How Old is the Priest?
Riddles are little poems or phrases that pose a question that needs answering. Riddles frequently rhyme, but this is not a requirement.
A very bright and sunny day
The priest did to the verger say:
"Last Monday met I strangers three
None of which were known to thee.
I ask'd them of their age combin'd
Which amounted twice to thine!
A riddle now will I give thee:
Tell me what their ages be!"
So the verger ask'd the priest:
"Give to me a clue at least!"
"Keep thy mind and ears awake,
And see what thou of this can make.
Their ages multiplied make plenty,
Fifty and ten dozens twenty."
The verger had a sleepless night
To try to get their ages right.
"I almost found the answer right.
Please shed on it a little light."
"A little clue I give to thee,
I'm older than all strangers three."
After but a little while
The verger answered with a smile:
"Inside my head has rung a bell.
Now I know the answer well!"
Now, the question is:
How old is the priest?
Answer
The puzzler tried to take the test;
Intriguing rhymes he wished to best.
But "fifty and ten dozens twenty"
Made his headache pound aplenty.
When he finally found some leisure,
He took to task this witty treasure.
"The product of the age must be
Twenty-four hundred fifty!"
Knowing that, he took its primes,
Permuted them as many times
As needed, til he found amounts
Equal to, by all accounts,
Twice the verger's age, so that
He would have that next day's spat.
The reason for the lad's confusion
Was due to multiple solution!
Hence he needed one more clue
To give the answer back to you!
Since only one could fit the bill,
And then confirm the priest's age still,
The eldest age of each solution
By one could differ, with no coercion.
Else, that last clue's revelation
Would not have brought information!
With two, two, five, seven, and seven,
Construct three ages, another set of seven.
Two sets of three yield sixty-four,
Examine them, yet one time more.
The eldest age of each would be
Forty-nine, and then, fifty!
Hide
Comments
Piffle 
Aug 09, 2002
| my goodness. did you write that yourself? |
Canary05
Aug 09, 2002
| Wow. "Ten dozens twenty" had me stuffed. |
Dreamphysique
Aug 09, 2002
| Wow!!! Gosh, you have talent!!! make more riddles, cuz i like being stumped like that!!! lol.. |
george1978
Aug 12, 2002
| i do not know said thine one before as he headed to go out the door, then turned around and said once more this rhyme has made my mind go sore.
very good it was an excellent job. |
pizzahead
Aug 13, 2002
| Did you mean "With two, two, five, seven, and seven," or "With two, FIVE, five, seven, and seven,"? |
pizzahead
Aug 13, 2002
| Nice job, btw. |
zangel3000
Aug 14, 2002
| ummm...was I supposed to understand that? |
Overlord
Aug 15, 2002
| You are good, riddler, but shalt thee solve my riddles-nay! |
sheepdrew
Aug 17, 2002
| that was brill
but not quite a thrill
next time to be sain
and try and test my tiny brain
good 1 |
Sir_Col
Sep 07, 2002
| Okay, someone please explain how this can be solved? By writing 2450=2x5x5x7x7, we produce seven realistic sets of ages: 2,25,49(adding to 76);2,35,35(76);5,10,49(64);5,14,35(54);7,7,50(64);7,10,35(52);7,14,25(46). We could have 5,7,70(82), but it is unlikely the priest is older than 70. So given this information how do we deduce the age of the priest? |
beppie_stark
Sep 09, 2002
| Aren't your pronouns are mixed in the first clue? Thine = yours. Since the priest is speaking of his own age, not the verger's, shouldn't the pronoun be "mine"? |
pachacutec
Sep 10, 2002
| To Sir_Col: The 'poetic' answer says the verger finds muliple solutions, hence he needs one last clue.
The only duplicate answer is 49+10+5 = 64 = 50+7+7. Since the last clue says that the priest is older than all three, the verger now knows which of the two solutions is correct, and we also know the priests age.
