"UNSOLVABLE" Logic Puzzle: How Old Is The Priest?

Thanks to Peter for sending me this problem! This is a really fun one and I encourage you to work it out. A priest and a cantor have a conversation about the product and sum of three visitors. After a few exchanges, the cantor figures out the visitors ages. But wait, how old is the priest? This is logically solvable from the clues. Can you figure it out?
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[@blackmamba1261]

The best part was seriously considering the possibility where one of the visitors was 2450 years old

[@TheMisternono94]

What's your problem with old people ? You should have respect for people who fought life for such a long time ! You're just another guy who show to everyone how unopen minded he is.

[@kordellcurl7559]

It could have been god (or anything old) that visited

[@niebuhr6197]

That isnt possible, if one has 2450, then the others have 1 and 1, making it 2452/2=1226 years old

[@0cheeseburga]

why can't the cantor be 1226 years old?

[@surajsapkal1293]

;)

[@AshesOfEther]

Plot twist: The cantor, like most people, was unable to do all this math in his head and/or just didn't want to bother, and pretended to know the ages at the end to get the priest to stop so he could get on with his life.

[@babyli9205]

But then the priest would have asked him the ages and he wouldn’t have known what to say

[@javidfarhan1675]

Wolfram alpha :)

[@Maxawa0851]

@@babyli9205 simple. 'That information is left as a trivial exercise for the lostener'

[@calebmurray4438]

Yeah like I can tell that the product is 2450 and the sum is even and they’re probably all below 90 but I still don’t wanna bother

[@jackmclane1826]

And hopefully not getting molested in the process... ;)

[@johnporonga44]

Class: 2+2=4 Test:

[@sparshgupta499]

Wow not even a single reply here

[@noobgamedev8621]

I am here

[@noobgamedev8621]

@@sparshgupta499 world is a lonely place.

[@kaushalandtulsigoswami6394]

@@noobgamedev8621 no it is not

[@noobgamedev8621]

@@kaushalandtulsigoswami6394 I know it was just for comment purpose 😅

[@muhammadawais1487]

so, the cantor is more mathematical than presh...he did all calculation in his mind

[@mrchin7562]

Nope. I did it in my head, as well. + the cantor knew his own age.

[@aadikhandekar2907]

Yep

[@joibgbg123]

Well it was stated that the cantor did calculations, but I didn't see it mentioned that he did it in his mind

[@marioluigi9599]

Muhammad... Please don't assume things you don't know, okay?

[@peterromero284]

The cantor was smart, but he didn’t know his own age.

[@henk-ottolimburg7947]

Imagine a 50-year old, saying about a young prick of 49 'when he reaches my age..."

[@diandradeeke]

yes that was really misleading...

[@GDGAI]

4:01 Now it will take SUM effort

[@Blurro]

The thumbnail question leads to a different answer. The clue in the thumbnail is that the cantor didn't eat cake, and therefore must be the same age or older than the priest, and also older than the oldest visitor. The only case in which the cantor is older than the visitors is when the visitors are ages 2, 35, 35, added to get 72, halved to get 36 (the cantor's age). And since the cantor can't be younger than the priest, the priest is also 36.

[@djarogames]

I got the same answer!

[@Zharque]

The thumbnail only said the cantor didn't eat cake at the PRIEST'S birthday, not at his birthday when he was at the priest's age. Still could go either way so it's a pretty useless hint imo.

[@saskatoon5699]

But then the cantor would know the answer before the last clue from the priest, knowing that only 2+35+35 corresponded with his own age at 36. So the fact that he needed the last clue means that this cannot be the answer.

[@phantomlogic6940]

This was certainly fun to work out. I'm kind of surprised that the triplets with impossible ages weren't immediately eliminated, though. Still, I loved figuring this out! Would love to see more like this.

[@DarkVoidIII]

What do you mean 930 years old is impossible? If Adam could have lived that long, according to Jewish tradition, then so could any of us, if we had the common sense and knowledge needed to get us to that age. I'm not Jewish, just pointing out that there's a flaw in your logic by calling a particular age range "impossible". 😊🙃👍

[@AdelaeR]

You don't know what the maximum age of the people talking is. That is not logical to assume from what is given.

[@ondrakucera4532]

I don't understand the triplets thing. Can someone explain it to me please?

[@createyourownfuture3840]

@@ondrakucera4532 which one?

[@hasanmuttaqin464]

@@ondrakucera4532 all the other answer don't have equal sum

[@factsverse9957]

That is really sad that there are only 3 visitors.

