Solve Enjoyable Enigmas with Mr.E's Challenge

[Enigman]

128V batteries are used in some laptops...but you wouldn't find them in hardware stores(at least not those which the enigma's talking about.) and I doubt that there are 1V batteries similar to laptop batteries...

If collinsmark hadn't mention home improvement shops USB flash drives would have worked as they would be found in computer hardware shops...don't know if they would fit the prices mentioned though.

 

[consciousness]

Pretty sure I got it. Don't read this if you don't want to spoil it.

Spoiler

They are the digits of the home addresses stuck on walls/front door, usually made of metal. Example- 16 refers to the digits 1 and 6 (equally priced).

BTW there is another answer to the bored corporate man Enigma!

 

[collinsmark]

Pretty sure I got it. Don't read this if you don't want to spoil it.

Spoiler

They are the digits of the home addresses stuck on walls/front door, usually made of metal. Example- 16 refers to the digits 1 and 6 (equally priced).

Correct.

Spoiler

http://www.homedepot.com/p/The-Hillman-Group-Distinctions-4-in-Brushed-Nickel-Flush-Mount-House-Number-1-843321/202982617#.UlbSFBBl09M

 

[zoobyshoe]

The Three Stooges, Moe, Larry, Curly, and Shemp, decide to have a custard pie party. Each is to bring however many custard pies they can manage to make on short notice. Shemp says he has no time at all to make pies, but will reimburse the others for his share if they make extra.

Moe arrives with a certain number of pies. Then Larry shows up, and he has brought one more pie than Moe. Curly enters, and it turns out he has brought one more pie than Larry.

Shemp is discovered sleeping under the carpet. He's been there the whole time, which is funny if you've seen the whole episode, but has no bearing on the enigma at hand. He says the pies all look tasty and they should divide them up so everyone gets the same number of pies. Then he will pay a dollar for each pie he gets. Everyone agrees there are a dozen pies, and they all agree a dollar is a fair price. Except Moe. He thinks his pies are much better than the others and should be worth more.

Passions escalate until the inevitable pie storm occurs. For some reason, a society matron enters the room and gets caught in the thick of it. She gets hit in the face with a pie, and starts throwing pies herself. If you keep track of all the pie-to-face impacts, the total is, strangely, more than a dozen. That has nothing to do with this enigma, though.

The question actually is: if the pie fight had not happened, how much would Shemp owe and to whom?

 

[zoobyshoe]

If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but Richard is 10, how much is Jennifer by the same logic?

 

[Enigman]

Spoiler

if everyone except Shemp eats only their own pies then it should be Larry- $1 and Curly- $2.

That's not a baker's dozen by any chance is it?

 

[zoobyshoe]

OH
OH
OH
OH
----
NO!

This is an addition. The letters represent numbers. One digit per letter. The same letter always represents the same number. Replace the letters with numbers to make the addition correct.

 

[Enigman]

oh
oh
oh
oh
----
no!

This is an addition. The letters represent numbers. One digit per letter. The same letter always represents the same number. Replace the letters with numbers to make the addition correct.

Spoiler

h=3 o=2 n=9

 

[zoobyshoe]

Billy had a coin purse with fifty coins, totaling exactly $1.00. Unfortunately, while counting them, he dropped one coin behind the radiator. What is the probability it was a penny?

For furriners:

$1.00

= 100 pennies
= 2 half dollars
= 4 quarters
= 10 dimes
= 20 nickels

 

[Office_Shredder]

Pennies:

Spoiler

He has a quarter, two dimes, two nickels and 45 pennies, so the probability that the coin he dropped a penny is 90%. Is there a way to solve this beyond try a bunch of coin combinations and come up with one that works?

 

[zoobyshoe]

Enigman is correct on the two he answered. Office Shredder, however, is not correct.

 

[lendav_rott]

The probability it's a penny is real, because there is no other way of making a dollar with 50 coins.
He had no 50 cent coins, that's for sure also the number of pennies has to be a x5 or a x0

Spoiler

He could have 40 pennies, 8 nickels and 2 dimes - 4/5 = 80%

 

[zoobyshoe]

The probability it's a penny is real, because there is no other way of making a dollar with 50 coins.
He had no 50 cent coins, that's for sure also the number of pennies has to be a x5 or a x0

Spoiler

He could have 40 pennies, 8 nickels and 2 dimes - 4/5 = 80%

This is a difficult problem. I got this out of a MENSA puzzle book. Only 30% of MENSA members who took the test got this one right.

