# Word Number Puzzle

If letters are factors of equations:

zero = 0

one = 1

two = 2

three = 3

four = 4

five = 5

six = 6

seven = 7

eight = 8

nine = 9

ten = 10

twelve = 12

fifteen = 15

twenty = 20

thirty = 30

forty = 40

sixty = 60

hundred = 100

thousand = 1000

million = 1000000

What is the product for answered = ?

• To clarify, you mean for example, z×e×r×o = 0 for some z, e, r, o? – greenturtle3141 Oct 7 '16 at 2:57
• @greenturtle3141: Formatting ate your asterisks. Here are some multiplication characters you can copy: ×·∗ – Deusovi Oct 7 '16 at 2:58
• yes (z)(e)(r)(o) = (r)(o)(z)(e)=0 – TSLF Oct 7 '16 at 3:05
• I'm not sure if I understood well. Every results are from the multiplication of every letters only? So we must deduce what value every letters have so that we can deduce what the word "answered" would equal too if we were to multiply all its letters? – stack reader Oct 7 '16 at 3:13
• @Numberknot I don't think non-integers are excluded anywhere, so $s,x$ could be $2,\frac13$. – Jonathan Allan Oct 7 '16 at 8:30

±3,401,222,400

Here I want to thank Matsmath to remind me that a negative solution do exist.

I use induction to find out the law behind the puzzle, with 11 steps below:

Step 1:

Let's take a look at sixty first. Here we can cut it into two pieces: six and ty. As six is known as 6, we can found that:

ty = 10

ten as 10, ty as 10, so there is a equation:

y = en

Then we can compare ty to other "tys". In forty we found:

for = 4

four as 4, also for as 4, so the letter u can be found as:

u = 1

In twenty we found that twen as 2, the same value as two, and another euqation is here:

en = o

Along with y = en we summarize that:

y = en = o In thirty, thir represent the value 3. Compare thir with three, we found that:

i = ee

Step 2:

Focus on one, and the equations we have in step 1:

y = en = o

Then we found the equation one can be seen as oo. Since the value is 1, we found that:

o = ±1

Then apply y = en = o with value o, we found:

y = en = o = ±1

Step 3:

Here we look at ten, with the fact that en is ±1, the value of t can be found:

t = ±10

Step 4:

Variable t has found, then we focus on equation two, along with the value of o, w is solved:

w = 1/5

Step 5:

Put equation en = ±1 into consider when we see nine, a new equation is found:

ni = ±9

Since ni need to multiply ne to have a positive 9, and we can assume that n could be a negative number.

Then combine i = ee with the equation above, we found that:

een = ±9

As we know en = ±1 and n can be negative, then e, n and i have been solved:

e = 9, n = ±1/9, i = 81

Step 6:

Back to equation thir, we join the value of t and i, and we found that:

81 * (±10) * hr = 3
27 * (±10) * hr = 1
hr = ±1/270

Step 7:

Take a look at hundred, here we join the value of u, e, and n (with both positive and negative), and:

hundred = hndrd = 100

We have found the value of hr in step 6, the equation become like this:

hndrd = hr * n * dd = (±1/270) * (±1) * dd = 100
dd = 27,000

Step 8:

For equation fifteen, put the value of i, t, e, n into it, and fifteen becomes:

fifteen = 81 * (±10) * 9 * (±1) * ff = 15

And ff is:

ff = 1/486

Then the f itself:

f = ±1/(9√6)

Step 9:

With the value of f, we can turn to four and solve r. First, join o = ±1, u = 1, we can see that:

four = f * (±1) * 1 * r

Then, join the value of f:

f * (±1) * 1 * r = ±1/(9√6) * (±1) * 1 * r
= 1/(9√6) * r = 4

And we can found r is (with square version):

r = 36√6
rr = 7,776

Step 10:

Move our step to thousand and hundred, we can apply a division here:

thousand / hundred = tosa / red = 1,000/100 = 10

Join t, o, and e s' value into it, and here it goes:

(±10) * (±1) * sa /9 * rd = 10
sa /9 * rd = 1
sa = 9 * rd

Step 11:

Finally, focus on the equation answered. Just a bit more to reach the solution!
Join the value of n, w, and e.

answered = a * (±1/9) * s * (1/5) * 9 * r * 9 * d
= (±1) * (1/5) * 9 * as * rd

We have known that as = 9*rd from step 10:

(±1) * (1/5) * 9 * as * rd = (±1) * (1/5) * 9 * (9 * rd) * rd
= (±1) * 81 * (1/5) * rrdd

Join the value of rr and dd from step 9 and 7 respectively:

(±1) * 81 * (1/5) * rrdd = (±1) * 81 * (1/5) * 7,776 * 27,000
= (±1) * 81 * 5,400 * 7,776

Note that there's a ±1 inside, and we found equation answered is actually:

• I am almost certain that you have lost a sign somewhere along the way. – Matsmath Oct 7 '16 at 9:19
• Oh, I found the lost sign in step 10, thanks! – Shane Hsu Oct 7 '16 at 9:24
• In step 2 couldn't it also be the case that en = o = y = -1 ? – Ivo Beckers Oct 7 '16 at 9:27
• It's may be reasonable for o = -1, since the way I solved it from step 2. But I don't really get that why y can be found as -1... – Shane Hsu Oct 7 '16 at 9:33
• I've improve the answer with the negative solution, I did miss something while going through step 2 ! – Shane Hsu Oct 8 '16 at 4:58

# Basic idea

First, we need to see that:

It is an equation system.

