrtsqrtsqrt...8$, with infinite roots
$=369times sqrtsqrtsqrt...8$.
edited 4 hours ago
user477343
3,4471745
answered 4 hours ago
binu kurian
312
New contributor
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
user477343
3,4471745
edited 4 hours ago
user477343
3,4471745
edited 4 hours ago
user477343
3,4471745
3,4471745
answered 4 hours ago
binu kurian
312
New contributor
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
binu kurian
312
answered 4 hours ago
binu kurian
312
312
New contributor
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
– tom
4 hours ago
1
I added roll over spoiler hider - so that people have to roll over to see your answer
– tom
4 hours ago
Thanks @tom. As a newbie I am still learning how to format my answers
– binu kurian
3 hours ago
no problem, but I think @user477343 sorted it out.
– tom
3 hours ago
add a comment |Â
Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
– tom
4 hours ago
1
I added roll over spoiler hider - so that people have to roll over to see your answer
– tom
4 hours ago
Thanks @tom. As a newbie I am still learning how to format my answers
– binu kurian
3 hours ago
no problem, but I think @user477343 sorted it out.
– tom
3 hours ago
Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
– tom
4 hours ago
Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
– tom
4 hours ago
1
1
I added roll over spoiler hider - so that people have to roll over to see your answer
– tom
4 hours ago
I added roll over spoiler hider - so that people have to roll over to see your answer
– tom
4 hours ago
Thanks @tom. As a newbie I am still learning how to format my answers
– binu kurian
3 hours ago
Thanks @tom. As a newbie I am still learning how to format my answers
– binu kurian
3 hours ago
no problem, but I think @user477343 sorted it out.
– tom
3 hours ago
no problem, but I think @user477343 sorted it out.
– tom
3 hours ago
add a comment |Â
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