Your answer is, unfortunately, incorrect.

 

[inotyce]

P(p=5)=1/10
P(p=10)=1/5
P(p=15)=3/5
P(p=20)=2/5
P(p=25)=1/2
...
P(p=45)=9/10

 

[Enigman]

Billy had a coin purse with fifty coins, totaling exactly $1.00. Unfortunately, while counting them, he dropped one coin behind the radiator. What is the probability it was a penny?

For furriners:

$1.00

= 100 pennies
= 2 half dollars
= 4 quarters
= 10 dimes
= 20 nickels

(EDIT1-And cross posted with inotyce now...)
a+b+c+d+e=50
0.01a+0.5b+0.25c+0.1d+0.05e=1
a,b,c,d,e belong to N
to find: a/50
There can be more than one solution for two equations.
a has to have a units digit of either 5 or 0(relying only on trial and error now)
a=45,c=1, d=2, e=2;

a=30, e=20
ed2-ARGGGHHHH! who cares? inotyce just got it...add'em up ino and you got the answer...
And put it in spoilers!

ed3-Okay just making sure the combinations are possible...
a=40,e=8,d=2;
a=35,nada don't get anything...
a=25,not here too...
a=20,...brain jam...
too much effort... for now I'm going with 1/3(45/50+30/50+40/50)
125/150=5/6...

 

[lendav_rott]

This is a difficult problem. I got this out of a MENSA puzzle book. Only 30% of MENSA members who took the test got this one right.

Your answer is, unfortunately, incorrect.

The problem is poorly worded, I can't see any "hidden text" referring to something elusive. If there are different combinations of 50 coins to make 1 dollar then they can't ask what the probability of the lost coin being a penny is. This is not math anymore, this is linguistics 1o1 seems like.

 

[Enigman]

The problem is poorly worded, I can't see any "hidden text" referring to something elusive. If there are different combinations of 50 coins to make 1 dollar then they can't ask what the probability of the lost coin being a penny is. This is not math anymore, this is linguistics 1o1 seems like.

On the contrary, I found it quite cleverly worded. Given the information we have been given we can't assume a specific combination of coins to be present and hence we have to account for all possible solutions- it was this particular thing according to me that the question was aimed at...after this its only simple logic and some trial and error or as a cryptologist would say brute force attack.
After all this is a problem for people in the MENSA, there is bound to be some clever deception in the wording...
-that is of course assuming my and inotyce's logic hold...

 

[zoobyshoe]

I got the same answer as Office Shredder. The book says this is wrong, due to the fact there is more than one combination of coins possible.

If there are different combinations of 50 coins to make 1 dollar then they can't ask what the probability of the lost coin being a penny is.

The authors of the book obviously believe you can. The answer they give as correct is the average of the probabilities of all the possible combinations.

 

[collinsmark]

The problem is poorly worded, I can't see any "hidden text" referring to something elusive. If there are different combinations of 50 coins to make 1 dollar then they can't ask what the probability of the lost coin being a penny is. This is not math anymore, this is linguistics 1o1 seems like.

I think the wording okay. It's this type of logic that is necessary to explore (in terms of information theory) when designing/analyzing digital communication systems. Digital communication systems involve things like your cell phone and WiFi.

One has to do things like calculate the probability of bit error without knowing a priori what the original bit pattern was or even without a priori knowledge of which bit (or bits) in the sequence is in error.

If the receiving unit already knew what the original bit pattern was, there wouldn't have been any need to transmit it over the communication channel in the first place.

[Edit: And by the way, not all permutations of the original bit patterns are possible due to parity bits, cyclic redundancy code (CRC) bits, or other bits added for error correction/detection. And any of those bits might be in error too. The point is the receiver knows a little a priori information about the original bit pattern, just not everything.]

 

[Enigman]

I got the same answer as Office Shredder. The book says this is wrong, due to the fact there is more than one combination of coins possible.



The authors of the book obviously believe you can. The answer they give as correct is the average of the probabilities of all the possible combinations.

