If $A=1$, $B=2$, etc, then what word, treated as a product of its letters, has value closest to $1000000$?

Clash Royale CLAN TAG#URR8PPP

up vote
5
down vote

favorite

Suppose that a product $n$ is the product of the numbers corresponding to its letter, eg. $A = 1$, $B = 2$, etc.

What is the word that has a product close $1000000$?

Here's some examples:

$$beginalign
8 &= BAD = 2 times 1 times 4 = 8 \
6 &= CAB = 3 times 1 times 2 = 6 \
168000 &= ADJACENT
endalign$$

EDIT : Heres some what I did.
First, I chopped $1,000,000$ into $10^6$, or $10 times 10 times 10 times 10 times 10 times 10$.

Then, I factored each $10$ into $2 times 5$.

Then, I tried to to combine the $2$'s and $5$'s in different quantity. In short, I produced the letters : $A$, $B$, $D$, $E$, $H$, $P$, $J$, and other letters.

Then, I think I can't produce some word that has a meaning and makes sense, because it exceeds the limitation of $1,000,000$. How do I get it through?

algebra-precalculus puzzle products

share|cite|improve this question

edited Sep 7 at 14:36

Quasicoherent

11.7k22041

asked Sep 7 at 11:39

MMJM

546

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What working out have you done so far? If you could edit your question to show this that would be good.
  – MRobinson
  Sep 7 at 11:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a "word" with product exactly thousand, and that is $BEBEBEBEBEBE$ : note that $2 times 5 = 10$, so I repeated this exactly $6$ times. We may add as many $A$s as we like. Also, our choice of letters is restricted to $A,B,D,E,H,J,P,T,Y$. Note that the funny : $BABY-BABY-BABY$ also works out.
  – Ð°ÑÑ‚он вілла олоф мэллбэрг
  Sep 7 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Puzzling SE? . .
  – BCLC
  Sep 7 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you say this is linear algebra? This actually reminds me of the Homophonic Group on abstract algebra
  – BCLC
  Sep 7 at 13:04
  

  
  
  
  

add a comment | 

up vote
5
down vote

favorite

Suppose that a product $n$ is the product of the numbers corresponding to its letter, eg. $A = 1$, $B = 2$, etc.

What is the word that has a product close $1000000$?

Here's some examples:

$$beginalign
8 &= BAD = 2 times 1 times 4 = 8 \
6 &= CAB = 3 times 1 times 2 = 6 \
168000 &= ADJACENT
endalign$$

EDIT : Heres some what I did.
First, I chopped $1,000,000$ into $10^6$, or $10 times 10 times 10 times 10 times 10 times 10$.

Then, I factored each $10$ into $2 times 5$.

Then, I tried to to combine the $2$'s and $5$'s in different quantity. In short, I produced the letters : $A$, $B$, $D$, $E$, $H$, $P$, $J$, and other letters.

Then, I think I can't produce some word that has a meaning and makes sense, because it exceeds the limitation of $1,000,000$. How do I get it through?

algebra-precalculus puzzle products

share|cite|improve this question

edited Sep 7 at 14:36

Quasicoherent

11.7k22041

asked Sep 7 at 11:39

MMJM

546

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What working out have you done so far? If you could edit your question to show this that would be good.
  – MRobinson
  Sep 7 at 11:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a "word" with product exactly thousand, and that is $BEBEBEBEBEBE$ : note that $2 times 5 = 10$, so I repeated this exactly $6$ times. We may add as many $A$s as we like. Also, our choice of letters is restricted to $A,B,D,E,H,J,P,T,Y$. Note that the funny : $BABY-BABY-BABY$ also works out.
  – Ð°ÑÑ‚он вілла олоф мэллбэрг
  Sep 7 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Puzzling SE? . .
  – BCLC
  Sep 7 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you say this is linear algebra? This actually reminds me of the Homophonic Group on abstract algebra
  – BCLC
  Sep 7 at 13:04
  

  
  
  
  

add a comment | 

up vote
5
down vote

favorite

up vote
5
down vote

favorite

Suppose that a product $n$ is the product of the numbers corresponding to its letter, eg. $A = 1$, $B = 2$, etc.

What is the word that has a product close $1000000$?

Here's some examples:

$$beginalign
8 &= BAD = 2 times 1 times 4 = 8 \
6 &= CAB = 3 times 1 times 2 = 6 \
168000 &= ADJACENT
endalign$$

EDIT : Heres some what I did.
First, I chopped $1,000,000$ into $10^6$, or $10 times 10 times 10 times 10 times 10 times 10$.