And:

There are 20 variables: $a$, $d$, $e$, $f$, $g$, $h$, $i$, $l$, $m$, $n$, $o$, $r$, $s$, $t$, $u$, $v$, $w$, $x$, $y$ and $z$.
We are lacking the remaining 6 letters: $b$, $c$, $j$, $k$, $p$ and $q$.

So, we "just" need to solve it:

# Solving it, letter-by-letter, the fun part

1. $zero = 0$ (definition of zero)
$one = 1$ (definition of one)
$four = 4$ (definition of four)
Then:
If $o = 0$, we would have that $one = 0$ and $four = 0$, which are contradictions, so $o \neq 0$.
If $r = 0$, we would have that $four = 0$, which is also a contradiction, so $r \neq 0$.
If $e = 0$, we would have that $one = 0$, which is another contradiction, so $e \neq 0$.
Since we have that $o \neq 0$, $r \neq 0$ and $e \neq 0$, then $ero \neq 0$.
Thus:
$z = \frac{0}{ero}$
The division is ok, since the divisor is non-zero. So:
$\mathbf{z = 0}$ (equation 1)

2. $six = 6$ (definition of six)
$sixty = 60$ (definition of sixty)
Then:
$6ty = 60$
$ty = 10$ (equation 2)

3. $forty = 40$ (definition of forty)
$ty = 10$ (equation 2)
Then:
$10for = 40$
$for = 4$ (equation 3)

4. $four = 4$ (definition of four)
$for = 4$ (equation 3)
Then:
$4u = 4$
$\mathbf{u = 1}$ (equation 4)

5. $twenty = 20$ (definition of twenty)
$ty = 10$ (equation 2)
Then:
$10twen = 20$
$twen = 2$ (equation 5)

6. $two = 2$ (definition of two)
$twen = 2$ (equation 5)
Then:
$twen = two$
If we have that $tw = 0$, then $two = 0$, which is a contradiction. So, $tw \neq 0$. Thus we can divide both sides of the above by $tw$:
$en = o$ (equation 6)

7. $one = 1$ (definition of one)
So:
$o = \frac{1}{ne}$
But we have that:
$en = o$ (equation 6)
And then:
$o = \frac{1}{o}$ (equation 7a)
Finally:
$\mathbf{o = \pm 1}$ (equation 7b)
$\mathbf{o^2 = 1}$ (equation 7c)

8. $ten = 10$ (definition of ten)
$en = o$ (equation 6)
$o = \frac{1}{o}$ (equation 7a)
Then:
$ten = to$
$to = 10$
$t = \frac{10}{o}$
Applying the substitution from 7a:
$\mathbf{t = 10o}$ (equation 8)

9. $two = 2$ (definition of two)
$o^2 = 1$ (equation 7c)
$t = 10o$ (equation 8)
Then:
$10o^2 w = 2$
$10 w = 2$
$\mathbf{w = \frac{1}{5}}$ (equation 9)

10. $ty = 10$ (equation 2)
$o^2 = 1$ (equation 7c)
$t = 10o$ (equation 8)
Then:
$10oy = 10$
$oy = 1$
$oy = o^2$
$\mathbf{y = o}$ (equation 10)

11. $nine = 9$ (definition of nine)
$en = o$ (equation 6)
$o^2 = 1$ (equation 7c)
Then:
$nio = 9$
Let's multiply both sides by $o$, which equation 7b ensures is non-zero:
$nioo = 9o$
$ni = 9en$
Since $n = 0$ would imply $en = 0$, contradicting equations 6 and 7b, so we have that $n \neq 0$ and we can divide both sides by $n$:
$i = 9e$ (equation 11)

12. $thirty = 30$ (definition of thirty)
$o^2 = 1$ (equation 7c)
$t = 10o$ (equation 8)
$y = o$ (equation 10)
Then:
$10o × hir × 10o × o = 30$
$100o^3 × hir = 30$
Applying equation 7b: $100o × hir = 30$
$hir = \frac{3}{10} × o$ (equation 12)

13. $three = 3$ (definition of three)
$o = \frac{1}{o}$ (equation 7a)
$t = 10o$ (equation 8)
Then: $10o hree = 3$
$hree = \frac{3}{10} × \frac{1}{o}$ $hree = \frac{3}{10} × o$ (equation 13)

