Hi,

>

> I'm having a go at my second "killer". I managed to complete

> the one in yesterday's Times, however the one published today

> appears to be illogical ( although I accept it's more likely

> to be me who is missing something.)

>

> Let me explain the problem : -

>

> If I notate the 9 * 9 grid, the numbers and letters, giving

> the rows numbers 1 - 9, and the columns letters A - I, then

> each little square can be addressed uniquely ( e.g. A1 is

> the top LH little square, and I9 is bottom RH one).

>

> Now, the sum of the values in each 3 * 3 box should be 45 (

> the sum of digits 1 - 9).

> As G1+ G2 + H3 = 15, and H1 + I1 + I2 = 13.

> then it follows that G3 + H3 + I3 = 45 - (13 + 15) = 17.

>

> Now for 3 unique digits to sum to 17, only the digits 4, 5

> and 8 can be used , so G3 = 4,5 or 8 ( as is H3 and I3).

>

> We also now that I3 + I4 + I5 = 18.

> For 3 unique digits to sum to 18, only the digits 4, 5 and 9

> can be used , so I3 = 4,5 or 9 ( as is I4 and I5).

>

> But we already know that I3 can only be 4,5 or 8, so for both

> these conditions to be satisfied, I3 must be 4 or 5 (only)

> and I4 and I5 contain

> 4,5 or 9.

>

> As G3 + G4 = 9, and G3 is 4,5 or 8, then G4 must be 4,5 or 1.

> As H3 + H4 = 9, and H3 is 4,5 or 8, then H4 must be 4,5 or 1.

>

> What we are seeking to show now, is that the square G5 must

> have the value 8. We can conclude this, because none of the

> other squares (G4, H4, I4,H5, I5, G6, H6 or I6) can have the

> value 8. ( So far we have shown that G4 and

> H4 have values 4,5 or 1 , and also that I4 and I5 have the

> value 4,5 or 9).

> We can also see that as H5 + H6 = 8, neither H5 or H6 can

> have the value 8.

> This leaves squares G5, G6 or I6 as each possibly having the values 8.

>

> We will now show that G6 and I6 cannot contain 8, which

> leaves only G5 ( so that must be 8).

> Neither G6 or I6 can have the value 8, as D6 has this value.

>

> We can see that D6 = 8 because A4 + B4 + C4 + A5 + B5 + C5 +

> A6 + B6 + C6 = 45; however A4 + B4 + C4 + A5 + B5 + C5 + A6 =

> 36. Therefore B6 + C6 = 45

> - 36 = 9.

> Now B6 + C6 + D6 = 17, so D6 = 17 -9 = 8.

>

> So we can now see that G5 must = 8 (above).

> As F5 + G5 = 12, F5 = 4.

>

> Now F6 + G6 = 6, (which means that F6 and G6 must have a

> value 5,4,2 or 1).

> As we already know F5 = 4, this means F6 can only be 5,2 or 1

> - and consequently G6 = 5,4 or 1.

>

> This leaves three squares within a 3 * 3 grid with 5,4 or 1

> as the only possible values (G6, G4 and H4). As each of these

> square must contain one of these values, none of the other

> squares within this 3 * 3 grid can contain the values.

>

> Before we saw that I4 and I5 both had possible values 4,5 or

> 9. ( So if we discount the values 4 or 5), this means that

> I4 and I5 can only contain 9, which is not permissible.

>

> Can anyone tell me what I've done wrong please ?

Kind regards

Chris