ABSOLUTELY BRILLIANT, AND THE SOLUTION |
cristos
Sep 12, 2002
| the priest is 50, just cause i guessed, but yea, the thing with the 10 dozens 20 was a little too confusing |
Sir_Col
Sep 12, 2002
| Thanks for the reply, pachacutec. I'm still confuddled. I can reluctantly accept that you can interpret the wording of multiple solutions to imply that it is one of the duplicates (but strictly I read it as there are seven possible sets of which we have no way of determining which one it is). Given my acceptance of that, how, then, do we work out the priest's age as 50? Knowing that he is older does not mean he is 50? Am I still missing something? .... a brain maybe?  ) |
pachacutec
Sep 25, 2002
| Sir_Col: You have to keep in mind that the priest KNOWS the answer to the riddle. With that, let's do a simple SET comparison. SET A = ALL INTEGERS > 50; SET B = ALL INTEGERS > 49; Since SET A is a subset of SET B, and the only element of SET B not found in SET A is 50, the only way this final clue sets off the Bell in the Verger's head is if the priest is 50. |
DEL719
Oct 19, 2002
| i didn't even understand the english... |
bluelightnign
Oct 20, 2002
| I couldn't make sense of that riddle. After 5 minutes I gave up and looked at the answer, which I thought would be straight foward. But I was obviously mistaken. The Answer is harder to understand than the riddle. |
gingimo711
Oct 25, 2002
| uh huh......i'll trust you on that |
CapnChubby
Dec 13, 2002
| The only thing I can say after that is wow... |
megan
Dec 21, 2002
| Excellent job. Did you think of that yourself? |
excellerator
Dec 24, 2002
| omg im still confused and i will probably remain confused for at least 5 or 6 years good job |
todd1975
Dec 30, 2002
| I'm surprised everybody hasn't caught on.... the answer is wrong altogether.
Okay... we know that the 3 strangers ages multiplied together is 2450. Broken down, that gives us 7x7x5x5x2x1 in the simplist form. But keep in mind the priest is older than all of them yet is only half of all their ages combined. The only combination that allows this is 7x5, 7x5, 2x1. Or 35, 35 and 2. This would make the priest half of those ages combined, or 36, which is older than all 3.
Check it!
The answer said he was 50, but that would mean one of the strangers was 49. If that was the case, the ages would be 7x7, but with only 5x5x2x1 left for two guys, the largest sum of ages would be 50 and 1. Add 49, 50 and 1 to get 100, the priest would be half of that or 50. Problem being, he'd be the same age as one of them, not older than them all. |
NuttieKiwis263
Dec 31, 2002
| Oh my! That was hard! But you did a great job! Make more of those! I like being challenged like that! |
Chili-N1
Jan 08, 2003
| Todd1975, Your answer is the only one that I identified with and understood. Working with 2,5,5,7,and7, I tried to come up with the 3 most "even" numbers so that when added and divided by 2, you would get an answer still greater than each individual number. I used 14, 35, and 5 (sum = 54 and 1/2 =26. No good, 26 is not less than 35) I also tried 14, 25, and 7 (sum = 46 and half = 23. No good, 23 is not less than 25.) My error was that I kept wanting to compine the smallest prime with largets prime to get most even spread. It did not occur to me to not multiply smallest prime (2) to something else. But, that one works, 2 35 and 35 yield sum of 72 with half = 36 which is greater than largest number 35. I accept the ONLY solution to be the priest is 36 and the three individuals are 2, 35, and 35. (I also agree if solution is stating that priest is 50 and the three people are 1,49, and 50 that it does not qualify since rules stated priest was older than all not older than OR EQUAL to their individual ages.) In a nutshell, I like this riddle, I ALMOST had it (Ha HA) and Thanks Todd1975 for clearing this up and correctly solving.
Chili-N1 |
skoolkid 
Jan 17, 2003
| It's impossible. Its too hard! |
(user deleted)
Jan 22, 2003
| the answer is to long to read |
jemmagrace
Jan 28, 2003
| I thought your riddle was good and it certainly had my mind working well done |
Asbestos
Jan 29, 2003
| Chili and Tod, you are both assuming that the question is worded wrongly, which is not a good assumption to make. For some reason it has gotten around that the PRIEST is half of the ages. Rather, the other guy is.
