Brainteaster and Puzzle Sites for Children ?
Joseph Misko
in posted problems (at least privately). However, I find little material
for my children. Does anyone know if there are sites with problems of a
slower speed? I think it would be great to bend and stretch the minds of
the little ones. Our family has children in all age ranges.
Many thanks for any suggestions,
Joe
Rainer Rosenthal
"Joseph Misko" wrote
> Our family has children in all age ranges.
Here is a nice puzzle:
Aunt Helga is coming for a sunday visit. She
brings a lot of old coins, all of same value.
She opens the pocket, puts the coins on the
table ... and the three children Anne, Berta
and Christopher immediately grasp as much as
they can get.
"No, no", says aunt Helga. "Anne is the oldest
and she did receive so many gifts from me that
she should get only 1/6 of the coins. And 1/3
is for Berta, the second oldest. The youngest,
Christopher, shall get half of the coins."
I always wanted to collect 500 of these coins
but it's not that much. So here you are, enjoy!
Father's shouting: "Hey children, didn't you
understand? Quick, quick - give the money back,
so that aunt Helga can divide it as she likes."
But the mother disliked the harsh tone. She said
to Anne: "Dear Anne, just give me 1/6 of what you
took aside". Then to Berta: "Please give me 1/3
of the coins you have now". And from Christopher
she got half of his coins. She did put all this
money together and divided it into three heaps
of equal size and value. Each child got one of
these heaps back.
"This must be nonsense", father thought (and nearly
said). But - alas! - when he counted later in the
evening, the funny method of mother did indeed have
the result, which aunt Helga wanted in the beginning.
How many coins did aunt
Helga have in the pocket,
when she arrived?
This riddle I received from Ingmar Rubin, who maintains
a website full of math problems of every kind. You can
learn German as well, if you visit his site :-))
It's http://www.matheraetsel.de/
--
Rainer Rosenthal r.ros...@web.de _______________________
| _ | http://chephip.free.fr/ie/pb117_en.html |
| (_) | Given A, P and a circle. Find B, C on the |
| A P | circle with P on BC and area(ABC)=maximum. |
|__________|___(Ingmar Rubin in de.sci.mathematik) ________|
Ed Murphy
[spoiler space]
> Aunt Helga is coming for a sunday visit. She brings a lot of old coins,
> all of same value. She opens the pocket, puts the coins on the table ...
> and the three children Anne, Berta and Christopher immediately grasp as
> much as they can get.
A1 = ?
B1 = ?
C1 = ?
A1 + B1 + C1 = T
> "No, no", says aunt Helga. "Anne is the oldest and she did receive so
> many gifts from me that she should get only 1/6 of the coins. And 1/3 is
> for Berta, the second oldest. The youngest, Christopher, shall get half
> of the coins." I always wanted to collect 500 of these coins but it's
> not that much. So here you are, enjoy!
A2 = T/6
B2 = T/3
C2 = T/2
T < 500
> Father's shouting: "Hey children, didn't you understand? Quick, quick -
> give the money back, so that aunt Helga can divide it as she likes."
>
> But the mother disliked the harsh tone. She said to Anne: "Dear Anne,
> just give me 1/6 of what you took aside". Then to Berta: "Please give me
> 1/3 of the coins you have now". And from Christopher she got half of his
> coins. She did put all this money together and divided it into three
> heaps of equal size and value. Each child got one of these heaps back.
S = A1/6 + B1/3 + C1/2
A3 = 5*A1/6 + S/3
B3 = 2*B1/3 + S/3
C3 = C1/2 + S/3
> "This must be nonsense", father thought (and nearly said). But - alas! -
> when he counted later in the evening, the funny method of mother did
> indeed have the result, which aunt Helga wanted in the beginning.
A3 = A2
B3 = B2
C3 = C2
> How many coins did aunt
> Helga have in the pocket,
> when she arrived?
16*A1 + 2*B1 + 3*C1 = 3*T
A1 + 14*B1 + 3*C1 = 6*T
A1 + 2*B1 + 12*C1 = 9*T
A1 + B1 + C1 = T
13*A1 - B1 = 0
-5*A1 + 8*B1 - 3*C1 = 0
-8*A1 - 7*B1 + 3*C1 = 0
B1 = 13*A1
C1 = 33*A1
T = 47*A1
T is a multiple of 6 (otherwise A2 = T/6 is not an integer). 47 has
no prime factors in common with 6, so A1 must be a multiple of 6. The
only value of A1 that satisfies T < 500 is A1 = 6, so T = 282.
Joseph Misko
might satisfy what I need for the older ones.
Thanks,
Joe
Rainer Rosenthal
"Joseph Misko" wrote
> Well . . . I was thinking also of puzzles
> for younger children -- but this might
> satisfy what I need for the older ones.
I strongly believe that each puzzle for the
older ones will contain a lot of little problems
for the younger also.
In my example it is not trivial for very young
people to see that aunt Helga does divide *all*
of the money in the pocket when she says: one sixth
for Anne, one third for Berta and one half for
Christopher. So even the very young may participate!
During a longer trip by train I did solve the
puzzle (which was elegantly solved by Ed Murphy)
in a slightly different variation:
1/3 for Berta as before.
But 1/4 for Anne.
And the rest for Christopher.
Mother collects the money in exactly these proportions,
i.e. 1/3, 1/4 and 5/12 from the respective children.
Well, it took me quite a time to find my way thru all
the many stupid errors, which I always make when I do
a calculation of more than two lines ...
You will have to use an upper bound of 5000 coins or
so in order to have a single solution :-)
Kind regards,
Rainer Rosenthal
r.ros...@web.de