 # Newspaper puzzle venda Posers and Puzzles 28 Feb '21 18:33
1. 24 Jun '22 08:09
Correct in every respect Pondy
2. 27 Jun '22 15:35
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.
3. 28 Jun '22 22:02
@venda said
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.
"The equation 81-34=47 is true but the reverse equation 74--43 =18"

Is that supposed to be only one minus [-] sign in front of the 43?
4. 29 Jun '22 11:471 edit
@bigdogg said
"The equation 81-34=47 is true but the reverse equation 74--43 =18"

Is that supposed to be only one minus [-] sign in front of the 43?
Yes
The equation must be in the format x-y =z
5. 29 Jun '22 15:06
@venda said
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.
There 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...
6. 30 Jun '22 00:28
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.
7. 30 Jun '22 13:23
@bigdogg said
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.
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!
There isn&#039;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
8. 30 Jun '22 18:36
@venda said
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
Ooh, 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.
9. 01 Jul '22 07:21
@venda said
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
ha I had played around with a similar equation an d didn't see the

divisible by 11 part
10. 24 Aug '22 09:14
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
11. 24 Aug '22 16:101 edit
@venda said
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
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.
12. 25 Aug '22 12:52
@bigdogg said
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 thought so as well.
I'll post the answer as in the paper.
I didn't really follow the logic but perhaps you will:-
7 people.I didn&#039;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
13. 28 Aug '22 00:48
@venda said
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
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 + .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.
14. 28 Aug '22 08:47
@bigdogg said
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.
Well done my friend.
15. 20 Sep '22 08:57
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

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