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numerical challenge

R

rickman

Jan 1, 1970
0
Yup. It's my home page. The sticky note manages my life.

That is scary. There are a handful of apps that I rely on, such as
Eudora for email. I had to use Microsoft Outlook at a job once and
hated it. More than hated it, I found it very disorganized and
disfunctional. I would be severely impacted if I could no longer use
Eudora for my email. Using anything web based from an outfit like
Google is uncertain at best. I think it is not so much *if* they change
it, but *when*. If you rely on this, you might want to switch preemptively.

Last time I checked, they were killing it *and* still pushing people
to develop new apps for it.

Yeah, well, that's Google!

I think the problem with iGoogle is that it doesn't include ads.

Evil.

But then you have eyes wide open, no?

Rick
 
N

Nobody

Jan 1, 1970
0
In modulo arithmetic, I thought the modulo had to be a prime number.

It depends upon what you mean by "modulo arithmetic". If the modulus is a
prime number, then you have a field, i.e. division is defined.

For all x and y (with y non-zero), there exists z such that x = y * z,
i.e. z = x / y is always defined. If the modulus isn't a prime number,
then this isn't the case; e.g. 1/2 modulo 2 isn't defined: there is no z
such that 2*z = 1 modulo 2.

But if you just want the result of additions "modulo N" (multiplication
being repeated addition and exponentiation being repeated multiplication),
then all intermediate calculations can be performed modulo N regardless of
whether or not N is prime.

If you calculate in base N (or a base R such that N=R^K for some integer
K), then remember that the carry always propagates from right to left. Any
given digit in the operands can affect that digit in the result, and the
digits to the left of it, but will never affect the digits to the right of
it. So the last K digits of the result will only be affected by the last K
digits of the operands. This holds for addition, multiplication, and for
the base operand of exponentiation, but not for division or the exponent.

This is why, when generating pseudo-random numbers using a linear
congruential generator, you keep the most significant bits and not the
least significant bits (the low bits have short periods as the last N bits
of the output are only affected by the last N bits of the input).
Is there anything special about 171 and 172?

There's nothing particularly special about 172, but Sylvia noted that the
fact 171=900-9^3 can be used to simplify the calculation.
 
That's a bunch of rigamarole with little to do with a working solution. The crux of the calculation using modular arithmetic is:

((a mod c) x (b mod c)) mod c = (ab) mod c , which is little more than a mental calcilation.

They really put people through the wringer for front desk clerk jobs these days...
 
S

Sylvia Else

Jan 1, 1970
0
That's a bunch of rigamarole with little to do with a working solution. The crux of the calculation using modular arithmetic is:

((a mod c) x (b mod c)) mod c = (ab) mod c , which is little more than a mental calcilation.

They really put people through the wringer for front desk clerk jobs these days...

Nobody's response seemed reasonable to me in the context of RichD's
question about modulo arithmetic.

Sylvia.
 
Nobody's response seemed reasonable to me in the context of RichD's

question about modulo arithmetic.



Sylvia.

Well then you have something over the people who have made a career in cryptography, since the method of choice IS repeated squaring in modular arithmetic- at least according to a CRC handbook that claims to survey all the background algebra used therein.
 
S

Sylvia Else

Jan 1, 1970
0
Well then you have something over the people who have made a career in cryptography, since the method of choice IS repeated squaring in modular arithmetic- at least according to a CRC handbook that claims to survey all the background algebra used therein.
Such people probably spend little of their time performing the
calculations manually, but are rather more concerned with finding the
most efficient algorithm that can be applied to the general case by a
computer.

Such people are also rarely interested in performing calculations modulo 10.

Just because an algorithm is the best solution for the general case on a
computer does make it the best solution for a particular case when being
calculated by a human modulo 10.

Sylvia.
 
Such people probably spend little of their time performing the

calculations manually, but are rather more concerned with finding the

most efficient algorithm that can be applied to the general case by a

computer.



Such people are also rarely interested in performing calculations modulo 10.



Just because an algorithm is the best solution for the general case on a

computer does make it the best solution for a particular case when being

calculated by a human modulo 10.



Sylvia.

As yet you have not found an efficiency improvement hinging of the numbers 171 and 172, probably because they were just random selections of the Taiwanese puzzle nut who posted the problem. He has a stake in refusing to believe the solution is other than straightforward calculation because he's addicted to the "aha" moment, well he's just not going to get that this time.
 