Then, I factored each $10$ into $2 times 5$.

Then, I tried to to combine the $2$'s and $5$'s in different quantity. In short, I produced the letters : $A$, $B$, $D$, $E$, $H$, $P$, $J$, and other letters.

Then, I think I can't produce some word that has a meaning and makes sense, because it exceeds the limitation of $1,000,000$. How do I get it through?

algebra-precalculus puzzle products

share|cite|improve this question

edited Sep 7 at 14:36

Quasicoherent

11.7k22041

asked Sep 7 at 11:39

MMJM

546

Suppose that a product $n$ is the product of the numbers corresponding to its letter, eg. $A = 1$, $B = 2$, etc.

What is the word that has a product close $1000000$?

Here's some examples:

$$beginalign
8 &= BAD = 2 times 1 times 4 = 8 \
6 &= CAB = 3 times 1 times 2 = 6 \
168000 &= ADJACENT
endalign$$

EDIT : Heres some what I did.
First, I chopped $1,000,000$ into $10^6$, or $10 times 10 times 10 times 10 times 10 times 10$.

Then, I factored each $10$ into $2 times 5$.

Then, I tried to to combine the $2$'s and $5$'s in different quantity. In short, I produced the letters : $A$, $B$, $D$, $E$, $H$, $P$, $J$, and other letters.

Then, I think I can't produce some word that has a meaning and makes sense, because it exceeds the limitation of $1,000,000$. How do I get it through?

algebra-precalculus puzzle products

share|cite|improve this question

edited Sep 7 at 14:36

Quasicoherent

11.7k22041

asked Sep 7 at 11:39

MMJM

546

share|cite|improve this question

share|cite|improve this question

edited Sep 7 at 14:36

Quasicoherent

11.7k22041

edited Sep 7 at 14:36

Quasicoherent

11.7k22041

edited Sep 7 at 14:36

Quasicoherent

11.7k22041

11.7k22041

asked Sep 7 at 11:39

MMJM

546

asked Sep 7 at 11:39

MMJM

546

asked Sep 7 at 11:39

MMJM

546

546

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What working out have you done so far? If you could edit your question to show this that would be good.
  – MRobinson
  Sep 7 at 11:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a "word" with product exactly thousand, and that is $BEBEBEBEBEBE$ : note that $2 times 5 = 10$, so I repeated this exactly $6$ times. We may add as many $A$s as we like. Also, our choice of letters is restricted to $A,B,D,E,H,J,P,T,Y$. Note that the funny : $BABY-BABY-BABY$ also works out.
  – Ð°ÑÑ‚он вілла олоф мэллбэрг
  Sep 7 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Puzzling SE? . .
  – BCLC
  Sep 7 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you say this is linear algebra? This actually reminds me of the Homophonic Group on abstract algebra
  – BCLC
  Sep 7 at 13:04
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What working out have you done so far? If you could edit your question to show this that would be good.
  – MRobinson
  Sep 7 at 11:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a "word" with product exactly thousand, and that is $BEBEBEBEBEBE$ : note that $2 times 5 = 10$, so I repeated this exactly $6$ times. We may add as many $A$s as we like. Also, our choice of letters is restricted to $A,B,D,E,H,J,P,T,Y$. Note that the funny : $BABY-BABY-BABY$ also works out.
  – Ð°ÑÑ‚он вілла олоф мэллбэрг
  Sep 7 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Puzzling SE? . .
  – BCLC
  Sep 7 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you say this is linear algebra? This actually reminds me of the Homophonic Group on abstract algebra
  – BCLC
  Sep 7 at 13:04
  

  
  
  
  

2

2

What working out have you done so far? If you could edit your question to show this that would be good.
– MRobinson
Sep 7 at 11:40

What working out have you done so far? If you could edit your question to show this that would be good.
– MRobinson
Sep 7 at 11:40

1

1

There is a "word" with product exactly thousand, and that is $BEBEBEBEBEBE$ : note that $2 times 5 = 10$, so I repeated this exactly $6$ times. We may add as many $A$s as we like. Also, our choice of letters is restricted to $A,B,D,E,H,J,P,T,Y$. Note that the funny : $BABY-BABY-BABY$ also works out.
– Ð°ÑÑ‚он вілла олоф мэллбэрг
Sep 7 at 11:48

