Kiddie problem.

[cheesehead]

Originally Posted by wblipp

Letting the 3 digit number be ABC and the base be "b",

and, further, the ratio between terms be "r",

we have

Ab^2+Bb+C=5(Bb+C)
Bb+C=5C

Ab^2+Bb+C = r(Bb+C)
Bb+C = rC

From which we can get
C=Ab^2/20

We know C<b. From this we get
Ab<20.

C = Ab^2 / (r(r-1))
Ab < r(r-1)

The gets a small number of possible values for A and b. The only cases with
A<b and C<b and C an integer is the base 10 solution.

Now I'll let someone else have the glory of figuring out what happens for r < 5 and r > 5.

[mfgoode]

QUOTE=cheesehead]I'm not baffled.

]The digit "0" is used as a placeholder after the decimal point. So?
Your puzzle rules mention taking away digits, but not adding/inserting them.
If we had a placeholder that was not a digit, we could use that ... but we don't.
Inserting the digit "0" as a placeholder is outside the rule requiring "take away that taking away the digit "2" from .25 produces .05 instead of .5
then you need to revise the puzzle's wording.
Remember, you said .0125 wouldn't do.]If that "0" isn't allowed, then neither is the one in .05[/spoiler

I’m one post behind but nevertheless I’ll have my say.

Thank you cheesehead for showing interest in my thread and boldly stating your convictions.
I am grateful to you for drawing my attention to matters I took for granted
In fundamental and basic arithmetic which were framed by the ancients so meticulously.

The simple phrase ‘taking away’ (or scratching out) is the point of contention which needs our attention. I’ve realised that when something is ‘taken away’a void is left, and tho’ it is nothing, still, a place is left vacant which cannot be ignored. It nevertheless has to be symbolised. In Maths it is the zero. In electronics it’s the ‘hole’

[Fine ... but "0" is not a sunya].

Yes it is! the Sanskrit ‘sunya’ is the same as our modern zero and IS the zero in any calculation. When a digit is taken from a similar digit the zero is left behind. This is the case whether the digit is to the left or right of the decimal point. If we move the decimal point to the right or left of a .
digit it is the operation of multiplying or dividing by ten resp.
In the problem this operation is not allowed and the decimal pt. remains and retains it original position i.e. it stays put.

[If you intend that taking away the digit "2" from .25 produces .05 instead of .5]
then you need to revise the puzzle's wording.]

I invite you to reword the problem provided you keep the decimal point
Fixed and immovable.

[Remember, you said .0125 wouldn't do. If that "0" isn't allowed, then neither is the one in .05[/QUOTE]


Also please read this simple problem and the answers right from the beginning. It says 3 digits and .0125 is 4 digits.Its not the case of adding or removing a zero but in keeping the 3 digit rule
Whereas I am willing to change the rules to suit your pattern of thought,
I am not willing to compromise or connive at the flouting of basic rules of Arithmetic.

It’s the question of trading the Danish Blue (2) with a cottage product (0)
But so far you are not willing.
Mally

[Wacky]

Originally Posted by mfgoode

It says 3 digits and .0125 is 4 digits.Its not the case of adding or removing a zero but in keeping the 3 digit rule

Mally,

I fear that I must agree with Cheesehead. If .0125 is 4 digits and .125 is 3 digits, then .025 is also 3 digits.

By common usage, "If you take away the 1st digit" (from a 3 digit number), then the remaining number must have only 2 digits. Thus the result would be ". 25" which is not the same as ".025".

I will even suggest that "125i" is not a number, but rather an expression. I do not consider it any more valid as "an answer" than is "125 X".

May I suggest that it is time to drop this pendantic discussion since it is not really proving anything other than the fact that the problem is not stated in such a rigorous manner as to exclude the misuse of language to contrive "solutions" which are outside the concepts which are acceptable to the "kiddies".

Richard

[mfgoode]

Originally Posted by Wacky

Mally,
May I suggest that it is time to drop this pendantic discussion since it is not really proving anything other than the fact that the problem is not stated in such a rigorous manner as to exclude the misuse of language to contrive "solutions" which are outside the concepts which are acceptable to the "kiddies".
Richard

I,ll take your advice and call this needless bickering off Richard.
We are all here for the pursuit of truth and thats our goal.

cheesehead: Thank you for generalising this problem further by introducing the ratio 'r'.
I must confess that after I entered into a labryinth of facts and figures and filled pages of calculations but I cannot go futher than your conclusion that
Ab < r(r-1) both in pure theory or trial and error.
I will be highly obliged if anyone could enlighten me further.

