Family Members
Four male members of the Smith family went racing one afternoon at a local track. The racetrack was short and each one them had a go at breaking the record time of 19 seconds for one lap. None of them managed it but on a positive note none of them embarrassed themselves by taking longer than a minute to complete a lap. When the Stewart collection the lap times (to the nearest second) he revealed that there was at least five seconds gap between each racer. He also gave out a series of clues to see if they could guess their lap times:
He told Bill that he his time was exactly two thirds of Alex’s.
He told Calvin that two racers were precisely 20 seconds apart and that the 45-year old was not one of them
He told Derek that one of the family members older than him was exactly 30 seconds quicker than him.
He told the only member of the family who was not old enough yet to buy an alcoholic drink, that his time was double his age.
He told the 30 year old, that his lap time was not a prime number but one of the other three racers lap time was.
He told the 25 year old that he does not share either digit of his number (i.e. lap time) with another family member.
Can you work out which family member had which time?
-- Antn'y Jacobs
He told Bill that he his time was exactly two thirds of Alex’s.
He told Calvin that two racers were precisely 20 seconds apart and that the 45-year old was not one of them
He told Derek that one of the family members older than him was exactly 30 seconds quicker than him.
He told the only member of the family who was not old enough yet to buy an alcoholic drink, that his time was double his age.
He told the 30 year old, that his lap time was not a prime number but one of the other three racers lap time was.
He told the 25 year old that he does not share either digit of his number (i.e. lap time) with another family member.
Can you work out which family member had which time?
-- Antn'y Jacobs
Labels: SharedPuzzle
posted by Rajesh Lal at
9:09 AM
18 Comments:
NOTES: i) For the clue 4, take legal drinking age as 21
ii) For clue 6 interpret as if the 25-year old has a time 15, then no other family member can have the times 10,11,12,13,14,15,16,17,18,19,21,25,31,35,41,45,50,51,52,53,54,55,56,57,58,59 – of course the 25 year old couldn’t have had a time of 15 seconds as non family member was quicker than 20 but this is just an example ;)
Hi Antn'y. I expect Cam or Ragknot (where is he when he's needed) will have a bash at your diophantine extravaganza.
Hi Antn'y ... I need clarification on one of the clues:
"He told Calvin that two racers were precisely 20 seconds apart and that the 45-year old was not one of them"
Perfectly clear that two racers were 20 seconds apart
But... is he naming Calvin as the 45-year old, who is not one of them ... or simple stating that another racer 45-years old was not one of the ones 20 seconds apart?
He is simply stating that the 45-year old one was not one of the guys 20 seconds apart.
Calvin may or may not be the 45 year old. Sorry for the confusion.
Thanks ...
Hmmmm ... gave it a shot, but struck out:
Alex - 45yrs. - ?
Bill - 30 yrs.- ?
Calvin - 20yrs. - 40 sec.
Dereck - 25yrs. - ?
I'm pretty sure there's an error, because I ran into conflicts, and could not complete it.
Family Members
Given:
-4 members
-each time 5 seconds apart
-times are >=20 and <=60
-B=2/3*A
-Two racers 20 seconds apart (not 45 year ould)
-Time of a racer older than Derek is D-30
-Youngest racers time is double his age, but he is <21
-One racer has a prime number,( not the 30 year old)
-25 year old does not share his number digits
since time are 20-60 D must have a time, from 20+30=50 to 60.
since the youngest racer has a time double his age but he is <21, his time must be 20 to 2*20=40, and his time must be an even number (can not prime)
D can not be 59 as 59-30=20 (59,29 both prime) or 53 as 5330=23 (53,23 both prime)
As such, D and D-30, yield no prime possibilities
and 2/3 a prime number >3 cannot yield an integer. nor can it yield a prime (number must be even).
2/3 number can range from 20 to 60*2/3=40
Since subtracting 30 is 10*3 it won't change the mod 3 of a number
Two possible case number is 2/3 of D or D-30, or number is 2/3 of youngest. (i.e D-30=2/3 youngest)
Assume number is 2/3 of D or D-30, then D must be divisible by 3
D=60,57,54,51
And D-20 or D-30+20 (D-10) = prime number
D=60 , D-20=40, D-10=50 no primes
D=57 , D-20=37, D-10=47 , 37 and 47 are prime
D=54 , D-20=34, D-10=44 no primes
D=51 , D-20=31, D-10=41, 31 and 41 are prime
So for D=57
We have : 57,27,(37 or 47) and (2/3*57 or 2/3*27)
2/3*27=18 is <20 so discard, 2/3*57=38
57,27,(37 or 47),38
fo one number to not share any digits with the other 3, we must choose 47 as the prime
so: 57,27,47,38 is a possible candidate
Now assume number is 2/3 of youngest. (i.e D-30=2/3 youngest)
D,D-30 yield no prime possibilities, youngest can't be prime, so find all prime possibilites for D-20,D-30+10.