Remember also that the other guy knows his own age. Therefore, when he stays up all night, he finds that there is more than one possiblility. Since he knows his own age, this means that there must be two or more answers that add up to the same number. There are two that add up to 64, therefore he is 64.
Now, he still doesn't know which of those two it is. But he ALSO knows the priest's age. Therefore, when he hears that the priest is older than all of them, he works it out.
Therefore, we know that the solution can't be the one with the oldest guy in it, and that the priest must be no older than the oldest guy in the two possibilities. If he is older than 49, and no older than 50, he is 50.
Hopefully this will make it a little clearer!  |
Asbestos
Jan 29, 2003
| oops, when I said "therefore he is 64" above, I meant "therefore he is 32." The rest of the answer is correct, though! |
griphook 
Feb 09, 2003
| The product of the ages is indeed 2450 because Ten Dozens Twenty could not reasonably be interpreted as 120 plus 20 in the absence of any conjunction.
The Verger knows his age and that the sum of the ages is twice that of his. With this he could have gotten the solution to the problem given to him(determine the ages of the three) after deriving the nine sets of ages a. 49+25+2=76, b.49+10+5=64,c.35+35+2=72,d.35+14+5=54,e.70+7+5=83,f.7+10+35=52,g.7+14+25=46,h.7+7+50=64,and 5+5+98=108(some guys live that long?)but he was not able to do so and needed another clue. This tells us that there must be more than one set of ages equal that twice of his age, in this case 49+10+5 and 7+7+50. Now we know that the Verger is 32.
When he was told that the priest is older than all three, he was able to determine which set of ages is correct (49+10+5) because he knows the age of the priest i.e 50.the other set would then be wrong becase the priest would be of the same age as the oldest in the group.
In addition, if the priest is older than 50 then the Verger would not be able to get the right solution because the two sets of ages would still be both possible(which is not the case in the Riddle). THIS FACT( that Vergel was able to answer the problem given to him after the last clue) tells us that the priest is 50 and not any older. GREAT GREAT RIDDLE. |
inudbz
Mar 19, 2003
| all i can say is wow |
icehaven
Apr 06, 2003
| huh!?!? |
smart432890
Apr 29, 2003
| BEST ONE I SEEN LATELY U OWN DUDE |
krishnan
Jun 09, 2003
| I have seen the same puzzle in a book. Though it is not original, it is a very good puzzle. |
beanie89  
Aug 10, 2003
| Um.........ok |
soccercow10
Sep 06, 2005
| I liked the teaser....though not as much as the rhymes
i keep on readin it but its too many times
i may not ever get it and i may not forget
how much this amused me i wish we had met? 
that was a bad poem i just did.....anyways great teaser |
(user deleted)
Oct 05, 2005
| Here's the rest of the answer, though I see why the poster didn't include it. But the two numbers in the last line were the maximum ages of the 2 solutions with identical sums. It should have continued:
With lack of proper rhyme and meter,
I've tried to be the first completor
of this poem and a puzzle;
my poetry, you'd try to muzzle!
And lest you think my wit is thrifty,
The answer, of course, must be fifty!
If dispute, you wish to tender,
note my addresss, as the sender!
--
Kevin Nechodom
But ...
The verger said, "I'm not a lad!
At 32, your rhymes are bad!"
But undeterred, the Puzzler wrote
the explanation for his note.
And Pizzahead, ever circumspect,
said "two, FIVE, five", indeed correct.
The posted email is no more,
so leave your question on the door!
knechod |
(user deleted)
Oct 05, 2005
| Sorry, but the email address got stripped because it was in angle brackets. (knechod@stacc.med.utah.edu) |
maingle-_-
Oct 30, 2006
| i didn't get it till i read some of the comments, but It was very well written! |
lmurray 
Sep 18, 2007
| THIS IS VERY GOOD!!
THANK YOU SIR!! 
I WOULD LIKE 2 C MORE. |
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