To get a 5 digit answer, which starts with the same digit as the 3 digit number that TWO starts with, we need to check the values of the squares of 3 digit numbers. Trying a few examples systematically shows that the only possibilities have 100 < TWO < 141.

We know that the digit associated with O cannot be 0, 1 or 5, because the last digit or THREE must be different to the last digit of TWO.

Therefore 100 < TWO < 140.

Since T is 1, W cannot be 1, so this means that W can only be 0, 2 or 3.

However we can also exclude W = 0, because if W = 0, then TWO = 100 + a, where a = 2, 3, 4, 7, 8, or 9.

But (100 + a)² = 10000 + 200a + a² = 102 x 100a + a², so a² must give the two last digits of THREE. However, none of our values of a have a² with the same two digits, so W cannot be 0.

We have reduced the number of options for TWO to only 12 numbers! They are

TWO = 120 + a, where a = 2, 3, 4, 7, 8, or 9
TWO = 130 + a, where a = 2, 3, 4, 7, 8, or 9

By calculating the last two digits of (120 + a)² and (130 + a)² when a = 2, 3, 4, 7, 8, or 9 we can be certain of finding all the possible numbers that TWO could represent.

Note that (120 + a)² = 14400 + 240a + a² = 100(144 + 2a) + 40a + a², so the last two digits will depend only on 40a + a².

Similarly, (130 + a)² = 16900 + 260a + a² = 100(144 + 2a) + 60a + a², so the last two digits will depend only on 60a + a².

The table shows the last two digits for each squared number

a	2	3	4	7	8	9
(120 + a)²	76	29	56	49	84	41
(130 + a)²	24	49	56	69	44	21

We see that the only expression that gives the required result is (130 + 8)².

Check: 138 x 138 = 19044 😊

So T = 1, W = 3, O = 8, H = 9, R = 0, E = 4.

Challenge 354: An Unexpected Sum