[@KAF128]

But on the plus side, two are quite young, so that is promising, from the point of view of attracting a younger/larger audience/congregation, moving forward !

[@TGears314]

KAF128 could also be interpreted as they’re indoctrinating young. Just depends on the viewpoint.

[@KAF128]

I don't think we need to worry too much - it was only a jokey comment on a fringe, meaningless stat that's not critical to the puzzle !

[@Rekko82]

And two of them were kids which probably never wanted to be there.

[@Rekko82]

I have to add, if the priest gives puzzles like this to them, they stop visiting the church very soon.

[@Mo-ev9vl]

Plot twist: The priest was never alive in the first place.

[@badam9656]

I thot the samd

[@jermainerace4156]

I love it when I have to use unstated implications to draw assumptions about a problem with one right answer...

[@danielhughes3758]

That's why I like hearing the solution to puzzles like this but I don't bother trying to solve. Too many odd assumptions you have to make.

[@vaibhavjain3234]

Presh: Can YOU figure it out? Me: What ME? Nahhh...

[@NordicMe]

How could you skip cake on your 50th birthday? That's a milestone!

[@anandk9220]

The priest looks to follow abstinence of such temptations. So not eating cake is just fine with him. Actually these logic puzzles are just out of bounds for me and many such minds. So comments is the only place to have some solace. 😊

[@JM-us3fr]

I did the same thing! Though I ruled out the pathological cases where someone was 170 or older (just so I could fit it on my whiteboard).

[@Gwunderi25]

And a 1 year old eating cake is - if not pathological - very premature :D

[@NestorAbad]

I knew the easier version (two daughters, the oldest plays the piano) but certainly this one is even funnier and trickier! Thanks for sharing!

[@omatbaydaui722]

compared to two daughter one, this one uses 2 times the trick of people "knowing" the ages but not the reader. To be honest, I doubt anyone who didn't come across the easier version can solve this one, because you HAVE to know that the people involved in the puzzle know an information you do not as the solver. And you HAVE to know that this is not a puzzle you solve with math alone but using a logic based on a realistic situation. I'm not even sure those kind of puzzles can be considered puzzles, because you have to extract informations that are out of the bound of the puzzle himself. You can't train with these kind of puzzles, or improve your problem solving skills .

[@AMR-555]

@omat Not to brag, but I solved this one without knowing the version you are talking about. However, now I am curious to see the "easier version", could you please copy and paste it here or link me to that problem?

[@dlevi67]

One version of that goes along the lines of: I have three kids, the product of their age is 40. The youngest has blue eyes. How old are they? (10, 2, 2, or 8, 5, 1 - if 10, 2, 2 there is no youngest). ETA FWIW, I disagree with @omat baydaui that these puzzles "cannot be solved blah blah". Puzzles are about logic and creativity first and foremost; then some (not this one, sorry!) require a fair amount of mathematical knowledge and ingenuity to be solved. The information is NOT out of the bounds of the puzzle itself; it is hidden in ways that are not obvious (e.g. statements about "cannot know" as opposed to simply "I don't know").

[@NestorAbad]

@@AMR-555 this one for example: www.techinterview.org/post/465654945/the-oldest-plays-the-piano/

[@pietervannes4476]

@@dlevi67 theres still a youngest person although theyre both 2 yo

[@nsnick199]

Got the right answer, somewhat by luck, since I totally forgot about 1 as a factor! Only had to check 11 sets of ages that way!

[@simer7878]

yep, I also got it right because I calculated the world equation, thus hindering me from making a mistake. Classical

[@alessandrobottero6544]

Nice one. The thumbnail is wrong/misleading, though: is the priest who didn’t eat the cake at his birthday, not the cantor.

[@patrickgould8415]

It's a variation of the puzzle with a different solution. Bonus content

[@christopherstoney4154]

@@patrickgould8415 I solved the 'thumbnail' version first, assuming that the Cantor and Priest were two of the three whose ages multiplied to 2450. I quickly deduced that the age of the Cantor must be the sum of the other two ages, and after going through the possible factorizations determined that the cantor was 50, the Priest 49, and the third individual (possibly a baptismal candidate?) was 1 year old. When I finally clicked on the thumbnail and got to the 'real' puzzle, I was able to solve it as demonstrated above, but not without taking some time and a pen and paper.

[@sakshamtyagi6527]

Let's consider a 2450 year old monster and 1 year old babies are attending his party...

[@PlayerMathinson]

If you're reading the comments before solving, remember the key is that Cantor *DID* find the ages.

