TWO x TWO = THREE

Each letter stands for one and only one digit, and no digit is represented by more than one letter. Can you work out what digits the letters in the above multiplication stand for so that the identity above is actually correct?

The person who is going to answer has to explain how he did think

138

t=1
w=3
o=8

138*138 will give 19044

t=1
h=9
r=0
e=4
e=4

well, i didnt excatly solve it but i think what i found
is solvable in math so in a way i did solve it

two * two = three

well, you said digit which is kinda misleading
you shoulda said decimal

i rewrote it to

((t*(10^2))+(w*(10^1))+(o*(10^0)))
* ((t*(10^2))+(w*(10^1))+(o*(10^0)))
-----------------------------------
= ((t*(10^4))+(h*(10^3))+(r*(10^2))+(e*(10^1))+(e*(10*0)))


processed can look like

10,000*(t^2)+100*(w^2)+(o^2)+2000*t*w+200*t*o+20*w*o
= 10,000*t+1000*h+100*r+11*e


six unknows
t,w,o,h,r,e
am too lazy to solve that

it is easier to solve by considering some limits.

for instance what is the smallest and largest three digits number whose square is 5 digits??
the answer is 100 and 317

so two has to fall between them.
now consider that the 100s digit in TWO equals the 10,000 digit in THREE, it is obvious that T must equal one, since 200^2 == 40,000 and 300^2 == 90,000.

so now we know that TWO falls between 100 and 141 (141 being the largest integer whose' square is less than 10,000).

now W cannot be 0 or 1 since 0 and one will always yeild 0 and 1 in the units of the result and the units digit of the result equals E not W.

so now we have TWO between 102 and 139.

numbers close to 100 will yield repeated zeros in their square and only the units and 10s digits are repeated and we already agreed that they can't be zero, so it follows that TWO has to be at least 104 (4 being the first digit whose square is more than one digit).

we already have the answer narrowed down to [104 - 139] - { 110, 111, 120, 121, 130, 131 } even brute force won't take much time.

but it can be solved more easily by observing the relation ship between the two least digits in the result.
namely that
E = units_of(O^2)
and
E = tens_of(O^2) + units_of(2OW)

with some odd vs even rules you can quickly tell that only
(o=2, w=1) and (o=8, w=3) solves this.

but since 112^2 yields 12544 with H == W we discards this solution and only 138 remains.

is there a simpler way to solve this??

cheers,
Alaa

well done alaa
but I thought in another way which shrotened the path
THREE has a minimum and a maximum range from 12344 to 98766 with their square roots (=TWO) from ~ 111 to 317
but it is known from TWO that all the digits must be different; then TWO ranges from 123 to 317
as u noticed that T appears in the result then neither 200s nor 300s work here; then we r sure now that the number is lower than 20000 (TWO < ~ 141)
or TWO is from 123 to 140
what about 140? it results in 19600

I assumed that T W O H R and E are different digits so 140 would not work since O == 0 and E == 0

in fact you said no digit is represented by more than one letter.

oops then Im wrong or actually I forgot this rule