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January 12

100 sum

How do I write 100 using all the digits from 1 to 9, not necessarily in their natural order, with only one written symbol which denotes an operation?

Also I wish to know how many ways are there to write statements which equals to 100 using all the digits in their natural order such as:

100=123-45-67=89

(pls list down the ways)

this is not homework, i saw this question on a maths magazine.Invisiblebug590 (talk) 10:51, 12 January 2008 (UTC)[reply]

9+8+7+6+5+4+3+2+1+\sum _{k=1}^{55}1

. Welp, —Preceding unsigned comment added by Damien Karras (talk • contribs) 15:41, 12 January 2008 (UTC)[reply]

That uses 1 twice, 5 three times, 10 operation symbols, and gives the value 1585. Algebraist 15:53, 12 January 2008 (UTC)[reply]

That's what they want you to think!!! 81.153.211.247 (talk) 17:02, 12 January 2008 (UTC)[reply]

81.153.211.247 changed

\sum _{k=1}^{55}k

to

\sum _{k=1}^{55}1

after the comment by Algebraist. PrimeHunter (talk) 19:12, 12 January 2008 (UTC)[reply]

It depends on the rules. I would say exponentiation written as a^b requires no "written symbol which denotes an operation", but I don't know whether 100 can be reached with exponentiation and a single written operation. If a is a sequence of digits then a_b can mean a interpreted as a base b number, for example 100₂ = 4. If this is allowed then there are many easy solutions. PrimeHunter (talk) 16:28, 12 January 2008 (UTC)[reply]

Oh nuts, and I thought I was so clever having found 321489base567. You beat me to it SpinningSpark 16:42, 12 January 2008 (UTC)[reply]

That, of course, is not a solution to the problem as stated which required the digits to be in there natural order. In fact, their are no solutions of the form abaseb. SpinningSpark 16:50, 12 January 2008 (UTC)[reply]

Many easy solutions PrimeHunter? I have only a small single digit (base ten) number for total solutions. SpinningSpark 17:18, 12 January 2008 (UTC)[reply]

I was referring to the first question where one written operation is allowed and any order of digits is allowed. 98+2_b gives 6! possible values of b. And there are other possible patterns. PrimeHunter (talk) 17:30, 12 January 2008 (UTC)[reply]

Ah, I see. I was counting the change of base as being the allowed operation. SpinningSpark 17:59, 12 January 2008 (UTC)[reply]

By the way, is there a prize in the magazine for solving this, and if you win it, will you share with Wikipedia? SpinningSpark 16:53, 12 January 2008 (UTC)[reply]

Raffle

Say we have a raffle with 100 tickets and 3 prizes. One and one ticket is sold until all the prizes are won. How many tickets can you expect to sell?

I can find the answer in R with Monte Carlo simulation:

> a<-replicate(1e6,max(sample(100,3)))
> mean(a)
[1] 75.76514

And the pdf:

> table(a)/length(a)
a
       3        4        5        6        7        8        9       10 
0.000009 0.000017 0.000031 0.000061 0.000095 0.000124 0.000173 0.000240 
      11       12       13       14       15       16       17       18 
0.000271 0.000341 0.000415 0.000490 0.000592 0.000621 0.000702 0.000852 
      19       20       21       22       23       24       25       26 
0.000935 0.001054 0.001223 0.001284 0.001379 0.001580 0.001800 0.001801 
      27       28       29       30       31       32       33       34 
0.002051 0.002104 0.002342 0.002460 0.002743 0.002983 0.003021 0.003263 
      35       36       37       38       39       40       41       42 
0.003510 0.003664 0.003782 0.004103 0.004300 0.004583 0.004832 0.005045 
      43       44       45       46       47       48       49       50 
0.005392 0.005691 0.005824 0.006133 0.006435 0.006764 0.007052 0.007013 
      51       52       53       54       55       56       57       58 
0.007601 0.007802 0.008294 0.008482 0.008876 0.009179 0.009520 0.009832 
      59       60       61       62       63       64       65       66 
0.010120 0.010691 0.010846 0.011362 0.011675 0.012062 0.012457 0.012694 
      67       68       69       70       71       72       73       74 
0.013018 0.013750 0.014184 0.014458 0.014838 0.015364 0.015773 0.016187 
      75       76       77       78       79       80       81       82 
0.016798 0.017136 0.017520 0.018039 0.018802 0.019151 0.019553 0.020083 
      83       84       85       86       87       88       89       90 
0.020735 0.021073 0.021243 0.022291 0.022502 0.023258 0.023667 0.024278 
      91       92       93       94       95       96       97       98 
0.024394 0.025330 0.025982 0.026277 0.027101 0.027910 0.028286 0.028522 
      99      100 
0.029578 0.030246

It would be fun to have the exact values, and a formula for m tickets and n prizes.