[@Care99]

Now it makes sense! I was extremely troubled when Phresh picked the ones with 64 out of 'nothing'

[@vidicate3963]

I love that Cantor shows up here. :) I solved it going from clue 1 -> 3 -> 2. I found two sets where the oldest in each had a difference of 1, and then figured out the younger visitors' ages that would give an equal sum to complete each set: {49,10,5},{50,7,7}

[@leviosarwingardium]

Never thought i would be so dedicated watching an impossible philosophical mathematics video about a 2450 years old person, cantors, and priests.

[@ffggddss]

Ages: cantor, C; priest, P; 3rd person, A CPA = 2450 = 2·5·5·7·7 Noted: This is a "rectangular" number; i.e., the product of 2 consecutive integers, 49·50 C + P + A = 2C; C = P + A, meaning that C is oldest. How many ways to factor 2450 into 3 positive integers? Not all that many; many of them are ruled out because one of the factors is too big. All the divisors (2·3·3 = 18 of them): 1, 2, 5, 7, 10, 14, 25, 35, 49, then all their cofactors: 50, 70, 98, 175, 245, 350, 490, 1225, 2450. Now we can draw an upper age-limit at 98, and look for any of those 12 #s being the sum of two smaller ones; while making sure that the chosen largest #, times the other two, is 2450. C = 98? 2450/98 = 25 = PA. Impossible for P + A to = 98. C = 70? 2450/70 = 35 = PA. Impossible for P + A to = 70. C = 50? 2450/50 = 49 = PA = 49·1 Hey! It can only be 50·49·1 = 2450; 50 = 49 + 1, making the priest 49, because ordination is not performed on infants. No other combination satisfies the conditions. ADDENDUM: I misunderstood the problem, which I got solely from the thumbnail. The priest & the cantor weren't among the "visitors." Nevermind!! Except that this modification of the problem is kind of interesting in its own right. Also, I missed the correct answer by just 1, albeit totally by coincidence. Fred

[@stephen3390]

I think the real conversation would be : Priest - 'you fancy a couple of beers'? Cantor - ' yeah'

[@simplebutpowerful]

My mental journey through this WONDERFUL puzzle: 1) Prime factorization of 2450, obvi. I broke it down to 49×50 and left it at that for the moment since I was comfortable with rearranging those numbers' factors if needed. 2) Prepare to churn through a bunch of cases, realizing the classic trick that the sum will be shared between two sets of factors. 3) Recognize that I can spot the true one by which has the smallest maximum, since cantor will know priest's age. 4) Reread the problem: How old is the PRIEST?! 5) Momentary existential crisis. 6) Realize the biggest trick of all: the two sets of factors will have maxima that differ by 1, hence we the reader can deduce the priest's age. 7) I've already found two large-ish factors of 2450 that differ by 1: 49 and 50, alas! 8) So I need two numbers that multiply to 50, whose sum is 1 greater than the sum of two numbers that multiply to 49. Figure out 10 & 5 and 7 & 7, checkmate.

[@BigDBrian]

nice

[@dlevi67]

5) I had to laugh. Thank you!

[@noobgamerlive2220]

The fifth point tho 😂🤣🤣

[@haruhisuzumiya6650]

That completely went over my head

[@awookieandagerman]

I went down the rabbit hole of thinking the cantor and the priest were actually two of the visitors in question.

[@YashPatel97335]

What teacher teaches in the class : 2+2=4 Question in exam : Prove that conjecture 3x + 1 will not end in 4,2,1. Best of luck!!

[@lunardancer6047]

Took me a while, but got there in the end. Thankfully you were kind to give a number without too many prime factors.

[@bificommander7472]

I got the two possible ages quickly enough, most time was spend writing them all out. I got stuck on which to pick until I noticed that the question wasn't for the people's ages but for the priests. When I realized the priest's age was supposed to be solvable, it clicked.

[@HensosGeometryDash]

One of the few riddles I've gotten right on here, a nice change of pace from the brain torture Alice and Bob gave me on the previous riddle

[@edgarallenhoe3518]

I didn't realize Christians had cantors! You learn something new every day. Edit: I looked it up and apparently the role in a church is slightly different than a cantor in a synagogue. Cool!

[@spencerhiginbotham7538]

I just wish there was a way that the ages of the visitors could've been 1, 1, an9 2450. Then the cantor would be 1226, and the priest would be older than 2450. That would prove the priest is John the Beloved.

