28 Jun 22
@venda said"The equation 81-34=47 is true but the reverse equation 74--43 =18"
Thankfully, we seem to have gone off the 700 kick:-
The equation 81-34=47 is true but the reverse equation 74--43 =18 is not true
Find an example of such an equation using three 2 digit numbers and involving a subtraction which is true in both directions -or prove there isn't one.
Is that supposed to be only one minus [-] sign in front of the 43?
29 Jun 22
@venda saidThere is no pair that it works for. I can't stringently prove it beyond showing that it won't work for any combination of the second digits...
Thankfully, we seem to have gone off the 700 kick:-
The equation 81-34=47 is true but the reverse equation 74--43 =18 is not true
Find an example of such an equation using three 2 digit numbers and involving a subtraction which is true in both directions -or prove there isn't one.
30 Jun 22
I'm not skilled enough at math to prove that there is no solution.
So I wrote a program and brute forced it. There is no solution.
Just for giggles, I allowed '0' to be one of the digits. The only solutions found all have at least one '0' in them. This is invalid for the problem, because a 2-digit number ending in 0 will not remain that way once reversed.
But it does give me some confidence that the program is correct.
30 Jun 22
@bigdogg saidYou are correct my friend.I'll type out the proof(answer) as given in the paper.
I'm not skilled enough at math to prove that there is no solution.
So I wrote a program and brute forced it. There is no solution.
Just for giggles, I allowed '0' to be one of the digits. The only solutions found all have at least one '0' in them. This is invalid for the problem, because a 2-digit number ending in 0 will not remain that way once reversed.
But it does give me some confidence that the program is correct.
Hope you enjoyed the challenge.I wouldn't know where to start!
There isn't one.The equations can be written as 10a+b-10c-d=10e+f and 10f+e=10d+c-10b-a where the letters represent single digits.Subtracting the second equation from the first gives11(a+b-c-d)=9(e-f).This requires e-f to be divisible by 11which is impossible unless e=f then either a and b are both bigger than c and d or one of b or d is 0
- Joined
- 18 Jan 07
- Moves
- 12473
30 Jun 22
@venda saidOoh, I was so close and so far... I had it down to a multiple of 9 also being a multiple of 11, and never considered that, with these being digits, it would imply a number less than 10 being divisible by 11... Shame on me, really.
You are correct my friend.I'll type out the proof(answer) as given in the paper.
Hope you enjoyed the challenge.I wouldn't know where to start!
Hidden content removed
01 Jul 22
@venda saidha I had played around with a similar equation an d didn't see the
You are correct my friend.I'll type out the proof(answer) as given in the paper.
Hope you enjoyed the challenge.I wouldn't know where to start!
Hidden content removed
divisible by 11 part
@venda saidEven assuming each tumbler has the same amount of liquid, there could be different total amounts of each. It doesn't seem like there's enough information to solve.
At a drinks gathering everyone had one tumbler of scotch and soda of different strengths.I drank 15% of the scotch drunk and 13% of the soda drunk.How many of us were there
25 Aug 22
@bigdogg saidI thought so as well.
Even assuming each tumbler has the same amount of liquid, there could be different total amounts of each. It doesn't seem like there's enough information to solve.
I'll post the answer as in the paper.
I didn't really follow the logic but perhaps you will:-
7 people.I didn't drink the same percentage of scotch and of soda,so I must have drunk above the average amount of scotch,and less the average amount of soda.This can only be true if there were 7 people present
28 Aug 22
@venda saidHere's some ugly algebra that arrives at their answer.
I thought so as well.
I'll post the answer as in the paper.
I didn't really follow the logic but perhaps you will:-
Hidden content removed
Expressing the amounts of each in units of glasses [or people]:
My glass is:
.15 sc + .13 so = 1
Solve for sc:
sc = (1-.13so)/.15
Everyone else's glasses (n = total number of glasses/people):
.85 sc + .87 so = n - 1
Substitute the equation above:
(.85-.1105so)/.15 + .87so = n - 1
5 2/3 - .736667so + .87so + 1 = n
6 2/3 + 0.13333so = n
6 2/3 + 2/15 so = n
There can't be negative soda, so the 6 2/3 on the left means at least 7 people.
With n=7, soda = 2.5 and scotch = 4.5.
With n=8, soda = 10 and that overfills my glass, so 7 people it is.
28 Aug 22
@bigdogg saidWell done my friend.
Here's some ugly algebra that arrives at their answer.
Expressing the amounts of each in units of glasses [or people]:
My glass is:
.15 sc + .13 so = 1
Solve for sc:
sc = (1-.13so)/.15
Everyone else's glasses (n = total number of glasses/people):
.85 sc + .87 so = n - 1
Substitute the equation above:
(.85-.1105so)/.15 + .87so = n - 1
5 2/3 - .736667so + .87 ...[text shortened]... 7, soda = 2.5 and scotch = 4.5.
With n=8, soda = 10 and that overfills my glass, so 7 people it is.
20 Sep 22
Divide 69 cakes between 70 people so that everyone has an identical set of pieces,whilst making as few slices as possible.
I'm away for a few days so the answer is "hidden"
cut 35 cakes in half ,14 cakes into 5 pieces and 20 cakes into 7 pieces.Everyone has 4 pieces one half,one fifth and 2 sevenths