14. $hir = \frac{3}{10} × o$ (equation 12)
$hree = \frac{3}{10} × o$ (equation 13)
Then:
$hree = hir$
If $hr = 0$ then $hir = \frac{3}{10} × o$ would be a contradiction (and we know that $o \neq 0$), so we have that $hr \neq 0$ and we can divide both sides by $hr$:
$ee = i$
$e^2 = i$ (equation 14)

15. $i = 9e$ (equation 11)
$e^2 = i$ (equation 14)
Then:
$e^2 = 9e$
$\mathbf{e = 9}$ (equation 15a)
$i = 9^2$
$\mathbf{i = 81}$ (equation 15b)

16. $en = o$ (equation 6)
$e = 9$ (equation 15a)
Then:
$9n = o$
$\mathbf{n = \frac{o}{9}}$ (equation 16)

# Solving it, letter-by-letter, the hard part

Now we start producing some much harder to grasp numbers. There is a lot of square roots and complicated fractions ahead.

1. $fifteen = 15$ (definition of fifteen)
$o^2 = 1$ (equation 7c)
$t = 10o$ (equation 8)
$e = 9$ (equation 15a)
$i = 81$ (equation 15b)
$n = \frac{o}{9}$ (equation 16)
Then:
$f × 81 × f × 10o × 9 × 9 × \frac{o}{9} = 15$
$7290 f^2 o^2 = 15$
$7290 f^2 = 15$
$f^2 = \frac{15}{7290}$
$f = \pm \sqrt{\frac{15}{7290}}$
Argh! Let's try to simplify it:
$f = \pm \sqrt{\frac{3 × 5}{2 × 3^6 × 5}}$
$f = \pm \sqrt{\frac{1}{2 × 3^5}}$
$f = \pm \frac{1}{9} \sqrt{\frac{1}{6}}$
Let's introduce a new variable ($\phi$ to represent the sign of the $f$):
$\mathbf{\phi = \pm 1}$ (equation 17a)
$\mathbf{\phi = \frac{1}{\phi}}$ (equation 17b)
$\mathbf{\phi^2 = 1}$ (equation 17c)
$\mathbf{f = \phi × \frac{1}{9} \sqrt{\frac{1}{6}}}$ (equation 17d)

That square-root-of-a-fraction will appear frequently from now on, so keep in mind that:

1. For every $k \neq 0$, we have that:
\begin{align} \frac{1}{\sqrt{\frac{1}{k}}} & = \frac{\sqrt{1}}{\sqrt{\frac{1}{k}}} \\ & = \sqrt{\frac{1}{\frac{1}{k}}} \\ & = \sqrt{1 \div \frac{1}{k}} \\ & = \sqrt{1 × k} \\ & = \sqrt{k} \end{align}
(equation 18)

2. $for = 4$ (equation 3)
$o = \frac{1}{o}$ (equation 7a)
$o^2 = 1$ (equation 7b)
$f = \phi \frac{1}{9} \sqrt{\frac{1}{6}}$ (equation 17d)
Then:
$\phi \frac{1}{9} \sqrt{\frac{1}{6}} × o × r = 4$
$\phi \frac{1}{9} \sqrt{\frac{1}{6}} × r = \frac{4}{o}$
$r = o\phi × \frac{4}{\frac{1}{9} \sqrt{\frac{1}{6}}}$
$r = o\phi × \frac{4}{\frac{1}{9}} \frac{1}{\sqrt{\frac{1}{6}}}$
Applying the equation 18:
$r = o\phi × \frac{4 \sqrt{6}}{\frac{1}{9}}$
$r = o\phi × 4 \left( \sqrt{6} \div \frac{1}{9} \right)$
$r = o\phi × 4 \sqrt{6} × 9$
$\mathbf{r = 36 \sqrt{6} × o\phi}$ (equation 19)

3. $five = 5$ (definition of five)
$e = 9$ (equation 15a)
$i = 81$ (equation 15b)
$\phi = \frac{1}{\phi}$ (equation 17b)
$f = \phi \frac{1}{9} \sqrt{\frac{1}{6}}$ (equation 17d)
Then:
$\phi \frac{1}{9} \sqrt{\frac{1}{6}} × 81 × v × 9 = 5$
$81 \sqrt{\frac{1}{6}} \phi v = 5$
$v = \frac{5}{81} \frac{1}{\sqrt{\frac{1}{6}}} \frac{1}{\phi}$
Applying the equation 18:
$\mathbf{v = \frac{5}{81} \sqrt{6} × \phi}$ (equation 20)

4. $seven = 7$ (definition of seven)
$o = \frac{1}{o}$ (equation 7a)
$e = 9$ (equation 15a)
$n = \frac{o}{9}$ (equation 16)
$\phi = \frac{1}{\phi}$ (equation 17b)
$v = \frac{5}{81} \sqrt{6} × \phi$ (equation 20)
Then:
$s × 9 × \frac{5}{81} \sqrt{6} × \phi × 9 × \frac{o}{9} = 7$
$\frac{5}{9}\sqrt{6} so\phi = 7$
$s = 7 × \frac{1}{o} × \frac{1}{\phi} × \frac{9}{5} \frac{1}{\sqrt{6}}$
$\mathbf{s = o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}}}$ (equation 21)