[@RobloxKid123]

2:31 As a 2450 year old, I agree that this is a possible solution

[@rubixman7x7]

I'm happy that I figured it out. I didn't solve all of the factors since many of them wouldn't make sense but I knew what I was looking for.

[@jasonpatterson8091]

I did figure it out, but I think the analysis would have been better if it had also considered reasonable combinations of ages. Additionally, the problem would be MUCH better if there were 3 sets that added together to the same sum, but one of them could be eliminated for human reasons rather than purely mathematical reasons. It doesn't work, but suppose we had also had 16, 16, and 32 as a set of ages - the cantor would have been able to tell the difference between a 5/7 year old and a 16 year old, even if the math couldn't. I wonder if a similar puzzle exists with that sort of limitation and sensible ages. To be clear, in this puzzle we have 3 people, the youngest is either 5 or 7, a reasonable uncertainty, the next is either 7 or 10, again a reasonable uncertainty, and finally 49 or 50. Had it been 5 or 14 for the youngest, the puzzle might have worked mathematically but it wouldn't have worked in a human sense.

[@AhsimNreiziev]

Absolutely wonderful comment. A lot of Mathematicians and Logicians seem to often forget that there is more to life than just Mathematics and Logic, and so I *LOVE* it whenever I see someone reminding them of that.

[@zanti4132]

That would be like the solution to the very similar puzzle with the very similar title "UNSOLVABLE logic puzzle: What are their ages," where one combination is eliminated because two ages are identical.

[@ganjagoblin1939]

Instead of doing the math, I just guessed 50

[@Scorpionwacom]

Presh, many thanks for mentioning Wolfram! I have totally forgotten about it. Might be useful for my next story. And no, I didn’t figure this out. I should listen to the part about the priest’s age several times-maybe I’ll get it then.

[@Tfin]

Here I am trying to figure it out from the thumbnail, which left me wondering: 1. Was the priest one of the visitors? (no) 2. Which person or people do the pronouns refer to? (the priest) The video answers these, making the question solvable.

[@MA-gg9in]

When I started solving this problem I was the age of the cantor, and when I ended I was the age of the priest

[@skillcheese]

I totally didn’t think about the combo where some of them could be 1 year old! Cool to be reminded that prime factorization does include 1, even though it’s technically not a prime. Great videos! I’ve started doing these in my head when I go to sleep, I find it improves my next day a lot! Especially if I fall asleep while doing it, usually wake up with the answer!

[@t71024]

I remember solving this puzzle told by my high school calculus teacher in the 80's. His story around it was less explicit and I didn't realize the two dialoguers must knew each others' ages until I had an epiphany a couple of days later. Great puzzle!

[@bowenzhou2360]

After careful and precise calculations I have finally figured out that there were 3 visitors.

[@mouaztabboush5571]

Way to go man. Love your videos

[@TheOtherGuys2]

Okay, without watching the video, just from the question as presented in the thumbnail, here's my thinking. 2450 could be the result of 50x49x1, meaning those are the three ages. Therefore, the Priest is 49 years old, because he's not 1 years old and a priest. The cantor is 50, and some random baby is 1. The sum of the three ages is 100, which is 2x50, the cantor's age. Who ate cake at the birthday party doesn't matter. Done, such math, very logic.

[@denizberkinmis]

I figured this out because of a similar question i saw earlier on a mathematical problems book.But first one was quite simplified comparing to this.

[@bb2fiddler]

Man, I'm terrible at these lol Please keep uploading more puzzles of this kind!

[@Anita95_original]

Oldie but goodie. My teacher gave me that problem 1982. Hard one, but the solution have helped me solving other kinds of problems by forcing me to disassemble the problem and use strict logic first and foremost.

[@Blahblah97555]

I got this right using the excellent method of looking at the prime factorization of the numbers, thinking “49 seems like a good age for a visitor”, and figuring the priest had to be one year older, since otherwise setting up the logic for the puzzle would be too much of a pain.

[@eurovisioncyan9550]

it should be SEEMINGLY impossible

[@hysterheister0947]

Was it edited?

[@harshilsangal6226]

@@hysterheister0947 No, Eurovision Cyan says that the 'seemingly' should be capitalised and not 'impossible', as is the case with the title of the video.

[@andywong3095]

@EC, Relax, This is NOT a Grammar School Channel, please go to the "RIGHT Channel". I think, Mr.. P.T., will not care about grammar or pronunciation, just like me! ha. Here is my point of view: (May not be the correct one) Circular logic, The premises would not work if the conclusion weren't already assumed to be true: Assume, the Cantor has to know the answer, because, he said, he has solved it. No wonder, church executed, so many innocents, because the church said, they were witches. You have to trust the Church, don’t you dare, to question the finding of the church.