There is a "word" with product exactly thousand, and that is $BEBEBEBEBEBE$ : note that $2 times 5 = 10$, so I repeated this exactly $6$ times. We may add as many $A$s as we like. Also, our choice of letters is restricted to $A,B,D,E,H,J,P,T,Y$. Note that the funny : $BABY-BABY-BABY$ also works out.
– Ð°ÑÑ‚он вілла олоф мэллбэрг
Sep 7 at 11:48

Puzzling SE? . .
– BCLC
Sep 7 at 12:59

Puzzling SE? . .
– BCLC
Sep 7 at 12:59

Why do you say this is linear algebra? This actually reminds me of the Homophonic Group on abstract algebra
– BCLC
Sep 7 at 13:04

Why do you say this is linear algebra? This actually reminds me of the Homophonic Group on abstract algebra
– BCLC
Sep 7 at 13:04

add a comment | 

3 Answers
3

active

oldest

votes

up vote
7
down vote



accepted

The word TYPEY, a variant spelling of TYPY, works exactly.

(Link is to the Dictionary.com definition.)

share|cite|improve this answer

answered Sep 7 at 12:57

nickgard

1,6551414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice!!!!!!!!!!!
  – BCLC
  Sep 7 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome ! Exactly 1,000,000
  – MMJM
  Sep 7 at 13:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doh: internet anagram server didn't produce either typy or typey :)
  – rschwieb
  Sep 7 at 15:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @rschwieb Nor does the dictionary...
  – MRobinson
  2 days ago
  

  
  
  
  

add a comment | 

up vote
6
down vote

You're going to struggle to get 1,000,000 exactly, as there are only 9 factors less than 26. These are 1, 2, 4, 5, 8, 10, 16, 20, and 25.

Therefore your allowed letters are A, B, D, E, H, J, P, T, and Y.

I haven't been able to come up with a word from those that gets you 1,000,000 - but if you mess around with some fun words you can definitely get close:

HAUNTED = 940,800, JUMPY = 1,092,000

share|cite|improve this answer

edited Sep 7 at 12:12

answered Sep 7 at 12:03

MRobinson

42712

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I thought it would be very hard.
  – MMJM
  Sep 7 at 12:08
  

  
  
  
  

add a comment | 

up vote
4
down vote

I searched Mathematica's built-in dictionary...

value[char_] := ToCharacterCode[char] - ToCharacterCode["a"] + 1
wordValue[word_] := Times @@ (value /@ Characters[word])

words = ToLowerCase /@ 
 DictionaryLookup[("a"|"b"|"d"|"e"|"h"|"j"|"p"|"t"|"y").., IgnoreCase -> True];
Select[words, wordValue[#] == 1000000 &]

...and found nothing that works exactly. The closest we can get from above (by modifying the code to try to get as close as possible) is with CURING or NICARAGUA (1000188) and the closest we can get from below is with BANQUET (999600).

share|cite|improve this answer

answered Sep 7 at 14:09

Misha Lavrov

37.7k55093

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: "69"
;
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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2908533%2fif-a-1-b-2-etc-then-what-word-treated-as-a-product-of-its-letters-has-v%23new-answer', 'question_page');

);

Post as a guest

Name Email

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

The word TYPEY, a variant spelling of TYPY, works exactly.

(Link is to the Dictionary.com definition.)

share|cite|improve this answer

answered Sep 7 at 12:57

nickgard

1,6551414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice!!!!!!!!!!!
  – BCLC
  Sep 7 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome ! Exactly 1,000,000
  – MMJM
  Sep 7 at 13:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doh: internet anagram server didn't produce either typy or typey :)
  – rschwieb
  Sep 7 at 15:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @rschwieb Nor does the dictionary...
  – MRobinson
  2 days ago
  

  
  
  
  

add a comment | 

up vote
7
down vote



accepted

The word TYPEY, a variant spelling of TYPY, works exactly.

(Link is to the Dictionary.com definition.)

share|cite|improve this answer

answered Sep 7 at 12:57

nickgard

1,6551414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice!!!!!!!!!!!
  – BCLC
  Sep 7 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome ! Exactly 1,000,000
  – MMJM
  Sep 7 at 13:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doh: internet anagram server didn't produce either typy or typey :)
  – rschwieb
  Sep 7 at 15:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @rschwieb Nor does the dictionary...
  – MRobinson
  2 days ago
  

  
  
  
  

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

The word TYPEY, a variant spelling of TYPY, works exactly.