5. $six = 6$ (definition of six)
$o = \frac{1}{o}$ (equation 7a)
$i = 81$ (equation 15b)
$\phi = \frac{1}{\phi}$ (equation 17b)
$s = o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}}$ (equation 21)
Then:
$o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}} × 81 × x = 6$
$x = 6 × \frac{1}{o} × \frac{1}{\phi} × \frac{5}{63} × \frac{1}{\sqrt{\frac{1}{6}}} × \frac{1}{81}$
Applying the equation 18:
$x = 6 × \frac{1}{o} × \frac{1}{\phi} \frac{5}{63} × \sqrt{6} × \frac{1}{81}$
$x = (2 × 3) × o\phi × \frac{5}{7 × 3^2} × \frac{1}{3^4} × \sqrt{6}$
$x = \frac{2 × 3 × 5}{7 × 3^6} × \sqrt{6} × o\phi$
$\mathbf{x = \frac{2 × 5}{7 × 3^5} × \sqrt{6} × o\phi}$ (equation 22, more convenient than the form below)
$x = \frac{10}{1701} \sqrt{6} × o\phi$

6. $hir = \frac{3}{10} × o$ (equation 12)
$o = \pm 1$ (equation 7b)
$i = 81$ (equation 15b)
$\phi = \frac{1}{\phi}$ (equation 17b)
$r = 36 \sqrt{6} × o\phi$ (equation 19)
Then: $h × 81 × 36 \sqrt{6} × o\phi = \frac{3}{10} × o$
$h = \frac{3}{10} × o × \frac{1}{81} × \frac{1}{36} × \frac{1}{\sqrt{6}} × \frac{1}{o} × \frac{1}{\phi}$
Knowing that $o \neq 0$ due to equation 7b, then $o$ and $\frac{1}{o}$ cancels out: $h = \frac{3}{2 × 5} × \frac{1}{3^4} × \frac{1}{2^2 × 3^2} × \sqrt{\frac{1}{6}} × \phi$
$h = \frac{3}{2^3 × 3^6 × 5} × \sqrt{\frac{1}{6}} × \phi$
$\mathbf{h = \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi}$ (equation 23, more convenient than the form below)
$h = \frac{1}{9720} × \sqrt{\frac{1}{6}} × \phi$

7. $twelve = 12$ (definiton of twelve)
$o = \frac{1}{o}$ (equation 7a)
$t = 10o$ (equation 8)
$w = \frac{1}{5}$ (equation 9)
$e = 9$ (equation 15a)
$\phi = \frac{1}{\phi}$ (equation 17b)
$v = \frac{5}{81} \sqrt{6} × \phi$ (equation 20)
Then:
$10o × \frac{1}{5} × 9 × l × \frac{5}{81} × \sqrt{6} × \phi × 9 = 12$
$10 \sqrt{6} lo\phi = 12$
$l = \frac{6}{5} \sqrt{\frac{1}{6}} × \frac{1}{o} × \frac{1}{\phi}$
$\mathbf{l = \frac{6}{5} \sqrt{\frac{1}{6}} × o\phi}$ (equation 24)

8. $million = 1000000$ (definiton of million)
$o = \frac{1}{o}$ (equation 7a)
$o^2 = 1$ (equation 7c)
$i = 81$ (equation 15b)
$\phi^2 = 1$ (equation 17c)
$l = \frac{6}{5} \sqrt{\frac{1}{6}} × o\phi$ (equation 24)
$n = \frac{o}{9}$ (equation 16)
Then:
$m × 81^2 × (\frac{6}{5} \sqrt{\frac{1}{6}} × o\phi)^2 × o × \frac{o}{9} = 1000000$
$m × 9^4 × \frac{1}{9} × \frac{36}{25} × \frac{1}{6} × o^4 × \phi^2 = 1000000$
$m × 9^3 × \frac{6}{25} = 1000000$
$m = 1000000 × \frac{1}{9^3} × \frac{25}{6}$
$m = 25000000 × \frac{1}{3^6} × \frac{1}{2 × 3}$
$m = 12500000 × \frac{1}{3^7}$
$m = \mathbf{\frac{12500000}{3^7}}$ (equation 25, more convenient than the form below)
$m = \frac{12500000}{2187}$