[@sirjaroid4725]

the cantor is 32, one visitor is 49, the others 10 and 5. the priest is 50. i first guessed 7, 7, and 50 for the visitors age, but then corrected it after.

[@jaywalker7721]

Very satisfying. I used a website to find all divisors and I ruled out anything where someone was over 98 years old to keep it cleaner.

[@Tehom1]

Thanks for the puzzle, Presh! First let's factor 2450, gives 2*5^2*7^2, and let's get the different possible individual ages: {1 2 5 7 10 14 25 35 49 50 70 98 175 245 350 490 1225 2450} Many of the larger numbers use up so many factors that there is only one possible combination of other factors, thus each of the larger numbers implies one possible sum, or just a few sums: 2450 -> (1,1) -> 2452 1225 -> (1,2) -> 1228 490 -> (1,5) -> 496 350 -> (1,7) -> 358 245 -> (2,5) -> 252 245 -> (1,10) -> 256 175 -> (2,7) -> 184 175 -> (1,14) -> 190 98 -> (1,25) -> 124 98 -> (5,5) -> 108 70 -> (1,35) -> 105 70 -> (5,7) -> 82 50 -> (1,49) -> 100 50 -> (7,7) -> 64 We know that the sum of their ages must be an even number that can be reached in two ways. By inspection, none of the numbers above was reached in two ways, and most are too large to be reached by any three of the smaller numbers. So we can rule out 2450, and dismiss the idea that a 2450-year-old visitor and two babies walked into church and the cantor thought the 2450-year-old might be younger than the priest. Similarly, we can also rule out everything above 98, since we know that any combination that includes an age of 50 or over won't match it, and any combination that does not can't exceed 147. Continuing the table: 49 -> (1,50) -> 100 49 -> (2,25) -> 76 49 -> (5,10) -> 64 We find our first two double-hits. One isn't valid, 100: it's not reach by two separate combinations, it's reach by the same combination as before. The other one, 49+10+5 = 50+7+7, is a valid double-hit. Now the table is getting a little thick, with multiple possibilities for each. But we've already considered everything above 35, so let's omit every combination where one product is a number we've already considered. So continuing: 35 -> (2,35) -> 72 35 -> (5,14) -> 54 35 -> (7,10) -> 52 25 -> (7,14) -> 46 For 14 and below, all combinations include some number greater than 14, so we need not consider them again. So there is only one sum that can be reached in two ways, 64, by {7,7,50} and {5,10,49}. From the final clue, the priest is older than everyone in one possible group, but not older than every member of the other possible group, and this must be sufficient to give his age. This gives us our answer: The priest is 50 (and the cantor is 32)

[@bobh6728]

Your sum of 105 is not correct because the sum of the possible ages is always going to be even because 2450 has only one factor of 2. So one of ages is even (has the factor of 2), the other two are odd. So even plus odd plus odd is even.

[@Tehom1]

So you've ruled out something I already ruled out for another reason.

[@inyobill]

Oh, they all had birthdays on the same day. Thanks a ton for hiding that clue.

[@nenadilic9486]

Somene can be, for example, 25 years and 7 months and 16 days old, but the AGE of that person is 25, until the next birthday.

[@Gowly_]

I actually solved this one by myself! It was super fun!

[@tonyhasler985]

I found a slight shortcut. Let us begin by assuming that there are 5 visitors, none of whom are 1, with an age product of 2450. The ages must be 7, 7, 5, 5, and 2. If we want to have 4 visitors, none of whom are 1, we can replace 2 of the original visitors by someone with an age equal to the product of the removed visitors. This will not affect the overall product but will increase the sum by the difference between the product of the two prime factors and their sum. So, for example, if we replace the two 7 year olds by a 49 year old, then the sum will increase by (7x7)-(7+7)=35. Combining other pairs of products gives us possible increases of: 35 (7x7)-(7+7) 23 (7x5)-(7+5) 15 (5x5)-(5+5) 3 (5x2)-(5+2) To reduce the number of visitors (none of whom are 1) from 5 to 3 we can either 1) replace two pairs of visitors or 2) replace 3 visitors by 1. The possible results of the latter are: 226 (7x7x5)-(7+7+5) 158 (7x5x5)-(7+5+5) 82 (7x7x2)-(7+7+2) 56 (7x5x2)-(7+5+2) 38 (5x5x2)-(5+5+2) We can see that all combined increases are unique except when the combined increase is 38. At first glance it appears that there are 3 solutions but then we realise that we cannot combine 7 and 5 and also 5 and 5 because there are only 2 fives. So option 1 is to replace two 7 year olds with a 49 year old(+35), and also a 5 year old and a 2 year old by a 10 year old (+3). The result is: 49, 10, and 5. The other option is to replace both 5 year olds and the 2 year old by a 50 year old. The result is 50, 7, and 7. All this assumes that there are no 1 year old visitors but it appears this does not mattter :-) Now we have two ways that the original sum of 26 can increase by 38 to give 64. We can now consider the priests age in the way that Presh explains.