only
D=51 , D-20=31, D-10=41, 31 and 41 are prime
D=57 , D-20=37, D-10=47 , 37 and 47 are prime
yield possibilities
if D-30 is 2/3 younger then younger is 1.5*(D-30)
So for D=51 D-30=21,
1.5*21=31.5 not valid
so for D=57, D-30=27
1.5*27=40.5 not valid
so: no valid solutions for D-30=2/3 youngest
Now analyze: 27,38,47,57
We let 57 belong to Derek at the start
only even number belongs to youngest: 38
but the time is double his age, so he must be 19. But it is the 25 year old that doesn't share his digits with another. (conflict 1)
38=2/3*57, but the time was to be 2/3 of Alex's not Derek's (conflict 2)
.......No valid solutions !
Possibly made a mistake here, but I can't see it.
Answer:
One cannot work out which family member had which time. No valid solutions exist.
Cam
Missing chunck of analysi re: D=51 below
D=51 , D-20=31, D-10=41, 31 and 41 are prime
So for D=51
We have : 51,21,(31 or 41) and (2/3*51 or 2/3*21)
2/3*21=14 is <20 so discard, 2/3*51=34
51,21,(31 or 41),34
both 31 and 41 share digits of other numbers, invalid solution
Cam
Your along the right lines Anonymous but have missed out at least one case.
How long am I supposed to wait until I post an answer? I may also right up a walkthrough of the solution as well.
Antn'y, you can post the solution whenever you like ... maybe a good way to decide is to wait until no one else is posting and no one has the correct solution.
I attempted it 3 times and ran into conflicts.
I'm ready to see the answer, but some others may not be.
Go ahead and post the solution. Hold off on the walkthrough until we can mull over the answer.
Cam
I wasn't going to try 'cos a lazy and that looked likely to be hard work.
I just ran a "brute force" program
testing each members time from 19 to 60 for each, (using 4 for next loops)and confirmed that 27,38,47,57 is the only set of numbers where:
-all numbers are at least 5 apart
-at least one number is 2/3 of the other
-at least one number is 20 minus another
-at least one number is 30 minus another
-at least one number is <=40 (for double our less than 21 year old)
-at least one number is prime
-at least one number has digits that do not match the others
Perhaps I botched the program. Otherwise one of the conditions I have listed must be invalid, for other solutions to exist.
I suspect that the solution will somehow bypass at least one of these constraints....
Cam
It should go without saying that an answer involving fractions would clearly be dirty pool
Cam
I spotted major flaws in my logic and my program.....
Back to the drawing board
Hold off on solution please...
Cam
Family Members (PART1)
Given:
-4 members
-each time 5 seconds apart
-times are >=19 and <=60
-B=2/3*A
-Two racers 20 seconds apart (not 45 year ould)
-Time of a racer older than Derek is D-30
-Youngest racers time is double his age, but he is <21
-One racer has a prime number,( not the 30 year old)
-25 year old does not share his number digits
since time are 19-60 D must have a time, from 19+30=49 to 60.
since the youngest racer has a time double his age but he is <21, his time must be 20 to 2*20=40, and his time must be an even number (can not prime)
D can not be 59 as 59-30=29 (59,29 both prime) or 53 as 53-30=23 (53,23 both prime)
As such, D and D-30, yield only one prime possibility D=49 D-30=49-30=19
2/3 a prime number >3 cannot yield an integer. nor can it yield a prime (number must be even).
2/3 number can range from 19 to 60*2/3=40
Since subtracting 30 is 10*3 it won't change the mod 3 of a number
4 numbers:
-as concluded previously, one number is D, which cannot be prime, it cannot be 2/3 any number as the minimum 49*1.5=73.5>60 is too large
-another number is D-30, it is not prime, unless D=49 and D-30=19,it is not 2/3 of D as x-30=2/3*x,x=3*30=90>60 is too large. It can be 2/3 of another non-prime number.
-another number is prime (already counted if D-30=19). It cannot be 2/3 of another number, nor can any number be 2/3 of it.
-another number is: 2/3 of (D or D-30), or 3/2*(D or D-30),or if D=49,D-30=19 2/3 of another unknown number
Possible cases for finding 2/3 number:
- number is 2/3 of D or D-30
-number is 3/2 of D or D-30
-if one number is 19, find 2 numbers such that one is 2/3 of the other, not violating other rules
Assume:
- number is 2/3 of D or D-30, then D must be divisible by 3
D=60,57,54,51
Two numbers left, one must be prime, the other 2/3.