9. $eight = 8$ (definition of eight)
$o = \frac{1}{o}$ (equation 7a)
$t = 10o$ (equation 8)
$e = 9$ (equation 15a)
$i = 81$ (equation 15b)
$\phi = \frac{1}{\phi}$ (equation 17b)
$h = \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi$ (equation 23)
Then:
$9 × 81 × g × \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi × 10 × o = 8$
$3^2 × 3^4 × g × \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × 2 × 5 × o\phi = 8$
$go\phi × \frac{3}{2^2} × \sqrt{\frac{1}{6}} = 2^3$
$g = \frac{2^2}{3} × \frac{1}{\sqrt{\frac{1}{6}}} × 2^3 × \frac{1}{o} × \frac{1}{\phi}$
Applying the equation 18:
$g = \frac{2^2}{3} × \sqrt{6} × 2^3 × \frac{1}{o} × \frac{1}{\phi}$
$\mathbf{g = \frac{32}{3} × \sqrt{6} × o\phi}$ (equation 26)

10. $hundred = 100$ (definiton of hundred)
$u = 1$ (equation 4) $o^2 = 1$ (equation 7c)
$e = 9$ (equation 15a)
$n = \frac{o}{9}$ (equation 16)
$\phi^2 = 1$ (equation 17c)
$r = 36 \sqrt{6} × o\phi$ (equation 19)
$h = \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi$ (equation 23)
Then:
$\frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi × 1 × \frac{o}{9} × d^2 × 36 \sqrt{6} × o\phi × 9 = 100$
$\frac{d^2}{2 × 3^3 × 5} × o^2 × \phi^2 = 100$
$d^2 = 100 × 2 × 3^3 × 5$
$d^2 = 2^3 × 3^3 × 5^3$
$d^2 = 30^3$
$d = \pm (30 \sqrt{30})$
It is important to note that there are two possible values for $d$ with opposed signs, this happens because we have $d^2$ in the equation, so there are two roots.
So, let's invent a new artificial variable for tracking $d$ 's signal more easily:
$\mathbf{\psi = \pm 1}$ (equation 27a)
$\mathbf{\psi = \frac{1}{\psi}}$ (equation 27b)
And then:
$\mathbf{d = 30 × \psi × \sqrt{30}}$ (equation 27c)

11. $thousand = 1000$ (definiton of thousand)
$u = 1$ (equation 4)
$o = \frac{1}{o}$ (equation 7a)
$o^2 = 1$ (equation 7c)
$t = 10o$ (equation 8)
$n = \frac{o}{9}$ (equation 16)
$\phi^2 = 1$ (equation 17c)
$s = o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}}$ (equation 21)
$h = \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi$ (equation 23)
$\psi = \frac{1}{\psi}$ (equation 27b)
$d = 30 × \psi × \sqrt{30})$ (equation 27c)
Then:
$10 × o × \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi × o × 1 × o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}} × a × \frac{o}{9} × 30 × \psi × \sqrt{30} = 1000$
$o^4 × \phi^2 × \frac{1}{2^3 × 3^5 × 5} × \frac{1}{6} × \frac{3^2 × 7}{5} × \frac{1}{3^2} × 30 × \psi × \sqrt{30} × a = 100$
$\frac{1}{2^3 × 3^5 × 5} × \frac{1}{2 × 3} × \frac{3^2 × 7}{5} × \frac{1}{3^2} × 2 × 3 × 5 × \psi × \sqrt{30} × a = 2^2 × 5^2$
$a = 2^3 × 3^5 × 5 × 2 × 3 × \frac{5}{3^2 × 7} × 3^2 × \frac{1}{2 × 3 × 5 × \psi} × \frac{1}{\sqrt{30}} × 2^2 × 5^2$
$a = 2^5 × 3^5 × 5^3 × \frac{1}{7} × \frac{1}{\psi} × \frac{1}{\sqrt{30}}$
$\mathbf{a = \psi × \frac{2^5 × 3^5 × 5^3}{7 \sqrt{30}}}$ (equation 28, more convenient than the form below)
$a = \psi × \frac{972000}{7 \sqrt{30}}$

# The final variable values

Ok. This should give the value of all the variables. The variables which depend on the $o$, $\phi$ and $\psi$ values have more than one possible final value. Here are the final values:

$a = \psi × \frac{2^5 × 3^5 × 5^3}{7 \sqrt{30}}$ (equation 28)
$d = 30 × \psi × \sqrt{30}$ (equation 27c)
$e = 9$ (equation 15a)
$f = \phi × \frac{1}{9} \sqrt{\frac{1}{6}}$ (equation 17d)
$g = \frac{32}{3} × \sqrt{6} × o\phi$ (equation 26)
$h = \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi$ (equation 23)
$i = 81$ (equation 15b)
$l = \frac{6}{5} \sqrt{\frac{1}{6}} × o\phi$ (equation24)
$m = \frac{12500000}{3^7}$ (equation 25)
$n = \frac{o}{9}$ (equation 16)
$o = \pm 1$ (equation 7b)
$r = 36 \sqrt{6} × o\phi$ (equation 19)
$s = o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}}$ (equation 21)
$t = 10o$ (equation 8)
$u = 1$ (equation 4)
$v = \frac{5}{81} \sqrt{6} × \phi$ (equation 20)
$w = \frac{1}{5}$ (equation 9)
$x = \frac{2 × 5}{7 × 3^5} × \sqrt{6} × o\phi$ (equation 22)
$y = o$ (equation 10)
$z = 0$ (equation 1)
$\phi = \pm 1$ (equation 17a)
$\psi = \pm 1$ (equation 27a)