[@lucaslucas191202]

It's kind of amazing you can figure out the exact ages of the visitors, the cantor and the priest from this simple information

[@nickymoore6827]

So close! Got stuck on the last step where I had the two possibilities. As soon as you said that the cantor KNOWS the priest's age, I had it. I got pretty lucky though. My first two guesses at the visitors' ages just happened to be the two correct options. They both summed to 64 so I figured it must be them as the answer could not be uniquely identifiable at that stage. Obviously There could have been more solutions if i had bothered to check but like I said, lucky...

[@kaushikram5095]

Whoa!! Epic question and solution.

[@milosmaric6412]

I solved it. I wrote down all denominators of 2450, while solving i realised there had to be atleast 2 equal sums of ages since cantor didnt guess it after second hint and the last one cleared it

[@ichdu6362]

I actually worked this out with by doing all the cases in Office Calc. I did prime factorization, then basically created all factors of 2450 from that, and for each factor considered all unique cases. This way, I essentially started from the largest factors downwards, which in retrospect was more work than if I had started from the smallest factors (like you did). Still fun to do, and I got real excited when I worked it out. ^^

[@octopus9001]

Very interesting problem. Instant sub. Edit: I did get to the part in the equation where there was two possibilities, but I could not figure out the last step.

[@CyberSlugGump]

I would argue that the "first clue" is that we are looking for three ages. Thus, we can eliminate any tuples with a number larger than 120ish (i.e., 98 is ok but 175 isn't). Then we only need to sum a dozen of the possibilities instead of all twenty!

[@AdelaeR]

That is not very mathematical of you, is it? You are assuming things, like the assumption that the priest and the cantor are human beings living on the planet Earth.

[@UnimatrixOne]

@@AdelaeR Logic puzzle or math puzzle...

[@1958vintage]

I didn't get the last step. Excellent work, I enjoyed the challenge.

[@cerealmug450]

I like these specific puzzles even though I already know the answer before it’s explained after a while

[@chrislankford7939]

When working this out, I forgot the cantor knows his own age as a piece of information. So I picked the triplet with the smallest maximum value of those with the right prime factors and parity (even, odd, odd). In this case, if I made no mistake, the ages would be 7, 14, and 25 and the priest would be at youngest 26.

[@NoNoahhhh]

Very nice puzzle. When the priest says 'when they reach my age' he knows that the cantor struggles for the 2 options. The priest already used wolphramalpha or wrote the combinations. Because if he didnt know the 2 options that the cantor is facing (after the priest said sum of the three visitors is double your age), he wouldnt mention 'when they reach my age' this is a very nice puzzle : math , perspective interpretation

[@thekidcores4319]

Not that difficult, but certainly fun. Initial response was ‘huh?! The priests age??’ But quickly realised there aren’t actually that many options of triplets to 2450, and as the cantor didn’t know after being told the sum was twice his age, it was easy to look for the only two triplets with sums that matched. Which took less than five minutes. Good stuff.

[@JohnMoseley]

I'M GOING TO SAY (don't look if you don't want to know) 50. I figured there had to be two ways of multiplying three numbers to get 2450 and also getting the same sum. That's why the cantor says it's not enough info even when he knows the sum. I found that... 10 x 5 x 49 = 2450 and the sum is 64 7 x 7 x 50 = 2450 and the sum is 64 So once he knows none of the visitors has reached the priest's age, he knows it has to be the first option. Therefore the priest has to be 50. I'm rather surprised I got it. I was about to give in, but Presh's encouragement to work through all the options kept me going.

[@eneaganh6319]

This is awesome I figured it out doing 1 by 1 and then I saw the video to check Amazing

[@valarmis]

I was almost able to do it on my own, I wasn't able to realize that since the cantor couldn't tell from his own age, that it must be a duplicate sum, but I was able to figure everything else, and I think most people would be able to get as far as I did, and would be able to know from the last two, so I figure the part I missed is the actual harder logic part of this puzzle

[@viniciusfernandes2303]

Thanks for the video!