Wblipp: This is right up your street--- My question is: Can there be any other base ratio and value of A which which satisfies this type of problem? Or is there a proof that this is not possible?
Mally

[wblipp]

Originally Posted by mfgoode

My question is: Can there be any other base ratio and value of A which which satisfies this type of problem?

Let x, y, and z by any positive integers.

r=(x+y)^2*z+1
b=(x+y)*z*r
A=x
B=x*(x+y)*z
C=x*z*r

For example,

(x,y,z)=(1,1,1) gives the standard base 10 solution.

(x,y,z)=(3,2,2) gives 3.30.306 base 510 with r=51.

William

[mfgoode]

Thank you William for your elegant and compact method which is brilliant.
I am a bit confused as to how you get these equations starting from first principles.

As I understand it you have taken x,y,z, as presumably known positive integers and worked out 5 unknowns. Kindly give me the master equations on what leads to this conclusion which is very sound indeed.

2) In your second example (x,y,z,) = (3,2,2) is it that way the resulting number is written in base 510 as you have given?

3)From the formula C=Ab^2/(r(r-1) by trial I got the foll: values.
A=7, b=8, r=8 , C=8 and B=7.
Has C to be less than b=8 or can it be equal to it? If it can then how will the number be written in base 8 ?
Kindly bear with me as I have so far been only familiar with binary numbers.
Mally

[wblipp]

Originally Posted by mfgoode

2) In your second example (x,y,z,) = (3,2,2) is it that way the resulting number is written in base 510 as you have given?

I don't know of any standard, but borrowing that techniqe from IP addresses seems like a good method.

Originally Posted by mfgoode

Has C to be less than b=8 or can it be equal to it?

There is no digit "8" in base 8, just like there is no digit "10" in base 10.

Originally Posted by mfgoode

I am a bit confused as to how you get these equations starting from first principles.

We start with the two equations and the 3 inequalities A<b, B<b, C<b. We also have our one known solution with b=10 and r=5. From these, we have a solution for any values of A, b, r for which
C=Ab^2/(r*(r-1)) is an integer
B=Ab/r is an integer
A<b
A<r
AB<r(r-1)

From the equation for "B," we know r divides Ab. For the most general solution, we would need to let "p" be the gcd(b,r), and then introduce new variables so that r=p*q, b=p*s, A=q*t and make substitutions. But know from our example that there is at least one solution with gcd(b,r)=r. So let's focus on that class of solutions and see if it generalizes. Setting b=r*s, we can see that there is a solution for any values of A, r, s which satisfy
B=As
C=Ars^2/(r-1) is an integer
A<rs
A<r
As<(r-1)

From which we can see that (r-1) must divide As^2. There are many ways for this to happen, but we know from our example that there is at least one solution with s^2=t*(r-1).

Substitutiing, we have a solution for any A, r, t with
B=A*sqrt(t(r-1)) an integer
C=Art
A<r
A^2*t<r-1

We need t*(r-1) to be a square. There are many ways for this to happen, but we know from our example that there is at least one solution with (r-1)=t*v^2. Substituting, we have a solution for any values of A, t, v with
B=A*t*v
C=A*t*(t*v^2+1)
A<t*v^2+1
A<v

we now have B and C as integers, but we need to force A<v. We get this by setting
A=x
v=x+y
t=z.

[mfgoode]

[QUOTE=wblipp].
There is no digit "8" in base 8, just like there is no digit "10" in base10./UNQUOTE]

Thank you William for your very lucid and logical account in deriving and generalising the Kiddie problem to other bases and their meaning.
Your reasoning was so clear and masterful that I managed to understand it at first reading leaving a detailed study for later on

This answer to digit 8 stumped me completely.
I suppose it was so obvious that you left it out. What then was the 8th number and what about 9?

As I was very hesitant and reluctant to ask you more silly questions I dug out my note book on binary arithmetic and Boolean algebra. It did me a world of good to revise it thoroughly once and for all.
Hence the delay in my response

The man who replaced the concept of ‘fluxion’with ‘limit’ Jean de Rond d’Alembert in the 18th century expressed great wisdom when he proclaimed his motto:
“Go on, and Faith will come to you”. I was happy to do just that
Thank you,
Mally

[wblipp]

Originally Posted by mfgoode

What then was the 8th number and what about 9?

10 base 8 means "One 8 and zero 1's," which is the same as 8 base 10.

11 base 8 means "One 8 and one 1," which is the same as 9 base 10.

[xilman]

Originally Posted by wblipp

10 base 8 means "One 8 and zero 1's," which is the same as 8 base 10.

11 base 8 means "One 8 and one 1," which is the same as 9 base 10.

Originally Posted by Tom Lehrer

Base eight is just like base ten really, if you're missing two fingers.

Paul

[mfgoode]

Thanks William for the clarification. Now there is absolutely no doubt
Mally