Submitted by jtedds on Wed, 07/20/2016 - 09:04
Forums:
Can I give an age for the minister that I can guarantee will be above the second oldest member of the group? I have a list of ages for the other people and a range of values for the minister, so without knowing which list is correct can I just give an age from the smallest range? Or is there any way to narrow down my lists and find an exact age for the minister?
a hint
Bear in mind that it helped the Bishop to know that the Minister was the oldest person there... this means the Bishop must have known the Minister's age, and that piece of information was useful!
Assuming that all possible
Assuming that all possible ages of bell ringers are feasible, my answer currently stands as 50. That said, I'll probably need to stare at my working for a little longer!
Impossible Auestion?!
........“Well,” the Minister said, knowing the Bishop had a penchant for numerical puzzles, “if you multiplied their three ages together, you’d get 2,450. But if you added them, you’d get twice your age.”..........
- Here should it say .... You'd get twice MY age??
Otherwise the Minster could be loads of different ages as he's the oldest!
The question is really confusing!
**Question
**Question
Options
There are quite a few options for the bell ringers ages, a good place to start is to systematically list these, i.e. find three numbers that multiply to give 2450 (finding the prime factorisation of this might help as well).
Start with this, and then if you need more help ask again!
No, the question is
No, the question is definitely right! It's a little strange - think about what information the Bishop must have. He can't work out the ages of the bell-ringers before finding out that the minister is oldest, but after he's been told that he can work it out. What must the bishop know?
I got the age to be 36 but I
I got the age to be 36 but I have no idea if that's right! It'd mean one of the bell ringers is 2 which is unrealistic!
What makes you think the age
What makes you think the age is 36?
I also got 36 with one of the
I also got 36 with one of the ringers being 2. I also got 23 as an option, 26 and 38 but 36 seemed most likely although I can't remember why.
Age of bellringers
Whilst one of the bellringers is very young (perhaps infeasibly so), they are not as young as 2. You can learn to ring from a fairly young age, you just need to stand on a tall box - it's much more about technique than strength!
So far...
So far I've done prime factorisation and I've got a list of 12 numbers which seem feasible ie)between 0-100 for the age.
I've also worked out that IF a,b,c bellringers are a+b+c=2y that y is the bishop this must mean that a+b+c equals an even number which does reduce the options slightly further due to the integer rules.
Is this direction correct based on the formulas? I'm pretty sure there is some clever catch to the wording of the puzzle with the bishop knowing something before the minister reveals all!
Bishop's age
The Bishop knows how old he is! So if he was 59 (say) then he could pick the set of ages that sum to 118. The fact that he cannot answer the question straight away therefore tells you something important!
That was my reasoning for my
That was my reasoning for my answer of 50. Based on all possible combinations of ages for the bell ringers, the possible ages of the Bishop are all unique except for one possible age that is repeated twice. Hence, he could not answer straight away.
My breakdown of the question :D
*Spoiler* Don't read unless you're really stuck!
Hi all! Haha I can see this forum is bursting with people. Anyway, I've got the age of our Minister to be 50.
So I got the prime factors of 2450: 2x5x5x7x7 then combined them to make different possible sets. (The closer the ages of the ringers, the younger the Bishop can be).
Let the ages of the ringers = x, y, z, the age of the Bishop = B and Minister = M.
Scenario 1) Let x=7, y=14, z=25 therefore B=23 (Oldest = 25)
Scenario 2) Let x=7, y=10, z=35 therefore B=26 (Oldest = 35)
Scenario 3) Let x=5, y=14, z=35 therefore B=27 (Oldest = 35)
Scenario 4) Let x=5, y=10, z=49 therefore B=32 (Oldest = 49)
Scenario 5) Let x=7, y=7, z=50 therefore B=32 (Oldest = 50)
Scenario 6) Let x=2, y=35, z=35 therefore B=36 (Oldest = 36)
Scenario 7) Let x=2, y=25, z=49 therefore B=38 (Oldest = 49)
Scenario 8) Let x=5, y=7, z=70 therefore B=41 (Oldest = 70)
Scenario 9) Let x=5, y=5, z=98 therefore B=54 (Oldest = 98)
Scenario 10) (It carries on; the Bishop get's pretty damn old and we have one year olds ringing bells.)
Our poor Bishop hasn't got enough information to know the ages of the ringers. This can only be the case if:
A) He has forgotten his age (feel free to be imaginative to why this may be the case)
B) There are two sets of ages that when added together are double the Bishop's age.
The latter is most probable therefore we are left with scenario 4 and 5. (I'll put them below)
Scenario 4) Let x=5, y=10, z=49 therefore B=32 (Oldest = 49)
Scenario 5) Let x=7, y=7, z=50 therefore B=32 (Oldest = 50)
As soon as the Minister says that he's the oldest lad in the room, the Bishop knows how old the ringers are. For this to be the case, the age of the Minister must leave only one option (Scenario 4). Hence M = 50
(If M>50 then both scenarios would still be feasible and the Bishop would require more hints.)
That's what I got! Awesome :D
That's what I got! Awesome :D