

kookastar
Honorable
Legendary Hero

posted August 10, 2014 02:17 PM 


Here's one I came across the other day...
What is the largest 3 digit number with the property that the number is equal to the sum of its hundreds digit, the square of its 10s digits, and the cube of its units digit?
no calculators writing computer programs to solve it!
____________
uhuh


Hayven
Famous Hero

posted August 10, 2014 04:35 PM 

Edited by Hayven at 16:37, 10 Aug 2014.

Difficult one
x = 100a + 10b + c
x = a + b^2 + c^3
100=< x =< 999
a =< 9
b =< 9
c =< 9
All numbers are natural.
100a + 10b + c = a + b^2 + c^3 /a
99a + 10b + c = b^2 + c^3
99a + 10b + c = bb + ccc /(10b + c)
99a = b(b10) + cc(c1)
a =< 9
b =< 9
c =< 9
99a = {99,198,297,396,495,594,693,792,891} (the first number must be higher than 0)
b(b10) = {0,9,16,21,24,25,24,21,16,9} = {0,9,16,21,24,25}
cc(c1) = {0,4,18,48,100,180,294,588,648}
We see that 99a can equal neither 693, 792 nor 891 because b(b10) + cc(c1) can't be higher than 648
99a = {99, 198, 297, 396, 495, 594}
NOW
99 is unachievable (1001)
198 is unachievable, too (29496)
297, 396 and 495 are also unachievable (58825 gives 563)
594 is unachievable, either.
Seems like if the number you're talking about didn't exist... or maybe I made a mistake here?
____________


Corribus
Hero of Order
The Abyss Staring Back at You

posted August 10, 2014 06:24 PM 

Edited by Corribus at 18:33, 10 Aug 2014.

There are numbers that satisfy this. One of them is 135. (1 + 3^2 + 5^3 = 1 + 9 + 125 = 135).
The highest one I found was 598. (5 + 9^2 + 8^3 = 5 + 81 + 512 = 598)
I used a little logic to narrow it down and then a little trial and error to find the highest working value. Maybe it's possible there's one higher, though.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. Mitch Hedberg


friendofgunnar
Honorable
Legendary Hero
able to speed up time

posted August 10, 2014 06:37 PM 


There's actually 4 numbers that can satisfy this puzzle, Cor correctly identified the highest one. The easiest way to solve this is to just write a computer program  it took me about 3 minutes. I have no idea though how to arrive at the answer mathematically and will be mighty impressed if anybody can do so...
edit:
Perhaps this is the way you could do it, by using multivariable calculus to find the solution to this:
0 = b(b10) + c(c1)  99a


Corribus
Hero of Order
The Abyss Staring Back at You

posted August 10, 2014 07:03 PM 


I figured the last digit realistically had to between 5 and 9, because anything lower than 5 is going to struggle to give a three digit number over 100 without having very high first and second digits.
If x = 100a + 10b + c and x = a + b^2 + c^3, you can do a little rearranging to write:
c^3  c = 99a + 10b  b^2.
I started with c values of 9 and worked my way down. Then I put in b values (of which 10b  b^2 only returns five unique values for the 9 possibilities  6 if you count 0  so that simplifies things) and saw which ones were evenly divisible by 99. There are non for c = 9 but one for c = 8. So I figured that was the highest one.
Like I said, some logic to narrow it down and then some trial and error with actual numbers. I don't know if there is an analytical solution. I'd have to think more about that.
A more interesting problem I think is whether there is a maximum number, period, that satisfies the more general case of a ndigit number equaling the sum of each of its n digits taken to the nth power. It's just a hunch, but I suspect there might be a maximum number that satisfies this relationship. Probably a computer program could easily tell you whether it converges or whether there are an infinite number of answers.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. Mitch Hedberg


Hayven
Famous Hero

posted August 11, 2014 01:17 PM 

Edited by Hayven at 13:19, 11 Aug 2014.

It seems like if I've made a mistake in my thoughts but was close, though.
By the way.
When I was about 10 years old I couldn't go asleep. That's why I thought a bit about maths to make myself sleepy
Anyway, I thought how does change between a fraction (a/b) and (a+z/b+z) look like. I found this one:
a+z a z(ba)
___  _ = ______
b+z b b(b+z)
Examples:
3/5  1/3 = (1+2)/(3+2)  1/3 = [2*(31)]/ (3*5) = 4/15
17/45  10/38 = (10+7)/(38+7)  10/38 = [7*(3810)]/ (38*45) = (7*28)/ (38*45) = (7*14)/ (19*45) = 98/855
5/2  4/1 = (4+1)/(1+1)  4/1 = [1*(41)]/ (1*2) = (3)/2 = (3/2)
Indeed
3/5  1/3 = 9/15  5/15 = 4/15
17/45  10/38 = (17*38)/(45*38)  (10*45)/(38*45) = [(17*38)(10*45)]/(38*45) = (646450)/1710 = 196/1710 = 98/855
5/2  4/1 = 2.5  4 = 1.5 = (3/2)
Probably it's obvious but I decided to post it anyway
____________


friendofgunnar
Honorable
Legendary Hero
able to speed up time

posted August 15, 2014 05:55 AM 


re Kooka's problem
3 digits = 4 solutions (135 175 518 598)
4 digits = 2 solutions (1676 2427)
5 digits = 0 solutions
6 digits = 0 solutions
There does appear to be an upper limit.


