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numerical challenge

R

rickman

Jan 1, 1970
0
Even if the applicant couldn't solve this problem (even at home) because
he just wasn't well enough versed in the math or just didn't think of
the solution, that doesn't mean he can think in other terms nor does it
mean he wouldn't be good at problem solving directly related to the job.

I don't see this as a good interview question. But then I didn't get the
answer... but I did learn from it.

Rick

Sorry, that should be "that doesn't mean he ***can't*** think in other
terms".

Rick
 
J

Jamie

Jan 1, 1970
0
John said:
John Larkin wrote:

On Wed, 24 Oct 2012 16:44:06 -0400, Spehro Pefhany



On Wed, 24 Oct 2012 16:27:02 -0400, Roberto Waltman



Spehro Pefhany wrote:


On Wed, 24 Oct 2012 15:27:12 -0400, Roberto Waltman



John Larkin wrote:
...



This is my first-pass qualifier question.

+10V
|
|
C
+5V--------------B
E
|
1K
|
gnd

(drawn like a real schematic, on a whiteboard)

What is the base voltage? (one answer: 0.6!)
What is the emitter voltage?
What is the collector voltage?

What is the base current?
What is the emitter current?
What is the collector current?

Any other comments?

Germanium or silicon?
PNP or NPN? ;)

Roberto Waltman
[Running for cover...]

For a (slightly) more advanced test..



-10V
|
1K
|
C
-5V--------------B NPN
E
|
gnd


One more question: What is the pressure of the magic smoke inside the
transistor?

If it's a jellybean NPN, the E-B junction is going to be just fine
(typically they break down around 8-9V and are rated for at least 5V),
and the answers are, as JL says, going to be about the same as his
example except for the higher base current due to typically low
reverse beta. But it fits a different visual pattern, being an NPN
with E grounded-- it's probably not obvious that it's a
"collector-follower".


That's way overkill for a first question interview. If some kid
doesn't know about reverse beta, I wouldn't chuck him out right away.

I measured a BFT25 as having reverse beta about 4.

And each time some one comments on that, I still try to think of an
application where that would prosper? Maybe something in speed switching
. etc..
Kind of reminds me of ECL logic in a way.

Jamie


I have used a BFT25 to ground an AC-coupled analog signal, in a
gain-switching application. You need a lot of base current to clamp
the negative part of the swing, since reverse beta is low.
If you're in need for a low saturated switch and don't want to use mos
then I guess this would work..

But, beware of what you place at the emitter, I would think you need
to stay belong the zenering of the emitter..

Jamie
 
Right- there's nothing I can do about
it, there are no text settings whatsoever,
GG completely ignores its 'Quote Original'
button, and it puts all this double space
junk in here just by hitting 'Post Reply'.

This is my reply posted via Google Groups.
Note how I edited the double spaced junk that Google
quoted into something readable.

When you hit "Post Reply" button, you get a box
with double spaced quoted text to which you can
add your comments to the bottom. The double
spaced quoted text is easily editable. Just
trim off as much as possible, leaving only a
few references, and your posting will be more
readable.
And if I don't watch it, it will send replies
to the author too.

Yep. I've seen that happen.

If you are unable to make your posting readable,
I suggest you subscribe to a usenet news provider
(I use SuperNews/Giganews), and a decent reader,
such as Forte Agent.
 
When you hit "Post Reply" button, you get a box
with double spaced quoted text to which you can
add your comments to the bottom. The double
spaced quoted text is easily editable. Just
trim off as much as possible, leaving only a
few references, and your posting will be more
readable.

I know that- but it takes all day and drives me nuts- I'll just block delete it.
 
I know that- but it takes all day and drives
me nuts- I'll just block delete it.

Well, it didn't take me all day and I think I'm
still sane.

You don't offer much of a choice. It's either 607
lines of quoted junk, or a one-liner. Neither
extreme is worthwhile. May I suggest a compromise.
Inscribe your one-line comment, trim, but continue on
to explain why you think so, substantiate your
allegations, and backup your claims. You already
do much of that, but scattered over multiple messages
where it's difficult to find or read. With 300+
messages per day in this newsgroup, it's impossible
for me to read everything.
 
R

Robert Baer

Jan 1, 1970
0
William said:
817. 5 divides every fifth number. 4089/5 = 817 + 4/5
Incorrect answer; a lot more than that (remember, i said FACTORIAL,
which means 4089 * 4088 * 4087 * ... * 3 * 2 * 1).
 
R

Robert Baer

Jan 1, 1970
0
William said:
Wrong. Hint: 25! has 6 terminal zeros.
5 divides every fifth number. 4089/5 = 817 + 4/5
Anoher 5 divides every 25-th number. 4089/25 = 163 + 14/25
Yet another 5 divides every 125-th number. 4089/125 = 32 + 89/125
etc.

