Didnt work when I typed in 666. Maybe its me? but does it work when all the numbers are the same? Or am I just a bit dumb? (I Pity the fool) - he said in his best Mr T voice!!
hum.....................ah...................................yes.............no...............weird..................
Yeah, something with multiples of nine. After taking one number from the other you can always add the digits together and get some multiple of nine, and so the number that's circled is whatever you need to add to the other two to make the next highest multiple of nine (55, you circle 8, 44, you circle 1 etc.) It always works out like this and so yes T-horn's subtracting skills suck
I know how it works!! *pulls her tongue out and does that "ner ner ni ner ner" thing* Go on then... I'll tell you.... When you jumble up any large number and subtract the smaller one from the biggest one then then answer is always a number where the individual digits add up to a multiple of 9. (eg 3168 gives 3+1+6+8 = 18, or 2799 gives 2+7+9+9 = 27). If you select any three digits and add them together, then subtract the total from the next biggest multiple of nine you are left with the value of the remaining digit. eg if you select 3, 6 and 8, add them together = 17, subtract from the next multiple of 9 which is 18 then the answer is 18-17=1! So there!! *sniggers*.....
Almost a complete answer, but you didn't say why it's always a multiple of 9. And a true BOC abhors nothing like an incomplete BOCcy answer... If you take an n-digit number, A, with digits a{1}, a{2}, a{3}, ..., a{n} then you can write it as 10^(n-1)*a{1}+10^(n-2)*a{2}+...+10*a{n-1}+a{n} If you muddle up the digits, you get a new n-digit number, B, with digits b{1}, b{2}, ..., b{n}, which can be written as 10^(n-1)*b{1}+10^(n-2)*b{2}+...+10*b{n-1}+b{n} However, because the bs are the same as the as, but in some other order, we can also write B as 10^m{1}*a{1}+10^m{2}*a{2}+...+10^m{n-1}*a{n-1}+10^m{n}*a{n}, where all of the ms are the integer powers from 1 to n-1, jumbled up to match the jumbling performed in changing A to B. Now, if we look at A-B (it doesn't matter which is larger - if something divides a number exactly, it will also divide minus that number exactly), then we see that we have the number (10^(n-1)-10^m{1})*a{1}+(10^(n-2)-10^m{2})*a{2}+...+(10^1-10^m{n-1})*a{n-1}+(10^0-10^m{n})*a{n} If we look at the ith term of this, we see that it is of the form (10^(n-i)-10^m{i})*a{i}. Just looking at the factor in brackets, we see that we have three cases: i) When n-i > m{i}, ii) When n-i < m{i} or iii) When n-i = m{i}. In case (i), the value of this can be written as 10^(n-i-m{i})*(10^m{i}-1). Looking at the bracketed bit in this, it is one less than a power of 10, i.e. a number of the form 9999...9, which evidently divides by 9 (giving the answer 1111...1). So the ith term of our expression divides by 9. In case (ii), it is similar, except that we write the factor as 10^(m{i}-(n-i))*(10^(n-i)-1). In case (iii), the factor is zero (10^x-10^x). So, every term is divisible by 9, and hence the whole difference of the two jumbled numbers is divisible by 9, for any number, of any number of digits, not just 3 or 4. Ahthankyewverymuch...