S

Sylvia Else

Jan 1, 1970
0
As yet you have not found an efficiency improvement hinging of the numbers 171 and 172, probably because they were just random selections of the Taiwanese puzzle nut who posted the problem. He has a stake in refusing to believe the solution is other than straightforward calculation because he's addicted to the "aha" moment, well he's just not going to get that this time.

The squaring approach appears to require 8 modulo 1000 multiplications.

Using the (170 + 1)^172 approach requires 4 modulo 1000 multiplications,
a modulo 1000 addition, a simple division by 2 and a simple addition.
Further, some of the factors in the modulo 1000 multipications are
multiples of 10 or 100, making them easier to handle.

How was the nationality of the person setting the puzzle a consideration?

Sylvia.
 
Using the (170 + 1)^172 approach requires 4 modulo 1000 multiplications,

a modulo 1000 addition, a simple division by 2 and a simple addition.

Further, some of the factors in the modulo 1000 multipications are

multiples of 10 or 100, making them easier to handle.

That is after some observations on the operand and other short cuts you took with the binomial coefficients. And it's supposed to be supperior to banging out 5x squares? Maybe you just have lots of time to kill.
How was the nationality of the person setting the puzzle a consideration?

It's as much a consideration as the blogger thought it was to include the information in his bio. If he said he was crossdresser living in Toledo , Ohio then fill in the blanks:"...because they were just random selections of the crossdressing Toledo puzzle nut..." Get it?
 
S

Sylvia Else

Jan 1, 1970
0
That is after some observations on the operand and other short cuts you took with the binomial coefficients. And it's supposed to be supperior to banging out 5x squares? Maybe you just have lots of time to kill.

I think that if someone reads that, and then reads your earlier comment
about people with careers in cryptography, they'd feel that you are
desperately trying to salvage an untenable position.
It's as much a consideration as the blogger thought it was to include the information in his bio. If he said he was crossdresser living in Toledo , Ohio then fill in the blanks:"...because they were just random selections of the crossdressing Toledo puzzle nut..." Get it?

So, not actually specified in the puzzle. Again, I ask, why was it a
consideration?

Sylvia.
 
I think that if someone reads that, and then reads your earlier comment

about people with careers in cryptography, they'd feel that you are

desperately trying to salvage an untenable position.

It happens to be a fact, the method of repeated squares is the most straight forward methodology for computing large powers.
So, not actually specified in the puzzle. Again, I ask, why was it a

consideration?

http://en.wikipedia.org/wiki/List_of_International_Mathematical_Olympiads

Taiwan has never won. The blogger specializes in IMO preparation, that is his slant, and the puzzles in his blog reflect that.
 
They still have three more problems with just hours remaining to deadline.

1) how many least significant zeroes in decimal expansion of 600! (factorial)

2) how many solutions to sin (theta)= -1/3 in interval [0, 25pi]

3) sum of all positive integers less than or equal to 61 divisible by 3 or 5

These are just high school problems for ages 12-18
 
On Tuesday, October 30, 2012 12:10:43 AM UTC-4, Sylvia Else wrote:

The three remaining problems are as trivial as the simple arithmetic exponent problem that you and others tried to blow up out of all proportion, but apparently not trivial enough for someone who thinks noting 171=170 + 1 is insight and then follows with a bunch of unrelated hot air ...
 
S

Sylvia Else

Jan 1, 1970
0
On Tuesday, October 30, 2012 12:10:43 AM UTC-4, Sylvia Else wrote:

The three remaining problems are as trivial as the simple arithmetic exponent problem that you and others tried to blow up out of all proportion, but apparently not trivial enough for someone who thinks noting 171=170 + 1 is insight and then follows with a bunch of unrelated hot air ...

Oh my - three replies from you to my one. Me thinks I've rattled
someone's cage.

Sylvia.
 
R

rickman

Jan 1, 1970
0
Nobody's response seemed reasonable to me in the context of RichD's
question about modulo arithmetic.

Sylvia.

I think Nobody should pick a better name. This had me going for a
moment. Reminds me of that story about Everybody, Somebody, Anybody and
Nobody...

Rick
 
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