[@TheGamerAdventurersX]

Who else looked at step 1 and 2 of the answer cause they were too lazy to calculate all the possible variables of xyz=2450 and the possible sums of x+y+z before pausing and deducing?

[@Nuoska]

I just guessed some reasonable ages and found that 64 was not a unique sum. Still had to watch steps 1 and 2 to confirm my solution.

[@dlxmarks]

I lost interest when I realized how much factoring was involved.

[@BigDBrian]

Yeah I understand how the reasoning in the puzzle is gonna work (that eventually gets you the priest's age), but I don't want to list all the factors and actually pinpoint the exact numbers.

[@kimf.wendel9113]

Well you could always deduce that 2450 is divisable by 50, which leaves 49. That's the square of 7, which gives you the numbers 2, 5 and 7. Now it is just a matter of multiplying those numbers, and you got you factors. I only did three and then I knew the answer

[@BigDBrian]

you get the prime factorisation of 2*5²*7² Given any combination of these 5 factors of these across 3 numbers, that's a lot of options to go through

[@georgiteodosiev9037]

Surely a great problem that I solved :)(took a long time for the last part though)

[@aquamarine245]

Give that Cantor a medal!

[@Sid-ix5qr]

Imagine this problem coming for your examination.

[@simsisthecommander]

This was in a scholarship exam for my University. Savage stuff

[@RandomPerson-gf6gd]

Omg. I first thought this in like a half a second and decided “I’m going with the first one I think about, 5x10x49” didn’t think I was actually right

[@JustAPersonWhoComments]

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?** The problem statement is vague on an important point - did the monk know the priest's age before the conversation? I think the answer is yes, but it's not clear to me. There are two people, the priest and the monk, in this conversation. Who knows what, before the conversation? The priest knows everything, he knows the ages of the strangers and of the monk and his own age. Presumably, the monk knows his own age. And obviously the monk doesn't know the strangers' ages. But I think he must know the priest's age in order for this puzzle to be solvable.

[@thomasaquinas2600]

I took a boring course in college which broke down numbers until you reached prime numbers. This is the 1st time I've used this. I got the factors and knew the cantor was 32 easily. I differ with the 'solution' shown; the priest poses a question which presumably is unknown at the time. If they reach his age in the future? Well, that would negate the answer given us. The priest, per the sense of what he was saying, must be older than any of the others, sitting there and waiting for that crucial knowledge, about cake consumption.

[@ni3070]

Too damn good..... Solving is pure genius

[@jagmarz]

Wow, I started reading the comments here and realized I'd neglected to think about visitors with age 1. But fortunately, this doesn't change the answer. I did struggle for rather a while on what that last clue provided, but then the light turned on! As with other similar puzzles, the first clue is that there are multiple ambiguous intermediate results, and the last clue helps you choose among those. Factoring 2450 (don't forget the 1's) you get 19 possible ages of the cantor, but all but one are unique. You have to realize that in order for the last clue to be meaningful in distinguishing between those results, the statement has to not be true for all but one case. And since the only age-related claim was that the visitors are all younger than the priest, we get that the priest must not be older than at least one visitor in the disqualified group(s), and so we are left with the priest being 50.

[@MissBlueberry2000]

That was an awesome puzzle, please upload more of this kind! I didn't get the right solution though because I didn't read the instruction carefully enough - thanks for clarifying :)

[@jermyeder2262]

2 25 and 49 also work but you just brushed it aside making your own non implicated deductions.

[@althafshameel4466]

Eeee. Even the cantor's responses were very much part of the question. Infact they contained the most important information. I was thinking like a casual response the way I would have given to the priest. 😂

[@Tensquaremetreworkshop]

This is a very interesting problem- but it is totally different to the one in the thumbnail!

[@Axalion71]

Ah!!! I'm so happy! This was the first one I was able to / took the time to complete on my own!

[@Coastfog]

Is it just me or is Presh Talwalkar a badass name?

[@rahulraidas2298]

Toughest question on your channel for me

[@lawrencedoliveiro9104]

The factors of 2450 are 2×5×5×7x7. Grouping them in various ways as factors of 3 integers, I came up with 1, 49 and 50, which sum up to 100 = 50×2.

[@StarTheTripleDevil]

I tried doing it but turns out I miscounted 7+7+50 and got stuck (because I couldn't find two sums that added up to be the same thing). Then when I started watching the solution (until the sums) and noticed my mistake, I almost instantly figured out the age was 50. (Also I only considered cases where the largest age was 98 or less.)