(Link is to the Dictionary.com definition.)

share|cite|improve this answer

answered Sep 7 at 12:57

nickgard

1,6551414

The word TYPEY, a variant spelling of TYPY, works exactly.

(Link is to the Dictionary.com definition.)

share|cite|improve this answer

answered Sep 7 at 12:57

nickgard

1,6551414

share|cite|improve this answer

share|cite|improve this answer

answered Sep 7 at 12:57

nickgard

1,6551414

answered Sep 7 at 12:57

nickgard

1,6551414

answered Sep 7 at 12:57

nickgard

1,6551414

1,6551414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice!!!!!!!!!!!
  – BCLC
  Sep 7 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome ! Exactly 1,000,000
  – MMJM
  Sep 7 at 13:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doh: internet anagram server didn't produce either typy or typey :)
  – rschwieb
  Sep 7 at 15:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @rschwieb Nor does the dictionary...
  – MRobinson
  2 days ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice!!!!!!!!!!!
  – BCLC
  Sep 7 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome ! Exactly 1,000,000
  – MMJM
  Sep 7 at 13:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doh: internet anagram server didn't produce either typy or typey :)
  – rschwieb
  Sep 7 at 15:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @rschwieb Nor does the dictionary...
  – MRobinson
  2 days ago
  

  
  
  
  

Nice!!!!!!!!!!!
– BCLC
Sep 7 at 13:02

Nice!!!!!!!!!!!
– BCLC
Sep 7 at 13:02

Awesome ! Exactly 1,000,000
– MMJM
Sep 7 at 13:27

Awesome ! Exactly 1,000,000
– MMJM
Sep 7 at 13:27

Doh: internet anagram server didn't produce either typy or typey :)
– rschwieb
Sep 7 at 15:22

Doh: internet anagram server didn't produce either typy or typey :)
– rschwieb
Sep 7 at 15:22

@rschwieb Nor does the dictionary...
– MRobinson
2 days ago

@rschwieb Nor does the dictionary...
– MRobinson
2 days ago

add a comment | 

up vote
6
down vote

You're going to struggle to get 1,000,000 exactly, as there are only 9 factors less than 26. These are 1, 2, 4, 5, 8, 10, 16, 20, and 25.

Therefore your allowed letters are A, B, D, E, H, J, P, T, and Y.

I haven't been able to come up with a word from those that gets you 1,000,000 - but if you mess around with some fun words you can definitely get close:

HAUNTED = 940,800, JUMPY = 1,092,000

share|cite|improve this answer

edited Sep 7 at 12:12

answered Sep 7 at 12:03

MRobinson

42712

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I thought it would be very hard.
  – MMJM
  Sep 7 at 12:08
  

  
  
  
  

add a comment | 

up vote
6
down vote

You're going to struggle to get 1,000,000 exactly, as there are only 9 factors less than 26. These are 1, 2, 4, 5, 8, 10, 16, 20, and 25.

Therefore your allowed letters are A, B, D, E, H, J, P, T, and Y.

I haven't been able to come up with a word from those that gets you 1,000,000 - but if you mess around with some fun words you can definitely get close:

HAUNTED = 940,800, JUMPY = 1,092,000

share|cite|improve this answer

edited Sep 7 at 12:12

answered Sep 7 at 12:03

MRobinson

42712

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I thought it would be very hard.
  – MMJM
  Sep 7 at 12:08
  

  
  
  
  

add a comment | 

up vote
6
down vote

up vote
6
down vote

You're going to struggle to get 1,000,000 exactly, as there are only 9 factors less than 26. These are 1, 2, 4, 5, 8, 10, 16, 20, and 25.

Therefore your allowed letters are A, B, D, E, H, J, P, T, and Y.

I haven't been able to come up with a word from those that gets you 1,000,000 - but if you mess around with some fun words you can definitely get close:

HAUNTED = 940,800, JUMPY = 1,092,000

share|cite|improve this answer

edited Sep 7 at 12:12

answered Sep 7 at 12:03

MRobinson

42712

You're going to struggle to get 1,000,000 exactly, as there are only 9 factors less than 26. These are 1, 2, 4, 5, 8, 10, 16, 20, and 25.

Therefore your allowed letters are A, B, D, E, H, J, P, T, and Y.

