1984 - take the digits 1,9, 8 and 4 and make 123 - Part III

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Part III ... 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 ... I think this is the final one from me on 1984

Warm up

Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. 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

The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, 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. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.

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 provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + sqrt4$ to get two $10$s, for example.

many thanks to the authors of the similar questions below for inspiring this question.

• 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

This is part III after the first and second in this series were solved

mathematics formation-of-numbers

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edited 56 mins ago

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tom

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Part III ... 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 ... I think this is the final one from me on 1984

Warm up

Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. 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

The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, 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. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.

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 provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + sqrt4$ to get two $10$s, for example.

many thanks to the authors of the similar questions below for inspiring this question.

• 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

This is part III after the first and second in this series were solved

mathematics formation-of-numbers

share|improve this question

edited 56 mins ago

asked 1 hour ago

tom

1,725324

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Part III ... 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 ... I think this is the final one from me on 1984

Warm up

Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. 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

The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, 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. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.

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 provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + sqrt4$ to get two $10$s, for example.

many thanks to the authors of the similar questions below for inspiring this question.

• 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

This is part III after the first and second in this series were solved

mathematics formation-of-numbers

share|improve this question

edited 56 mins ago

asked 1 hour ago

tom

1,725324

Part III ... 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 ... I think this is the final one from me on 1984

Warm up

Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. 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

The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, 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. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.

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 provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + sqrt4$ to get two $10$s, for example.

many thanks to the authors of the similar questions below for inspiring this question.

• 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

This is part III after the first and second in this series were solved

mathematics formation-of-numbers

mathematics formation-of-numbers

share|improve this question

edited 56 mins ago

asked 1 hour ago

tom

1,725324

share|improve this question

edited 56 mins ago

asked 1 hour ago

tom

1,725324

share|improve this question

share|improve this question

edited 56 mins ago

edited 56 mins ago

edited 56 mins ago

asked 1 hour ago

tom

1,725324

asked 1 hour ago

tom

1,725324

asked 1 hour ago

tom

1,725324

1,725324

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote

Is it

$(8-sqrt9)!+(4-1)$

And

$(8-4+1)! + sqrt9$

share|improve this answer

edited 21 mins ago

answered 32 mins ago

PotatoLatte

1,175228

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
  – tom
  25 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Here is two possibilities that I have found:

$sqrt9times(8!/4!+1)=123$

and the other one is (thanks to @PotatoLatte)

$(8-sqrt9)!+(4-1)=123$

Bonus:

$sqrt8+1+9!/4!=123$

share

answered 3 mins ago

Oray

14.4k435142

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Is it

$(8-sqrt9)!+(4-1)$

And

$(8-4+1)! + sqrt9$

share|improve this answer

edited 21 mins ago

answered 32 mins ago

PotatoLatte

1,175228

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
  – tom
  25 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

Is it

$(8-sqrt9)!+(4-1)$

And

$(8-4+1)! + sqrt9$

share|improve this answer

edited 21 mins ago

answered 32 mins ago

PotatoLatte

1,175228

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
  – tom
  25 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Is it

$(8-sqrt9)!+(4-1)$

And

$(8-4+1)! + sqrt9$

share|improve this answer

edited 21 mins ago

answered 32 mins ago

PotatoLatte

1,175228

Is it

$(8-sqrt9)!+(4-1)$

And

$(8-4+1)! + sqrt9$

share|improve this answer

edited 21 mins ago

answered 32 mins ago

PotatoLatte

1,175228

share|improve this answer

share|improve this answer

edited 21 mins ago

edited 21 mins ago

edited 21 mins ago

answered 32 mins ago

PotatoLatte

1,175228

answered 32 mins ago

PotatoLatte

1,175228

answered 32 mins ago

PotatoLatte

1,175228

1,175228

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
  – tom
  25 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
  – tom
  25 mins ago
  

  
  
  
  

I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
– tom
25 mins ago

I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
– tom
25 mins ago

add a comment | 

up vote
0
down vote

Here is two possibilities that I have found:

$sqrt9times(8!/4!+1)=123$

and the other one is (thanks to @PotatoLatte)

$(8-sqrt9)!+(4-1)=123$

Bonus:

$sqrt8+1+9!/4!=123$

share

answered 3 mins ago

Oray

14.4k435142

add a comment | 

up vote
0
down vote

Here is two possibilities that I have found:

$sqrt9times(8!/4!+1)=123$

and the other one is (thanks to @PotatoLatte)

$(8-sqrt9)!+(4-1)=123$

Bonus:

$sqrt8+1+9!/4!=123$

share

answered 3 mins ago

Oray

14.4k435142

add a comment | 

up vote
0
down vote

up vote
0
down vote

Here is two possibilities that I have found:

$sqrt9times(8!/4!+1)=123$

and the other one is (thanks to @PotatoLatte)

$(8-sqrt9)!+(4-1)=123$

Bonus:

$sqrt8+1+9!/4!=123$

share

answered 3 mins ago

Oray

14.4k435142

Here is two possibilities that I have found:

$sqrt9times(8!/4!+1)=123$

and the other one is (thanks to @PotatoLatte)

$(8-sqrt9)!+(4-1)=123$

Bonus:

$sqrt8+1+9!/4!=123$

share

answered 3 mins ago

Oray

14.4k435142

share

share

answered 3 mins ago

Oray

14.4k435142

answered 3 mins ago

Oray

14.4k435142

answered 3 mins ago

Oray

14.4k435142

14.4k435142

add a comment | 

add a comment | 

 

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