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 (2009-12-24 15:27:44)

Nice! That works perfectly.

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

Last edited by phrontister (2009-12-24 15:43:22)

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).

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.

Hi;

Problem #9:

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

Hi Jane;

Yes, you are right. Well done!

Hi phrontister;

Good work! It is what I have.

Hi phrontister;

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

I am in the same spot.

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?

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?

Very good, that is correct.

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!