# Final check

Let's do a final check if all the values match:

• zero:

$zero = 0 × 9 × \frac{1}{9} \sqrt{\frac{1}{6}} × o = 0 × o = 0$

• one:

$one = o × \frac{o}{9} × 9 = o^2 = 1$

• two:

$two = 10o × \frac{1}{5} × o = 2o^2 = 2$

• three:

\begin{align} three & = 10o × \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi × 36 \sqrt{6} × o\phi × 9 × 9 \\ & = \phi^2 × o^2 × \frac{1}{2^2 × 3^5} × 36 × 81 \\ & = \frac{36}{2^2 × 3} \\ & = \frac{36}{12} \\ & = 3 \end{align}

• four:

$four = \phi \frac{1}{9} \sqrt{\frac{1}{6}} × o × 1 × 36 \sqrt{6} × o\phi = \frac{36}{9} × o^2 × \phi^2 = 4$

• five:

$five = \phi \frac{1}{9} \sqrt{\frac{1}{6}} × 81 × \frac{5}{81} \sqrt{6} × \phi × 9 = 5 × \phi^2 = 5$

• six:

\begin{align} six & = o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}} × 81 × \frac{2 × 5}{7 × 3^5} × \sqrt{6} × o\phi \\ & = \frac{3^2 × 7}{5} × 3^4 × \frac{2 × 5}{7 × 3^5} × o^2 × \phi^2 \\ & = 9 × 81 × \frac{2 × 3^6 × 5 × 7}{3^5 × 5 × 7} \\ & = 2 × 3 \\ & = 6 \end{align}

• seven:

\begin{align} seven & = o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}} × 9 × \frac{5}{81} \sqrt{6} × \phi × 9 × \frac{o}{9} \\ & = \frac{63}{5} × 9 × \frac{5}{81} × o^2 × \phi^2 \\ & = \frac{63}{81} × 9 \\ & = \frac{63}{9} \\ & = 7 \end{align}

• eight:

\begin{align} eight & = 9 × 81 × \frac{32}{3} × \sqrt{6} × o\phi × \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi × 10o \\ & = 3^6 × 2 × 5 × \frac{2^5}{3} × \frac{1}{2^3 × 3^5 × 5} × o^2 × \phi^2 \\ & = \frac{2^6 × 3^6 × 5}{2^3 × 3^6 × 5} \\ & = 2^3 \\ & = 8 \end{align}

• nine:

$nine = \frac{o}{9} × 81 × \frac{o}{9} × 9 = \frac{81}{9} × o^2 = 9$

• ten:

$ten = 10o × 9 × \frac{o}{9} = 10 × \frac{9}{9} × o^2 = 10$

• twelve:

\begin{align} twelve & = 10o × \frac{1}{5} × 9 × \frac{6}{5} \sqrt{\frac{1}{6}} × o\phi × \frac{5}{81} \sqrt{6} × \phi × 9 \\ & = 2 × 81 × \frac{6}{5} × \frac{5}{81} × o^2 × \phi^2 \\ & = 2 × 6 \\ & = 12 \end{align}

• fifteen:

\begin{align} fifteen & = \phi × \frac{1}{9} \sqrt{\frac{1}{6}} × 81 × \phi × \frac{1}{9} \sqrt{\frac{1}{6}} × 10o × 9 × 9 × \frac{o}{9} \\ & = \frac{1}{9^3} × \frac{1}{6} × 9^4 × 10 × o^2 × \phi^2 \\ & = 9 × \frac{10}{6} \\ & = \frac{90}{6} \\ & = 15 \end{align}

• twenty:

$twenty = 10 × \frac{1}{5} × 9 × \frac{1}{9} × 10 × 1 = 10 × 10 × \frac{1}{5} = \frac{100}{5} = 20$

• thirty:

\begin{align} thirty & = 10o × \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi × 81 × 36 \sqrt{6} × o\phi × 10o × o \\ & = 10^2 × \frac{1}{2^3 × 3^5 × 5} × 81 × 36 × o^4 × \phi^2 \\ & = 2^2 × 5^2 × 3^4 × 2^2 × 3^2 × \frac{1}{2^3 × 3^5 × 5} \\ & = \frac{2^4 × 3^6 × 5^2}{2^3 × 3^5 × 5} \\ & = 2 × 3 × 5 \\ & = 30 \end{align}

• forty:

\begin{align} forty = & = \phi × \frac{1}{9} \sqrt{\frac{1}{6}} × o × 36 \sqrt{6} × o\phi × 10o × o \\ & = \frac{1}{9} × 36 × 10 × o^4 × \phi^2 \\ & = 4 × 10 \\ & = 40 \end{align}

