What do you think?

bobbym · 2009-12-23 14:19:20

Do you see the pattern that all the solutions have?

phrontister · 2009-12-23 15:35:20

I think so.

Let's say that:
y = the smallest answer x, and
z = 434782608695652173913

For any of the answers x, the next-highest answer is:

(x * 10^(the number of digits in (y * z))) + (y * z)

I don't think that wording's too clever - but I hope you know what I mean. I'll try to tidy it up with an edit if you don't beat me to it. Maybe using a subscript 'n' for x would do it, but I haven't worked out how to display that yet - or if that's the way to do it.

Last edited by phrontister (2009-12-23 16:27:44)

bobbym · 2009-12-23 16:26:09

Hi phrontister;

phrontister · 2009-12-23 16:38:02

Nice! That works perfectly.

It's the senior version of the junior version I used to find my first answer

Last edited by phrontister (2009-12-23 16:43:22)

bobbym · 2009-12-23 18:55:43

Hi phrontister;

Thats very clever, someone on another forum did exactly that.

phrontister · 2009-12-23 21:33:04

Hi Bobby,

...someone on another forum did exactly that.

Must be my twin brother - we think alikn (sic). I hadn't seen that site before - some interesting characters there!

Good question by the OP, too, asking for the problem to be solved without using a calculator!

No - I came up with my wonderful solution all on my own (must have been holding my calculator downside-up just at the right moment of inspiration to think of the inversion).

bobbym · 2009-12-23 23:50:35

Hi phrontister;

I hadn't seen that site before - some interesting characters there!

No one interesting over there.

phrontister · 2009-12-24 12:25:31

Hi Bobby,

No one interesting over there.

Yes...I'm losing interest. There seems to be only one genuine bogus imposter there - not very exciting.

bobbym · 2009-12-27 05:18:50

Hi;

Problem #8:

How high can you go? If you don't know the rules then go here:

http://nrich.maths.org/786

I've given you a starter puzzle, it sums to 66. You should be able to beat that,

bobbym · 2010-01-01 01:20:35

Hi;

Problem #9:

Try this one on for size, it is quite easy. Which is larger and why?

JaneFairfax · 2010-01-01 13:16:03

bobbym wrote:

Hi;
Try this one on for size, it is quite easy. Which is larger and why?

Last edited by JaneFairfax (2010-01-01 13:16:23)

bobbym · 2010-01-04 01:31:44

Hi Jane;

Yes, you are right. Well done!

phrontister · 2010-01-04 03:48:57

Hi Bobby,

76.

This one's just a variation of my pattern for the 6-number daisy that scored 46 in the Number Daisy and Proof? thread. I haven't tried any other options yet.

Last edited by phrontister (2010-01-04 19:48:22)

bobbym · 2010-01-05 15:23:19

Hi phrontister;

Good work! It is what I have.

phrontister · 2010-01-05 15:45:45

Hi Bobby,

I thought that basing the answer to this 7-number puzzle on the (currently) highest-scoring 6-number one would be a good place to start...but I have no way of proving maximums for either form.

I've tried several other options since, including a central 2, 3 and 4, but got nowhere.

Last edited by phrontister (2010-01-05 15:54:42)

bobbym · 2010-01-05 19:25:23

Hi phrontister;

but I have no way of proving maximums for either form.

I am in the same spot.

gibbo · 2010-01-06 21:17:06

Hi Bobby
to yield nines (problem 17?). Use multiples of 23. The clue is that the number is 21 digits long. It would have been more obvious if it had been 22 digits long. eg 1304347826086956521739

bobbym · 2010-01-07 00:01:25

Hi gibbo;

Welcome to the forum!

gibbo wrote:

to yield nines (problem 17?). Use multiples of 23.

Not all multiples of 23 will yield all nines.

The next number that yields all nines after 23 is:

23 000 000 000 000 000 000 00 23

Here is another problem:

This one is so, so.

An urn contains 15 balls. There are only 2 different colors the balls can have, red and orange. There are at least 3 of each color in the urn. Picking 3 balls without replacement. the probability that all of the picked ones are red is the same as the probability that exactly one of them is orange. How many of the 15 balls are red?

bobbym · 2010-01-11 11:12:36

Problem #10:

Two people pick six numbers from the set {1,2,3,4,5,...44} with replacement. What is the probability that they have the same six numbers?

Person A says the probability that they have the same six numbers is:

Person B says the probability that they have the same six numbers is:

The textbook says .0000102

Who is right?

bobbym · 2010-01-13 21:44:58

Problem #11:

In Clare's math class there are 2 times as many 11 th-grade students as there are 9 th and 10th grade students combined. Then there are 2 times as many 12th grade students as 9th grade students. The number of 11th grade students is 10 times the number of 12th grade students. Altogether there are 32 students in Clare's math class, How many are there of each?

Problem #12:

An urn contains 15 balls. There are only 2 different colors the balls can have, red and orange. There are at least 3 of each color in the urn. Picking 3 balls without replacement. the probability that all of the picked ones are red is the same as the probability that exactly one of them is orange. How many of the 15 balls are red?

phrontister · 2010-01-14 01:04:12

Hi Bobby,

Clare's maths class

Last edited by phrontister (2010-01-14 03:32:56)

bobbym · 2010-01-14 04:36:51

Very good, that is correct.

JaneFairfax · 2010-01-15 02:44:07

bobbym wrote:

An urn contains 15 balls. There are only 2 different colors the balls can have, red and orange. There are at least 3 of each color in the urn. Picking 3 balls without replacement. the probability that all of the picked ones are red is the same as the probability that exactly one of them is orange. How many of the 15 balls are red?

Suppose there are n red balls (). The probability of picking 3 red balls is . The probability of picking exactly one orange ball is . Hence, if they are equal, we have .

bobbym · 2010-01-15 03:44:57

Hi Jane;

So good to see you. Yes! There is no solution, n must be an integer.

This is a variant of a problem posed on another forum that was too easy. So I adjusted the numbers hoping to make it somewhat confusing. I can't point to the link because that page is not coming in right now. Anyway, because his problem was so easy it succumbed to a brute force attack. I believe the first one I tried was the answer. I don't like problems that the poser never even considers a simplistic method of solution. I posed mine to force the solver to do something other than trying one example.

Well done!

bobbym · 2010-01-15 04:20:10

Hi Jane;

Here is the link:

http://www.artofproblemsolving.com/Foru … p?t=323599

Too easy!!!

Since we are discussing pieces of the balls in the urn , you did make a slight arithmetic error. Do you see it?

Math Is Fun Forum

#26 2009-12-23 14:19:20

Re: What do you think?

#27 2009-12-23 15:35:20

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