And D-20 or D-30+20 (D-10) = prime number if D-20 != 2/3*D
D=60 , D-20=40, D-10=50, 40=2/3*60, so search for valid primes for
60,30,40,x (23,31,37,41,43,47,53) have non-unique digits (29,59) are <5 away from other digits. Leaving only 19.
19,30,40,60 is a candidate.
D=57 , D-20=37, D-10=47 , 37 and 47 are prime
D=54 , D-20=34, D-10=44 no primes
D=51 , D-20=31, D-10=41, 31 and 41 are prime
Test D=57
We have : 57,27,(37 or 47) and (2/3*57 or 2/3*27)
2/3*27=18 is <20 so discard, 2/3*57=38
57,27,(37 or 47),38
for one number to not share any digits with the other 3, we must choose 47 as the prime
so: 27,38,47,57 is a possible candidate
Test D=51
We have : 51,21,(31 or 41) and (2/3*51 or 2/3*21)
2/3*21=14 is <20 so discard, 2/3*51=34
51,21,(31 or 41),34
both 31 and 41 share digits of other numbers, invalid solution
Now Assume: number is 3/2 of D,D-30
-D is from 49 to 60, 49*1.5=73.5>60, so number can't be 3/2 of D
-D-30 is from 19 to 30
-other number, call it O, ranges from 28.5 to 45, and must be mod 3=0
-O=45,42,39,36,33,30
-O=45,D-30=30,D=60, no prime 20 apart from the others fits
-O=42,D-30=28,D=58, no prime 20 apart from the others fits
-O=39,D-30=26,D=56, 19,59 are prime and 20 apart from 39. But then no number has digits not shared.
-O=36,D-30=24,D=54, no prime 20 apart from the others fits
-O=33,D-30=22,D=52, 53 is prime and 20 apart from 33. But then no number has digits not shared.
-O=30,D-30=20,D=50,already have 20 apart number taken care of. Now need to select prime that satisfies, >=5 apart and no matching number rules. 19,23,29,31,47,53 fail closer than 5 condition. leaving 37,41,43,59. Only 41 passes no shared digit condition.
so:20,30,41,50 is a possible candidate
Cam 2b cont'd
(PART 2)
Now Assume: D=49 D-30=19,
-find 2 numbers such that one is 2/3 of the other, not violating other rules
-One of numbers must be 20 apart from 49 or 19 OR
-the numbers are 20 apart from eachother
Assume: one is 20 apart from 49,19
This leaves 49-20=29,49+20=69>60,19-20=9<19,19+20=39.
Need to test 29,39
19,29,49,x one number must be 2/3 of another. But no number is even so they can't be 2/3 of x. And none of the number are mod 3=0 so x can not be 2/3 of the number. 29 is invalid here.
19,39,49,x. no number is even but 39 is mod3=0, so x must be 2/3 of 39=26
19,26,39,49 obeys the rules and is a candidate.
Assume: Numbers are 20 apart from each other
x-20=2/3*x
x=3*20=60
X-20=40
19,40,49,60 numbers share at least one digit.
Candidates:
19,26,39,49
19,30,40,60
20,30,41,50
27,38,47,57
Analyze: 19,26,39,49
Youngest must be even:26
25 year old shares no digits: 26
conflict, invalid
Analyze: 19,30,40,60
25 year old must share no digits:19
B=2/3*A, B=40, A=60
D is 30 seconds slower.... 60=30+30, but this is A already
conflict,invalid
Analyze: 20,30,41,50
25 year old shares no digits:41
B=2/3*A, B=20, A=30
D is 30 second slower. D=30+20=50, D=50
Alex=30 Youngest (15)
Bill=20 45 year old
Calvin=41 25 year old
Derek=50 30 year old
-Analyze: 27,38,47,57
25 year old shares no digits: 38
Youngest must have even number: 38
conflict, invalid
Answer:
The only valid solution is:
Alex=30, Youngest (15)
Bill=20, 45 year old
Calvin=41, 25 year old
Derek=50, 30 year old
Cam
After massive levels of fail, I think I may have actually finally scored a win.
This time I also explored the possibility of 19 as a time, where they matched the record,but did not break it. This turned out to be a significant chunk of extra work for one little number. Alas, no valid solutions there.
After doing the hand solution I fixed my program, which was previously only accepting 47 as a prime number, and confirmed the 4 candidates.
Fairly dismal performance on my part on this one. I think I will sit in the corner with my dunce cap, until the next pot of coffee is ready.....
Cam
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