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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:
$$\begin{align} 8 &= BAD = 2 \times 1 \times 4 = 8 \\ 6 &= CAB = 3 \times 1 \times 2 = 6 \\ 168000 &= ADJACENT \end{align}$$
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?

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    $\begingroup$ What working out have you done so far? If you could edit your question to show this that would be good. $\endgroup$ – MRobinson Sep 7 '18 at 11:40
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    $\begingroup$ 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. $\endgroup$ – Teresa Lisbon Sep 7 '18 at 11:48
  • $\begingroup$ Puzzling SE? . . $\endgroup$ – BCLC Sep 7 '18 at 12:59
  • $\begingroup$ Why do you say this is linear algebra? This actually reminds me of the Homophonic Group on abstract algebra $\endgroup$ – BCLC Sep 7 '18 at 13:04
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The word TYPEY, a variant spelling of TYPY, works exactly.

(Link is to the Dictionary.com definition.)

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  • $\begingroup$ Nice!!!!!!!!!!! $\endgroup$ – BCLC Sep 7 '18 at 13:02
  • $\begingroup$ Awesome ! Exactly 1,000,000 $\endgroup$ – MMJM Sep 7 '18 at 13:27
  • $\begingroup$ Doh: internet anagram server didn't produce either typy or typey :) $\endgroup$ – rschwieb Sep 7 '18 at 15:22
  • $\begingroup$ @rschwieb Nor does the dictionary... $\endgroup$ – MRobinson Sep 10 '18 at 7:41
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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

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  • $\begingroup$ Thanks, I thought it would be very hard. $\endgroup$ – MMJM Sep 7 '18 at 12:08
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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).

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