I haven't been able to come up with a word from those that gets you 1,000,000 - but if you mess around with some fun words you can definitely get close:

HAUNTED = 940,800, JUMPY = 1,092,000

share|cite|improve this answer

edited Sep 7 at 12:12

answered Sep 7 at 12:03

MRobinson

42712

share|cite|improve this answer

share|cite|improve this answer

edited Sep 7 at 12:12

edited Sep 7 at 12:12

edited Sep 7 at 12:12

answered Sep 7 at 12:03

MRobinson

42712

answered Sep 7 at 12:03

MRobinson

42712

answered Sep 7 at 12:03

MRobinson

42712

42712

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I thought it would be very hard.
  – MMJM
  Sep 7 at 12:08
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I thought it would be very hard.
  – MMJM
  Sep 7 at 12:08
  

  
  
  
  

Thanks, I thought it would be very hard.
– MMJM
Sep 7 at 12:08

Thanks, I thought it would be very hard.
– MMJM
Sep 7 at 12:08

add a comment | 

up vote
4
down vote

I searched Mathematica's built-in dictionary...

value[char_] := ToCharacterCode[char] - ToCharacterCode["a"] + 1
wordValue[word_] := Times @@ (value /@ Characters[word])

words = ToLowerCase /@ 
 DictionaryLookup[("a"|"b"|"d"|"e"|"h"|"j"|"p"|"t"|"y").., IgnoreCase -> True];
Select[words, wordValue[#] == 1000000 &]

...and found nothing that works exactly. The closest we can get from above (by modifying the code to try to get as close as possible) is with CURING or NICARAGUA (1000188) and the closest we can get from below is with BANQUET (999600).

share|cite|improve this answer

answered Sep 7 at 14:09

Misha Lavrov

37.7k55093

add a comment | 

up vote
4
down vote

I searched Mathematica's built-in dictionary...

value[char_] := ToCharacterCode[char] - ToCharacterCode["a"] + 1
wordValue[word_] := Times @@ (value /@ Characters[word])

words = ToLowerCase /@ 
 DictionaryLookup[("a"|"b"|"d"|"e"|"h"|"j"|"p"|"t"|"y").., IgnoreCase -> True];
Select[words, wordValue[#] == 1000000 &]

...and found nothing that works exactly. The closest we can get from above (by modifying the code to try to get as close as possible) is with CURING or NICARAGUA (1000188) and the closest we can get from below is with BANQUET (999600).

share|cite|improve this answer

answered Sep 7 at 14:09

Misha Lavrov

37.7k55093

add a comment | 

up vote
4
down vote

up vote
4
down vote

I searched Mathematica's built-in dictionary...

value[char_] := ToCharacterCode[char] - ToCharacterCode["a"] + 1
wordValue[word_] := Times @@ (value /@ Characters[word])

words = ToLowerCase /@ 
 DictionaryLookup[("a"|"b"|"d"|"e"|"h"|"j"|"p"|"t"|"y").., IgnoreCase -> True];
Select[words, wordValue[#] == 1000000 &]

...and found nothing that works exactly. The closest we can get from above (by modifying the code to try to get as close as possible) is with CURING or NICARAGUA (1000188) and the closest we can get from below is with BANQUET (999600).

share|cite|improve this answer

answered Sep 7 at 14:09

Misha Lavrov

37.7k55093

I searched Mathematica's built-in dictionary...

value[char_] := ToCharacterCode[char] - ToCharacterCode["a"] + 1
wordValue[word_] := Times @@ (value /@ Characters[word])

words = ToLowerCase /@ 
 DictionaryLookup[("a"|"b"|"d"|"e"|"h"|"j"|"p"|"t"|"y").., IgnoreCase -> True];
Select[words, wordValue[#] == 1000000 &]

...and found nothing that works exactly. The closest we can get from above (by modifying the code to try to get as close as possible) is with CURING or NICARAGUA (1000188) and the closest we can get from below is with BANQUET (999600).

share|cite|improve this answer

answered Sep 7 at 14:09

Misha Lavrov

37.7k55093

share|cite|improve this answer

share|cite|improve this answer

answered Sep 7 at 14:09

Misha Lavrov

37.7k55093

answered Sep 7 at 14:09

Misha Lavrov

37.7k55093

answered Sep 7 at 14:09

Misha Lavrov

37.7k55093

37.7k55093

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2908533%2fif-a-1-b-2-etc-then-what-word-treated-as-a-product-of-its-letters-has-v%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email