Any suggestions?

--Δεζηθ (talk) 14:45, 12 January 2008 (UTC)[reply]

Working from the back end, so to speak, the probability that the 100th ticket is NOT a prize is;

$P_{1}={\frac {97}{100}}$

The probability that there is not a prize in the last two tickets;

$P_{2}={\frac {97}{100}}{\frac {96}{99}}$

and generally for r tickets from the end;

$P_{r}={\frac {97}{100}}{\frac {96}{99}}{\frac {95}{98}}...={\frac {(m-n)!}{(m-n-r)!}}{\frac {(m-r)!}{(m)!}}$

The expected number of tickets is the value of r making P_r closest to 0.5
SpinningSpark 17:49, 12 January 2008 (UTC)[reply]

A better strategy is to not announce the prizes till you have sold all the tickets. SpinningSpark 17:52, 12 January 2008 (UTC)[reply]

What's the probability of ending the raffle with the rth ticket's sell? In other words, what are the chances of giving away the third (or mth) prize with the rth ticket?

This is clearly the product of two probabilities: The probability of having given already m-1 prizes with the sell of r-1 tickets, times the probability of selling the last lucky ticket out of the n-(r-1) remanining numbers. Hence,

Q(r)=P(n,m,r-1,m-1)\times {\frac {1}{n-(r-1)}}

...where

P(n,m,r-1,n-1)

is the probability (under the hypergeometric model) of taking m-1 out of m marked items, from a total reference of n items, with r-1 extractions.

A simple expression for this product is given by

$Q(r)={\frac {m(n-m)!(r-1)!}{n!(r-m)!}}.$

Letting n=100 and m=3, values for the predicted chances for the OP's original inquiry can be computed. I must say that these values astonishingly match those estimated by the original poster; I praise the OP's simulation's performance.

The expected number of tickets sold can be computed with

{\tilde {r}}=\left[\sum _{r=m}^{n}r\times {\frac {(m(n-m)!(r-1)!)}{n!(r-m)!}}\right].

It comes as no surprise that, for n=100 and m=3, this amounts to 75.75. Pallida Mors 20:45, 13 January 2008 (UTC)[reply]

By the way, my silicon friend tells me that the expected number of tickets can be simplified into

{\tilde {r}}={\frac {m(n+1)}{m+1}}.

Pallida Mors 22:55, 13 January 2008 (UTC)[reply]

Dividing a slice of cake across the radius to end up with two equal volumes.

Your collective kind indulgence sought in solving this question that's been bugging me- and everyone I've bored with it.

Imagine a piece of cake, a segment, of radius R and angle at the apex (a). There is icing along the sector of circumference C. The fact that I like this icing, but my friend doesn't, causes us to agree to cut the cake on a line perpendicular to R at half a. We wish to each have equal volumes of cake sponge- The cake is of even height. The icing cannot be scraped off, nor can the cake be cut at half the height (above the plate). These sneaky suggestions have already been put forward! If we cut the cake half way between the apex and the icing, then clearly I get too much cake in my (almost) trapezius and my friend gets too little in their triangle. How far up or down the R, at half a, should we cut the cake to ensure equal amounts of cake? The icing is neglible (though tasty). —Preceding unsigned comment added by Geoffgraves (talk • contribs) 21:06, 12 January 2008 (UTC)[reply]

Well, I'd do it by integration. say we cut the cake down the middle at a/2, then by symmetry the area of cake (and hence volume) at the midpoint on one side is the midpoint of the whole thing. Integrate over this. I'm thinking it and will get back to you. Mmmm.... cake.... - mattbuck 21:16, 12 January 2008 (UTC)[reply]

OK, working with half the slice, we find that,

I=-R\int \limits _{0}^{\arccos(1-\cos {\frac {a}{2}})}\sin ^{2}ydy+{\dfrac {R^{2}}{4}}\sin a.

Use this to find the midpoint. - mattbuck 21:48, 12 January 2008 (UTC)[reply]

I don't think you need calculus for this. If a is small then one of the two pieces will be an isosceles triangle whose area you can work out with high school trigonometry. The area of the full slice is also easy to work out. Set the former equal to half the latter and solve for the position of the cut. I get

h=R{\sqrt {\frac {a}{4\tan(a/2)}}}

, where h is the distance from the apex to the plane of the cut, and a is in radians. -- BenRG (talk) 07:46, 13 January 2008 (UTC)[reply]

I also get that formula, and get

108.604^{\circ }

as the maximum a for which a straight cut can separate the slice into icingless and iced portions of equal area. --Tardis (talk) 17:38, 14 January 2008 (UTC)[reply]