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Make a formula to get the highest score

Find this Unique UVC Palindrome ( ignoring signs and decimal) from Given Fractional RelationshipUse 2, 0, 1 and 8 to make 109Use 2, 0, 1 and 8 to make 1991984 - take the digits 1,9, 8 and 4 and make 2461984 - take the digits 1,9, 8 and 4 and make 369Find 2018 with the least amount of numbersPalindromic number puzzle - make 505 from 20202Use 0, 5, 7 and 1 to make 89Use 6, 5 and 3 to make 57Find 108 by using 3,4,6Make 24 using exactly three 3s

3

$begingroup$

In this puzzle you need to obtain the highest score with the following rules

1) choose two digits from 1, 2, 3, 4, 5, 6, 7, 8 and 9. These digits are represented by two letters, let us say $T$ and $M$. You are also given the digit 0.

2) make up an equation using all three of your digits, $T$, 0 and $M$, once. The numerical result of the equation should be a three digit number which can be $TM0$, $T0M$, $MT0$, $M0T$, $0MT$ and $0TM$. You may not concatenate digits, but you may use as many as the following symbols as you would like $+$, $-$, $div$, $times$, !, $sqrt$, (, ). Note that multiple ! symbols are considered each to be factorial functions. See the note below.

3) Your score is your three digit number divided by the sum of your digits. So for example if you start with 0, 2, 4 then your equation could be

$$(0*2)+4!=024$$

which would score $024 div (2+4+0) = 24div 6 = 4$

---

This question was inspired by this question

Note that: $4!! = (4!)! = 24!$ this is allowed. Double, triple etc. factorials are not allowed; $4!! ne 4 times 2$

mathematics formation-of-numbers

share|improve this question

edited 8 hours ago

Rand al'Thor

72.6k14238481

asked 9 hours ago

tomtom

2,0761730

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to help you in a modest way
  $endgroup$
  – Uvc
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the use of the decimal point allowed?
  $endgroup$
  – hexomino
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hexomino no, but i will of course upvote a creative answer using one....
  $endgroup$
  – tom
  29 mins ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

In this puzzle you need to obtain the highest score with the following rules

1) choose two digits from 1, 2, 3, 4, 5, 6, 7, 8 and 9. These digits are represented by two letters, let us say $T$ and $M$. You are also given the digit 0.

2) make up an equation using all three of your digits, $T$, 0 and $M$, once. The numerical result of the equation should be a three digit number which can be $TM0$, $T0M$, $MT0$, $M0T$, $0MT$ and $0TM$. You may not concatenate digits, but you may use as many as the following symbols as you would like $+$, $-$, $div$, $times$, !, $sqrt$, (, ). Note that multiple ! symbols are considered each to be factorial functions. See the note below.

3) Your score is your three digit number divided by the sum of your digits. So for example if you start with 0, 2, 4 then your equation could be

$$(0*2)+4!=024$$

which would score $024 div (2+4+0) = 24div 6 = 4$

---

This question was inspired by this question

Note that: $4!! = (4!)! = 24!$ this is allowed. Double, triple etc. factorials are not allowed; $4!! ne 4 times 2$

mathematics formation-of-numbers

share|improve this question

edited 8 hours ago

Rand al'Thor

72.6k14238481

asked 9 hours ago

tomtom

2,0761730

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to help you in a modest way
  $endgroup$
  – Uvc
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the use of the decimal point allowed?
  $endgroup$
  – hexomino
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hexomino no, but i will of course upvote a creative answer using one....
  $endgroup$
  – tom
  29 mins ago
  
  

  
  
  
  

add a comment | 

$begingroup$

In this puzzle you need to obtain the highest score with the following rules

1) choose two digits from 1, 2, 3, 4, 5, 6, 7, 8 and 9. These digits are represented by two letters, let us say $T$ and $M$. You are also given the digit 0.

2) make up an equation using all three of your digits, $T$, 0 and $M$, once. The numerical result of the equation should be a three digit number which can be $TM0$, $T0M$, $MT0$, $M0T$, $0MT$ and $0TM$. You may not concatenate digits, but you may use as many as the following symbols as you would like $+$, $-$, $div$, $times$, !, $sqrt$, (, ). Note that multiple ! symbols are considered each to be factorial functions. See the note below.

3) Your score is your three digit number divided by the sum of your digits. So for example if you start with 0, 2, 4 then your equation could be

$$(0*2)+4!=024$$

which would score $024 div (2+4+0) = 24div 6 = 4$

---

This question was inspired by this question

Note that: $4!! = (4!)! = 24!$ this is allowed. Double, triple etc. factorials are not allowed; $4!! ne 4 times 2$

mathematics formation-of-numbers

share|improve this question

edited 8 hours ago

Rand al'Thor

72.6k14238481

asked 9 hours ago

tomtom

2,0761730

$endgroup$

share|improve this question

edited 8 hours ago

Rand al'Thor

72.6k14238481

asked 9 hours ago

tomtom

2,0761730

share|improve this question

edited 8 hours ago

Rand al'Thor

72.6k14238481

asked 9 hours ago

tomtom

2,0761730

edited 8 hours ago

Rand al'Thor

72.6k14238481

edited 8 hours ago

Rand al'Thor

72.6k14238481

asked 9 hours ago

tomtom

2,0761730

asked 9 hours ago

tomtom

2,0761730

1 Answer
1

active

oldest

votes

5

$begingroup$

I'm going to try:

$(7-2+0!)!=6!=720$.
Score 720/9=80.


This I believe is the best available, as the only combo's that could beat it are:

$410,;510,;610,;710,;810,;910,;820$ and $920$


and I haven't managed to make any of these.

share|improve this answer

edited 6 hours ago

answered 9 hours ago

JonMark PerryJonMark Perry

21.9k643103

$endgroup$

add a comment | 

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5

$begingroup$

I'm going to try:

$(7-2+0!)!=6!=720$.
Score 720/9=80.


This I believe is the best available, as the only combo's that could beat it are:

$410,;510,;610,;710,;810,;910,;820$ and $920$


and I haven't managed to make any of these.

share|improve this answer

edited 6 hours ago

answered 9 hours ago

JonMark PerryJonMark Perry

21.9k643103

$endgroup$

add a comment | 

5

$begingroup$

I'm going to try:

$(7-2+0!)!=6!=720$.
Score 720/9=80.


This I believe is the best available, as the only combo's that could beat it are:

$410,;510,;610,;710,;810,;910,;820$ and $920$


and I haven't managed to make any of these.

share|improve this answer

edited 6 hours ago

answered 9 hours ago

JonMark PerryJonMark Perry

21.9k643103

$endgroup$

add a comment | 

$begingroup$

I'm going to try:

$(7-2+0!)!=6!=720$.
Score 720/9=80.


This I believe is the best available, as the only combo's that could beat it are:

$410,;510,;610,;710,;810,;910,;820$ and $920$


and I haven't managed to make any of these.

share|improve this answer

edited 6 hours ago

answered 9 hours ago

JonMark PerryJonMark Perry

21.9k643103

$endgroup$

share|improve this answer

edited 6 hours ago

answered 9 hours ago

JonMark PerryJonMark Perry

21.9k643103

answered 9 hours ago

JonMark PerryJonMark Perry

21.9k643103

answered 9 hours ago

JonMark PerryJonMark Perry

21.9k643103