Was I correct? Or was inotyce?

 

[lisab]

Maybe the space behind the radiator *only* big enough to hold a penny. A dime would fall through and the other coins are too big.

Therefore the chance of the coin being a penny is 100%.

:tongue2:

 

[collinsmark]

Billy had a coin purse with fifty coins, totaling exactly $1.00. Unfortunately, while counting them, he dropped one coin behind the radiator. What is the probability it was a penny?

For furriners:

$1.00

= 100 pennies
= 2 half dollars
= 4 quarters
= 10 dimes
= 20 nickels

Spoiler

I'm going to guess that the probability is 0.85 (or if you wish, "85%").

There are exactly two permutations that are possible.
(a) 2 dimes, 8 nickels, and 40 pennies.
(b) 1 quarter, 2 dimes, 2 nickels, 45 pennies.

No other combinations are possible.

Now we need to make the first assumption about a priori probabilities: it is equally likely for pennies to fall behind the radiator than it is for coins of other denominations. In other words, we neglect the size/weight characteristics of the coins that might otherwise keep quarters from falling behind the radiator while pennies easily drop through. If we were given other information about these probabilities we would have to modify everything following. But for now we assume that it is just as easy for a penny to fall behind the radiator as it is for any other type of coin. I just want to point out that this is an assumption I'm making.

We have the conditional probabilities, that

If (a), the probability that a penny fell behind the radiator is 0.8.
If (b), the probability that a penny fell behind the radiator is 0.9.

Next we need to make a second assumption about the a priori probabilities: (a) is equally likely as (b). This is an assumption I'm making. If we found that the Billy worked in the 25 cent gumball machine industry and had a high propensity to carry around quarters (more than other types of coins), it would change the probabilities from here on out. If there was shortage of nickels in circulation, it would affect the probabilities from here on out. There is a rarity of half-dollars for example, but fortunately a configuration involving a half-dollar is not possible. So I just want to point out that my "(a) is equally as probable as (b)" is merely an assumption I'm making here. If more information was given about the relative distribution of coins in circulation -- particularly applied to Billy's habits -- one could modify everything from here after.

Since (a) is equally probable as (b), final answer is 85 coins out of 100 coins, or 0.85.

(I'm sorry if somebody else guessed this first, but I honestly wasn't able to quite understand what some of the other guesses were.)

 

[zoobyshoe]

Was I correct? Or was inotyce?

Spoiler

I think you said there was a 5/6 (8333...%) probability of it being a penny, and I think Intoyce said there was a 90% chance. If that's correct, the book says those are both wrong answers.

 

[zoobyshoe]

Maybe the space behind the radiator *only* big enough to hold a penny. A dime would fall through and the other coins are too big.

Therefore the chance of the coin being a penny is 100%.

:tongue2:

There could also be a gremlin behind the radiator turning every coin that falls there into an old, dry skittle.

 

[Enigman]

Okay, I think collinsmark got it. I just checked my calculations and one of the cases was wrong...
(30 pennies and 20 nickels...have I mentioned I'm an idiot?)

 

[zoobyshoe]

Spoiler

There are exactly two permutations that are possible.
(a) 2 dimes, 8 nickels, and 40 pennies.
(b) 1 quarter, 2 dimes, 2 nickels, 45 pennies.

No other combinations are possible.

Spoiler

Just for my edification, how did you arrive at the fact there was more than one possible combination, and that two combinations was the upper limit? (I assumed there must only be one possible combination and, as soon as I found it by trial-and-error, stopped and thought I had the answer. That, to me, was the trick part of this puzzle, that they passively allow you to assume there's probably only one combination.)

 

[zoobyshoe]

No one's got an idea about the enigma in post 565?

 

[zoobyshoe]

I was at the swap meet a couple weeks ago and saw a book titled something like, "238 Migraine-Inducing Classic Russian Math Puzzles." I didn't buy it because I figured I, personally, had no chance of solving any of them. Now I wish I had bought it because most of the purely mathematical enigmas posted here are way too easy for a lot of you. Maybe that seller will show up again sometime with that book.