AlexSpl
Responsible
Supreme Hero

posted August 15, 2014 04:46 PM 


Quote: 99a = b(b10) + cc(c1)
Here is the mistake, I suppose. It should be 99a = b(b  10) + c(c^2  1), or, in more intuitive way:
99a + b(10  b) = c(c  1)(c + 1)
Remember, a, b, c are all digits.


Corribus
Hero of Order
The Abyss Staring Back at You

posted August 16, 2014 05:38 AM 


friendofgunnar said: re Kooka's problem
3 digits = 4 solutions (135 175 518 598)
4 digits = 2 solutions (1676 2427)
5 digits = 0 solutions
6 digits = 0 solutions
There does appear to be an upper limit.
Well there you go. I'd like it to be shown analytically though.
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. Mitch Hedberg


AlexSpl
Responsible
Supreme Hero

posted August 16, 2014 05:54 AM 

Edited by AlexSpl at 06:19, 16 Aug 2014.

Quote: 99a + b(10  b) = c(c  1)(c + 1)
Here you can spot some kind of symmetry: if b is a solution then 10  b is also a solution. Then, the left side is always positive and greater or equal to 99 (as 1 <= a <= 9), so c cannot be 0, 1, 2, 3 or 4. Also, you can easily prove that the right side is always even, so both a and b should be even or odd at the same time.


Corribus
Hero of Order
The Abyss Staring Back at You

posted March 10, 2016 04:24 AM 

Edited by Corribus at 04:34, 10 Mar 2016.

*BUMP*
Haven't had one of these in a while.
Suppose your uncle likes to randomly give you $10. The probability of receiving $10 from your uncle on any given day is 3%, or roughly once a month. He gives you $10 today: Wednesday, March 9th, 2016. What is the most likely day (if any) for the next $10 gift from your uncle?
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. Mitch Hedberg


Elvin
Admirable
Omnipresent Hero
Rejuvenation process

posted March 10, 2016 11:39 AM 


Can you even pick a day? The percentage is still 3% regardless of the day you were given the money.
____________
Restoring the light, facing the dark.


JollyJoker
Honorable
Undefeatable Hero

posted March 10, 2016 12:57 PM 


Tomorrow. Always tomorrow.


master_learn
Legendary Hero
walking to the library

posted March 10, 2016 01:08 PM 


The most likely day is when he decides to give you a bigger sum of money(which will include the 10$).
____________
"They made a Heroes V? "OhforfSake


Corribus
Hero of Order
The Abyss Staring Back at You

posted March 10, 2016 01:41 PM 

Edited by Corribus at 13:43, 10 Mar 2016.

Nicely done, JJ. Indeed it is tomorrow! (Well, Thursday the 10th, which is now today. )
____________
I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. Mitch Hedberg


Elvin
Admirable
Omnipresent Hero
Rejuvenation process

posted March 10, 2016 01:54 PM 


I don't get it. How is the chance of tomorrow different from let's say in 10 days?
____________
Restoring the light, facing the dark.


JollyJoker
Honorable
Undefeatable Hero

posted March 10, 2016 01:56 PM 


For explanation, it was clear, that it was kind of a trick question because of the minimum of information given, especially with the minimum of PERTINENT information (the exact date and weekday isn't a useful info, probabilitywise). So asking for a day couldn't be a specific date, like the 21st of March. Likewise, no specific weekday, not even something along the line of "Easter" or "next birthday" (due to the missing additional info.
That left only the relative answer: with every day as an independent occurence having the same probability, while a chain of events as a whole would deliver no exact date either, both together deliver: the next day  tomorrow  has always the biggest probability.
Edit: "Tomorrow" as an answer is an unspecific date within a chain of days  it can be every day, going from day to day  a trick of the language, sotospeak.


Pawek_13
Supreme Hero
Maths, maths everywhere!

posted March 10, 2016 02:09 PM 



kiryu133
Hero of Order
Highly illogical

posted March 10, 2016 02:14 PM 


Pawek_13 said: probability trees.
*instantly gets it*
visualization helps a lot
____________
It is with a heavy heart that I must announce that the cis are at it again.


Elvin
Admirable
Omnipresent Hero
Rejuvenation process

posted March 10, 2016 02:29 PM 


Hah, nice!
____________
Restoring the light, facing the dark.