[@StarTheTripleDevil]

And, of course, the "two times" part doesn't rule out possibilities since all of the sums are even by default (there are 3 people and exactly one of them has an even age since 2450 is divisible by 2 but not 4).

[@youssefahmed8245]

Hello@@StarTheTripleDevil could you please explain how if a number is divisible by 4 or not teaches us that from a 3 factors of a number (eg 2450) one of them only should be even if their product is to be 2450 Thanks in advance

[@StarTheTripleDevil]

@@youssefahmed8245 This is based on how many times 2 is included in the prime factors. In this case, it is exactly once. So no matter how you multiply the prime factors, you're left with exactly one even number. (2, 5, 5, 7, 7) (10, 35, 7) (2, 25, 49) etc. Multiply numbers. If they're all odd, their product will also be odd because none of the original numbers' prime factors included 2 so neither does their product's. If there are 2 or more even numbers, the result must be divisible by 4, since 2 is included in the prime factors at least twice (2^2). Now, the only way to get a number divisible by 2 but not 4 is if exactly one of the numbers is divisible by 2 and not 4, because it's only then when the prime factors of that product contain the number 2 exactly once. To multiply numbers to get something divisible by 4, you can do it by having at least 2 even numbers or one even number that's divisible by 4.

[@youssefahmed8245]

@@StarTheTripleDevil So if I want to get a number that is divisible by 2 but not 4 I need to get a number that has only one even number (factor) that is not 4 or any of its multiples , Right? Also execuse my curiosity I just wanted to know if you had learned about this matter from a proper educational academy eg school,college etc Or just you thought about it specifically just for this puzzle, I don't really know but would really like to know cuz I wasn't teached such a case in school or at least came across me once , though it is not a complex matter at all that just need a little bit of thinking about it to get to a conclusion Yes I thought about it while you were replying to me but thanks a lot anyway... again :)

[@UnimatrixOne]

@@youssefahmed8245 ????

[@benjaminmorris4962]

The real question is why are two young children with three middle-aged people, two of which are members of the clergy? 🤔

[@chinareds54]

This is not solvable unless it's explicitly stated that the cantor is not a member of Blink-182.

[@atlas7425]

I don't get it, please explain.

[@ErikBongers]

Learned something about prime factorization today.

[@jessstuart7495]

If you keep track of "half years" during the calculations then an additional solution is possible (priest = 56). cantor = 36, visitors=(2, 35, 35), priest=56 where the alternate visitor ages that are eliminated in the last step are v_alt = (3.5, 12.5, 56) and cantor = 38, visitors=(2, 25, 49), priest=56 where the alternate visitor ages that are eliminated in the last step are v_alt = (2.5, 17.5, 56)

[@nikhiljangir9379]

Why did you chosse the triplet and why did you assume priest age as 50 or 51. I am confused please tell me.

[@badam9656]

The logic in this puzzle is that they're all dead by now!! 🤣😂

[@yvessioui2716]

I like the fact that you didn't put some combinations out in relation based on high age. Nothing in the problem says when that situation happened nor it says anything about the 'condition' of that situation. It could have happen in the time of Gilgamesh or in the prime of human from the biblical creation (if someone still believe that) when and where humans can live 900 years.

[@MD-vs9ff]

In other words, it could have been a fantasy setting.

[@edgarallenhoe3518]

Except that it's specified as taking place after a "church service." Unless the word "church" is being used VERY loosely, the story must take place in the last 2000 years.

[@djarogames]

I got a different solution because I interpreted it incorrectly, but IMO what I did would have made for a better puzzle. I did the following: I thought the priest said that the sum of their ages was 2 times his (the priest's) age, instead of the cantor's. So it was unknown. The priest's age is half that of the sum of the 3 visitors. I ruled out all "impossible" ages (anything that contained 100), and then for each triplet I calculated the age of the priest. There was only 1 triplet of ages where the priest was older than all of the 3 people, which would make the comment about the birthday work.

[@ted4682]

I love it I unfortunarely forgot my riddles😢

[@christopherswanson5849]

I figured out the process, but I didn’t assemble the complete list of possible sets of ages for the visitors, so was confused when none of the sums were repeated

[@factsheet4930]

Can't believe I got this right with just prime factorying and logic 😂

[@simondando8142]

One of these days Presh (legend by the way) will ask ‘did you figure it out?’ and I won’t answer no 🤣