 

[Enigman]

Spoiler

Just for my edification, how did you arrive at the fact there was more than one possible combination, and that two combinations was the upper limit? (I assumed there must only be one possible combination and, as soon as I found it by trial-and-error, stopped and thought I had the answer. That, to me, was the trick part of this puzzle, that they passively allow you to assume there's probably only one combination.)

a+b+c+d+e=50
0.01a+0.5b+0.25c+0.1d+0.05e=1
a,b,c,d,e belong to N
to find: a/50
There can be more than one solution for two equations.
a has to have a units digit of either 5 or 0(relying only on trial and error now)

a-no. of pennies
and so on...
You have 2 equations and 5 variables so there's bound to be infinite solutions but then there's the constraint that numbers can only be positive integers less than 50. After that's just trial and error. When you have got till 40 pennies its obvious any more solutions can't be found(- I did a mistake and took the 30 penny case and messed up the sums 50 and 100...)

 

[collinsmark]

Spoiler

Just for my edification, how did you arrive at the fact there was more than one possible combination, and that two combinations was the upper limit? (I assumed there must only be one possible combination and, as soon as I found it by trial-and-error, stopped and thought I had the answer. That, to me, was the trick part of this puzzle, that they passively allow you to assume there's probably only one combination.)

Spoiler

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Coin_Behind_Radiator
{
    class Program
    {
        static void Main(string[] args)
        {
            int F, Q, D, N, P;
            double f = 0.5;
            double q = 0.25;
            double d = 0.1;
            double n = 0.05;
            double p = 0.01;

            for (F = 0; F < 51; F++)
            {
                for (Q = 0; Q < 51; Q++)
                {
                    for(D = 0; D < 51; D++)
                    {
                        for(N=0; N < 51; N++)
                        {
                            for (P = 0; P < 51; P++)
                            {
                                if ((f * F + q * Q + d * D + n * N + p * P == 1)
                                    && (F + Q + D + N + P == 50))
                                {
                                    // We got one. Print it out.
                                    Console.WriteLine(F + " Fifty-Cent pieces, "
                                        + Q + " Quarters, "
                                        + D + " Dimes, "
                                        + N + " Nickels "
                                        + P + " Pennies.");
                                }
                            }
                        }
                    }
                }
            }
            Console.WriteLine("Press any key to quit.");
            Console.ReadKey();
        }
    }
}

 

[Enigman]

If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but Richard is 10, how much is Jennifer by the same logic?

I tried letter-number exchange and syllable-number exchange...but I'm drawing a blank...
So, Baseball?

 

[zoobyshoe]

I tried letter-number exchange and syllable-number exchange...but I'm drawing a blank...
So, Baseball?

Not baseball. You're on the verge of a real "DOH!" moment, I think.

 

[zoobyshoe]

I'm also afraid I didn't understand either of the explanations.

 

[Enigman]

that's cheating Collinsmark you can't use programming...and to think I didn't even use a pen and paper, typed all my thoughts out ...but that probably lead me to my 30/20 mistake...to atone for that I am giving up my avatar.

 

[collinsmark]

that's cheating Collinsmark you can't use programming...and to think I didn't even use a pen and paper, typed all my thoughts out ...but that probably lead me to my 30/20 mistake...to atone for that I am giving up my avatar.

Spoiler

Yes, I thought about that.

I got as far as proving that it is impossible to have a combination involving a half-dollar. That simplifies the math a little. It's pretty easy to prove that just by thinking about it. But then I just wrote the computer program; I figured, "ah, screw it." Here is what I was planning to do though in lieu of coding:

Now we have,

Quarters: Q
Dimes: D
Nickels: N
Pennies: P

(0.25)Q + (0.1)D + (0.05)N + (0.01)P = 1
Q + D + N + P = 50.

Then find valid combinations with Q = 1. Then prove that it's not possible if Q = 2 (or greater).
This will get you the first possible solution.

Then simplify the math more,

(0.1)D + (0.05)N + (0.01)P = 1
D + N + P = 50.

From here it's just brute force to find the other solution with the two dimes. (Once can solve the two simultaneous equations about 10 times to see if anything comes up, once for each value of D. It's a hit if N and P are both non-negative integers. [Edit: and you can reduce the number of times realizing that if D < 5 it's impossible to get sufficient coins. And also impossible to find a solution if D = 5, since things can't work out. That simplifies things a little more.]

Yes, I could have done that, but I figured coding would be easier.