Puzzles with numbers and things

[davis_greatest]

Special congratulations to new entrant Monique in the Puzzles with numbers and things Hall of Frame:

Oliver (my pet orang-utan)
Gordon (my pet gorilla)
Charlie (my pet chimpanzee)
snookersfun
abextra
davis_greatest (Oliver's, Gordon's and Charlie's pet something)
Vidas
chasmmi
elvaago
robert602
Sarmu
The Statman
austrian_girl
austrian_girl's dad
Semih_Sayginer
Snooker Rocks!
Ginger_Freak
April Madness
steveb72
rambon
Microsoft Excel
dantuck_7
berolina
Parklife Ricky
oddyseus
Monique

Do we want another question?

[dantuck_7]

I'll let everyone else judge:

!
.---.---.----(1)
!(3) /
. . .
! /(2)
. . .
/

This is pretty much how it looked. I decided it would get too messy to draw the last line!

- Yep, it doesn't come out too well!

[rambon]

Do we want another question?

Yes please.

(Do I win a point?)

[davis_greatest]

I'll let everyone else judge:

!
.---.---.----(1)
!(3) /
. . .
! /(2)
. . .
/

This is pretty much how it looked. I decided it would get too messy to draw the last line!

- Yep, it doesn't come out too well!

That looks like the picture that Oliver, my pet orang-utan, drew with his right paw. And he's left handed.

[dantuck_7]

Anyother question would be good davis - along the same lines as round 144/146?

[snookersfun]

I'll put one up again:
Round 148:

A professor states to his assistant about the age of three persons A,B,C: "A+B+C is twice your age, A*B*C is 2450. But I have to mention that I am older than A, B and C." How old is the professor?

Answers by PM please!

[snookersfun]

followed by:
Round 149:

Find a number, containing all digits from 1-9 exactly once, so that each number containing the first n digits divides by n for each n (1-9).

(so 563417892 for example would not work, as 563 already doesn't divide by 3 without remainder)

Answers by PM please!

[snookersfun]

an update to the last two puzzles.
D_g has solved both of them , although the logic in this one here is maybe a bit confusing. (but I did find it like that!)

Round 148:

A professor states to his assistant about the age of three persons A,B,C: "A+B+C is twice your age, A*B*C is 2450. But I have to mention that I am older than A, B and C." How old is the professor?

Answers by PM please!

So, let's try to shed some light on it: It is supposedly one of these 'perfect logician' puzzles. The assistant does not need to know the respective ages of A, B and C to solve this, but does need the 'I have to mention...' part to decide on one of the options of (A+B+C).

Does that help anybody????

[abextra]

I'll put one up again:
Round 148:

A professor states to his assistant about the age of three persons A,B,C: "A+B+C is twice your age, A*B*C is 2450. But I have to mention that I am older than A, B and C." How old is the professor?

Answers by PM please!

I have few questions:

"A+B+C is twice your age" - is it assistant's age? Or professor's age?

"I am older than A, B and C" - does it mean older than A+B+C? Or older than A and older than B and older than C?

Are A, B and C all different?

[snookersfun]

[QUOTE=abextra] I have few questions:

"A+B+C is twice your age" - is it assistant's age? Or professor's age?

That is the assistants age. (Prof. talking to his assistant). So, basically the assistant should know his own age (hint!)"

I am older than A, B and C" - does it mean older than A+B+C? Or older than A and older than B and older than C? yes, the second option (not older than the sum of them)

Are A, B and C all different? could be, but not necessarily

[snookersfun]

Abextra has joined d_g in solving round 148 and 149, despite the logic.
Well done to the both of you!

If anybody else would like a go, he/she can post the answers on this thread now.

[snookersfun]

No customers- so here are the answers to close the last two rounds:

Round 148:

A professor states to his assistant about the age of three persons A,B,C: "A+B+C is twice your age, A*B*C is 2450. But I have to mention that I am older than A, B and C." How old is the professor?

I bid the professor is 50 years old.
(Assistant is 32, A, B and C are 5, 10 and 49 years old).

I'm not sure I understood the logic. There are many ways to get A*B*C=2450, but I've found two sets of A, B and C, which add up to the same sum (5+10+49=64 and 7+7+50=64). If the assistant is 32 years old, he can't find the ages of A, B and C only by the sum, he needs more information. Professor's statement that he's older than A, B and C has meaning only if he's 50 years old, if he was older than 50, he's words wouldn't help the assistant to decide the exact ages of these persons.

Round 149:

Find a number, containing all digits from 1-9 exactly once, so that each number containing the first n digits divides by n for each n (1-9).

Round 149 - 381654729 ... making sure it was divisible by 7 was the tricky bit!

[rambon]

Now maybe I'm missing something here, but why can you say with any certainty that the professor is 50? If he were 51 he'd be older than the other 3. If he were 52 ditto. Also 53, 54, 55, 56, etc

Now am I being unbelievably dumb or have I missed something? The only thing I can think of is that the assistant knows the age of the professor, but that still doesn't say why he can't be any age between 50 and 64....

[snookersfun]

Yes, well, Rambon, we all had kind of the same logic problems. I copied that puzzle from a German site btw., but translated carefully and faithfully.
I guess, if the professor expects an answers giving only these informations it has to work only for one of the two possible solutions. So it is either older than 49 or older than 50, which leaves the 50 as unique solution....

kind of.... but don't worry, the logic wasn't too foolproof on this one...

...why he can't be any age between 50 and 64....

could be of course older than 64 btw....

[rambon]

Yes, well, Rambon, we all had kind of the same logic problems. I copied that puzzle from a German site btw., but translated carefully and faithfully.
I guess, if the professor expects an answers giving only these informations it has to work only for one of the two possible solutions. So it is either older than 49 or older than 50, which leaves the 50 as unique solution....

kind of.... but don't worry, the logic wasn't too foolproof on this one...


could be of course older than 64 btw....

Agreed, but part of me thinks if older than 64 then the other solution is more valid...

Anyway, moving on. I have a question that I'm working on for you. Should be able to post it later today with a bit of luck.