• sixty:

\begin{align} sixty & = o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}} × 81 × \frac{2 × 5}{7 × 3^5} × \sqrt{6} × o\phi × 10o × o \\ & = \frac{3^2 × 7}{5} × 3^4 × \frac{2 × 5}{7 × 3^5} × 2 × 5 × o^4 × \phi^2 \\ & = 9 × 81 × \frac{2^2 × 3^6 × 5^2 × 7}{3^5 × 5 × 7} \\ & = 2^2 × 3 × 5 \\ & = 4 × 3 × 20 \\ & = 60 \end{align}

• hundred:

\begin{align} hundred & = \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi × 1 × \frac{o}{9} \\ & \qquad\; × 30 × \psi \sqrt{30} × 36 \sqrt{6} × o\phi × 9 × 30 × \psi \sqrt{30} \\ & = \frac{36}{2^3 × 3^5 × 5} × o^2 × \phi^2 × \psi^2 × (30 \sqrt{30})^2 \\ & = \frac{2^2 × 3^2}{2^3 × 3^5 × 5} × 30^2 \times 30 \\ & = \frac{1}{2 × 3^3 × 5} × 30^3 \\ & = \frac{30^3}{30 × 3^2} \\ & = \frac{30^2}{3^2} \\ & = \frac{900}{9} \\ & = 100 \end{align}

• thousand:

\begin{align} thousand & = 10o × \frac{1}{2^3 × 3^5 × 5} × \sqrt{\frac{1}{6}} × \phi × o × 1 \\ & \qquad\; × o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}} × \psi × \frac{2^5 × 3^5 × 5^3}{7 \sqrt{30}} × \frac{o}{9} × 30 × \psi × \sqrt{30} \\ & = 10 × \frac{1}{2^3 × 3^5 × 5} × \frac{1}{6} × \frac{63}{5} × 30 \sqrt{30} × \frac{2^5 × 3^5 × 5^3}{7 \sqrt{30}} \\ & \qquad\; × \frac{1}{9} × o^2 × \phi^2 × \psi^2 \\ & = 10 × \frac{1}{2^3 × 3^5 × 5} × \frac{1}{6} × \frac{7}{5} × 30 × \frac{2^5 × 3^5 × 5^3}{7} \\ & = 10 × \frac{1}{2^3 × 3^5 × 5} × \frac{1}{6} × \frac{1}{5} × 30 × 2^5 × 3^5 × 5^3 \\ & = 10 × 2^2 × 5^2 \\ & = 1000 \end{align}

• million:

\begin{align} million & = \frac{12500000}{3^7} × 81 × \frac{6}{5} \sqrt{\frac{1}{6}} × o\phi × \frac{6}{5} \sqrt{\frac{1}{6}} × o\phi × 81 × o × \frac{o}{9} \\ & = \frac{12500000}{3^7} × 9^4 × \frac{1}{9} × (\frac{6}{5})^2 × (\sqrt{\frac{1}{6}})^2 × o^2 × \phi^2 \\ & = \frac{12500000}{3^7} × 9^3 × \frac{36}{25} × \frac{1}{6} \\ & = \frac{500000 × 25}{3^7} × 3^6 × \frac{6}{25} \\ & = \frac{500000 × 25}{3^7} × 3^6 × \frac{2 × 3}{25} \\ & = \frac{500000 × 25}{3^7} × \frac{2 × 3^7}{25} \\ & = \frac{500000 × 25 × 3^7 × 2}{3^7 × 25} \\ & = 500000 × 2 \\ & = 1000000 \end{align}

# The gran finale

1. $\lambda = answered$ (definition of the final answer)
$\lambda = \psi × \frac{2^5 × 3^5 × 5^3}{7 \sqrt{30}} × \frac{o}{9} × o\phi × \frac{63}{5} × \sqrt{\frac{1}{6}} × \frac{1}{5} × 9 × 36 \sqrt{6} × o\phi × 9 × 30 × \psi \sqrt{30}$
$\lambda = \psi × \frac{2^5 × 3^5 × 5^3}{7 \sqrt{30}} × \frac{o}{3^2} × o\phi × \frac{3^2 × 7}{5} × \sqrt{\frac{1}{6}} × \frac{1}{5} × 3^2 × 2^2 × 3^2 × \sqrt{6} × o\phi × 3^2 × 2 × 3 × 5 × \psi \sqrt{30}$
$\lambda = \frac{2^8 × 3^{14} × 5^4 × 7 × \sqrt{30}}{3^2 × 5^2 × 7 \sqrt{30}} × o^3 × \phi^2 × \psi^2$
$\lambda = 2^8 × 3^{12} × 5^2 × o$
$\lambda = 256 × 531441 × 25 × o$
$\lambda = 3401222400o$
$\mathbf{\lambda = \pm 3401222400}$ (equation 29)

You may be wondering why the variables $o$, $\phi$ and $\psi$ that have more than one possible value didn't messed up anything in the equations or produced multiple possible answers. The reason is:

Because they are paired.

