Fido Puzzle

[PJH]

Click Fido to proceed

 

[ClaireML]

Very annoying! How does he do it? I'm sure someone must have the mathematical answer

 

[imadoofus]

That's impressive!

How's it done?

I WANT TO KNOW!!

PJ

 

[PJH]

Very annoying! How does he do it? I'm sure someone must have the mathematical answer

I do. :bannana:

 

[imadoofus]

Well?

 

[andy_k]

I know as well.

Think cats

Andy

 

[Spinal]

WT.... that is impressive!

 

[imadoofus]

Think cats

Andy

Well I think I know what you mean, but I won't spoil it for anyone else....

Too thick to work out the details though....

PJ

 

[PJH]

Think cats

Andy

Yes in deed. I'll explain later.

 

[pammy]

tis a good 'un

 

[Cronos]

Very clever... still trying to work it out myself.

 

[PJH]

Here goes:- it's all to do with multiples of 9.....
Take your number, say 7843, mix up the digits and take the lowest from the highest
eg 7843 - 3478 = 4365
TIP
Then add up the digits in the answer ie 4+3+6+5 = 18
You can do the same with the answer ie 1+8 = 9

Back pedal to the 4365 bit, and circle any digit you like eg 6
Now add up the remaining digits ie 4+3+5 = 12
The circled number ie 6, is the difference between 12 and 18

It only fails if the answer for the subtraction is an exact hundred eg 900, when the circled number could be 9 or 0.

Incidently, all numbers whose digits add up to 9 are divisible by 9 eg 108, 1026, 2160 etc

 

[mergli]

But why does it always add up to 9????

 

[Spinal]

But why does it always add up to 9????

Pick ANY 3 digit number; (which we will call N for now) write it in the form:

N = 100x + 10y + z

scrambling the digits we get (N'):

N' = 100a + 10b + c
where:
a= y OR z
b= x OR z
c= x OR y

subtracting the two:
N-N' = 100(x-a) + 10(y-b) + (z-c)

if we add the digits together (say, M):
M = (x-a) + (y-b) + (z-c)

Now we already know a can be y or z, b can be... etc SO we need to prove for both cases. This said, if 2 or more digits are the same, it will not work (for a 3 digit number- 4 digit is a different matter):
a= y | z
b= z | x
c= x | y

Now we need to prove this for both cases, but I get:
M = x-x + y-y + z-z; so I MUST have made a mistake somewhere; I'm too dead to figure out where. Someone can correct me otherwise I'll check it again tomorrow morning.

Goodnight!
Michele

 

[mergli]

OK so now I'm more confused than I was before

 

[Spinal]

OK back... I think my mistake is that I did not account for all three numbers being the same (x=y=z) and for no two numbers being a 0 (x=y=0 || x=z=0 || y=z=0). But, at this point, I can't fingure it out, anyone care to help?
Michele