Number of zero's
= [4089/5] + [4089//25] + [4089/125] + [4089/625] + [4089//3125]
= 817 + 163 + 32 + 6 + 1 = 1019.
YES!!
Correct solution AND correct method.
 
R

Robert Baer

Jan 1, 1970
0
With that answer? Are you sure? ;-)


Been there. It's taken some convincing, at times, to get management to
understand that sometimes more people = less work. It's often too late to
improve a schedule, by *any* means.
Well, there IS a way...fire absolutely everybody and dump the company..
 
R

RichD

Jan 1, 1970
0
?

Perhaps not even then. All the offered solutions require knowledge of
congruences, which is part of basic number theory. Unless a person has
done a maths degree, or happens to have encountered it in passing (my
situation), they're unlikely to be sure that the required steps are
mathematically valid.

It's a firm that does financial modeling. They're recruiting people
to write code, for statistical analysis etc.; quants, not brokers.

It's a very good question for such a position. OF course, you want
someone unafraid of math, but better yet, it requires some original
thinking, a problem that wasn't on the exam.
 
G

George Herold

Jan 1, 1970
0
On Wednesday, October 24, 2012 7:53:38 PM UTC-4, Jeff Liebermann wrote:

Right- there's nothing I can do about it, there are no text settings whatsoever, GG completely ignores its 'Quote Original' button, and it puts all this double space junk in here just by hitting 'Post Reply'. And if I don'twatch it, it will send replies to the author too.

Hi Fred, I'm still using the old google groups. (Though google keeps
trying to swtich me over.. and swears that the old GG is going away
soon.) Can't you revert to the old GG? Perhaps only a temporary fix,
but....

George H.
 
I'm increasingly wondering about how it would have ended up as an
interview question. My best guess is the interviewer came across the
problem and assumed, without doing it themselves, that it was reasonably
simple - an intelligence test style question, rather than a maths question.

It took me a significant time to come up with an answer, and I would
almost certainly have not succeeded when subject to the pressure of an
interview - I function very badly under pressure :(.

Generally, other than to prove how smart the interviewer is, these sorts of
questions are asked to find out how a candidate goes about solving a problem.
The right answer isn't very important (it suggests he's seen it before ;-).
 
I saw this posed as an interview question:



(usual exponent notation)

What are the last 3 digits of 171 ^ 172?



Presumably, one is given pen and paper.



Is there a trick here? The second digit isn't too tough,

but the third is a lot of work.
 
R

RichD

Jan 1, 1970
0
What are the last 3 digits of 171 ^ 172?
Is there a trick here?

a = 171;  d = 1000
b = 172 = 128 + 32 + 8 + 4

Calculate:
a^2 mod d = 241
a^4 = (a^2)^2 mod d = 241^2 mod d = 81

a^8 = (a^4)^2 mod d
a^32 = (a^8)^4 = [(a^8)^2]^2 mod d
a^128 = (a^32)^4 = [(a^32)^2]^2 mod d

and use those results to calculate
a^172 = a^128 a^32 a^8 a^4 mod d.

Yes, that's probably what they expect, and would
get you the job. It's not too much work.

I did it by looking at periodicities in the second digit, as one
does the multiplications. But that became excessive, in
the third digit.

But I wonder if there are some obscure number theory short
cuts. Which might give you a leg up on other candidates,
though it's doubtful the interviewers expect that.
 
R

RichD

Jan 1, 1970
0
Don't know about a trick. After noting the (*) result below using a
caculator, I managed to get the right answer from scratch on a
whiteboard, but I doubt I'd have managed to do it in an interview
environment.

Anyway, since we're only looking at the last three digits, we can do
calculations mod 1000.

In modulo arithmetic, I thought the modulo had to be a prime number.
Also, note that 172 is 4 * 43.

Doing 171^2 by long multiplication, keeping only the last three digits
gives 241.
241^2 by long multiplication, keeping only the last three digits is 81.
So 171^4 = 81 mod 1000 (this is the * result) and 171^172 = 81^43 mod
1000 = 9^86 mod 1000.

Now 9 is 10 - 1, so we have 171^172 = (10 - 1)^86 mod 1000.

If we were to expand that using the binomial theorem, all the terms
would by multiples of 1000, and therefore unimportant, except the last
three.

The last three are

a)(86 * 85) / 2 * 10^2 * (-1)^84,
b) 86 * 10^1 * (-1)^85, and
c)(-1)^86.

a) can also be written 43 * 85 * 100, and since we're taking it modulo
1000, we need only multiply the 3 by the 5, giving 15, and keep only the
5, so a) is 500 mod 1000.

b) is -860

c) is 1

So the result is 500 - 860 + 1 = -359 mod 1000.
But that's negative, so add 1000 to make it positive, and the answer is 641.

Is there anything special about 171 and 172?
 
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