Observe that every word that contains letters with a $o$, $\phi$ or a $\psi$ factor contains an even number of such letters. This makes the words produce values that are always multiples of $o^2$, $\phi^2$ and $\psi^2$, which are equals to 1. The only exceptions are the words $zero$ and $answered$.

However, $zero$ has a $0$ multiplication factor, ending up being 0. So, $answered$, which have an odd number of o's, ended with a $\pm$ in its final form, being the only word in the list which have two possible values.

• Line 6 in part 23 is incorrect; an 'o' is missing from the right hand side. – Matsmath Oct 8 '16 at 1:14

I have solved the equations with a computer, and I obtained

answered = +-3401222400 (so there are two solutions).

Here are some of the individual values:

z=0 y=+-1 w=1/5 o=y, i=81, e=9,t=10y,n=y/9,x^2=200/964467, etc.

It seems, based on the answers that this problem was one tough nut. Here I offer another approach to it, first, throughout a warm-up example.

We assume that the problem is sensible, that is, there is indeed some solution. Then consider the algebraic expressions A[1], A[2], A[3], and A[4]:

A[1]:= z e r o -  0
A[2]:= o n e   -  1
A[3]:= f o u r -  4
A[4]:= t e n   - 10


Here these expressions all evaluate to zero, once a solution is substituted in place of the variables.

Consider the following magical weights W[1], W[2], W[3], and W[4]:

W[1]:= f n t u/40
W[2]:= f r t u z (10 - e n t)/400
W[3]:= - z/4
W[4]:= f r u z (o n e t - t - 10*o)/400


Now consider the weighted sum of the four algebraic expressions above:

A[1]*W[1]+A[2]*W[2]+A[3]*W[3]+A[4]*W[4]


We look at this expression from two points of view. From one point of view, we substitute in a solution in place of the variables, while from the other point of view we treat this big expression as multi-variate polynomial, unevaluated. So on the one hand, if a solution is indeed exist, then this expression is zero, as all the algebraic expressions A[.] are zero. On the other hand, you can treat this as a BIG algebraic expression in terms of all the involved variables e, f, n, o, r, t, u, z, and direct computation yields (without substituting in anything, that is, without using the fact that A[.] evaluates to zero):

A[1]*W[1]+A[2]*W[2]+A[3]*W[3]+A[4]*W[4]=z.


Wow, you must be astonished at this point! This means, that the left hand side is zero, whereas the right hand side is z, that is, we infer from these manipulations that z=0.

Next we address the problem posed by the OP. We assume, that their problem is sensible, that is, there is indeed some solution. Then we consider the algebraic expressions, given by in the OP, as

A[1] := z e r o       - 0
A[2] := o n e         - 1
A[3] := t w o         - 2
...
A[19]:= m i l l i o n - 1000000
A[20]:= (a n s w e r)^2-p


where A[20] is a new equation, and p:=(a n s w e r)^2 is a new variable. Observe, that if a solution indeed exist, then all of these algebraic expressions evaluate to zero. The task is now to find another expression involving p.

Now imagine, that a wizard gives you some magical weights W[1], W[2], ... , W[20] (the omitted expressions have something between 2000 to 20000 terms):

W[1] := 0
...
W[5] :=-((3099363912*a*d*(29160000 + 7*a*d)*h*n*s*t^3*y)/3125)
...
W[9] :=0
...
W[12]:=0
...
W[18]:=11568313814261760 + (347128758144*a*d)/125
W[19]:=0
W[20]:=-1


And then you consider the weighted sum of the 20 algebraic expressions above, in the exact same way as in the warm-up example, and you consider the BIG expression

 A[1]*W[1]+...A[20]*W[20]


from the two viewpoints. On the one hand, this expression must be zero, since all the algebraic expressions A[.] are zero. On the other hand, based on the magic weights, given to you by the wizard, direct computation will testify, that it simplifies to -11568313814261760000 + p. Therefore, you must be astonished once again, since this means that

0 = -11568313814261760000 + p,


and therefore p=(a n s w e r e d)^2=11568313814261760000.

Hence

a n s w e r e d = +-3401222400.

Note that this argument does not mean that both signs are possible. However, it shows that the expression a n s w e r e d cannot take any other value.

As for what are the magical weights, and how one should come up with formulae like that in the first place, well... you should ask these questions at our sister site Mathoverflow.SE.

• You should explained a bit more. Currently, it seems like you handwaved this answer :-) – IAmInPLS Oct 7 '16 at 9:10
• Yes, as of now I don't see a clear path how to do this in any intelligent way. – Matsmath Oct 7 '16 at 9:12
• The second I saw it couldn't be solved with integers, I gave up haha. I guess I'm just lazy. Making a program is the only way I imagine right now. – stack reader Oct 7 '16 at 9:15
• Great computation approach.. – TSLF Oct 11 '16 at 16:09