Brain Teasers
Justin's Marbles
Justin has a bag of marbles and nine friends. He gives Mike exactly 1/2 of the marbles. He gives Paul exactly 1/3 of the remaining marbles. He gives Dave exactly 1/4 of the remaining marbles. He gives Bert exactly 1/5 of the remaining marbles. He gives Kyle exactly 1/6 of the remaining marbles. He gives Andy 1/7 of the remaining marbles. He gives Rick 1/8 of the remaining marbles. He gives Fred 1/9 of the remaining marbles. He gives Will 1/10 of the remaining marbles. What is the smallest quantity of marbles that Justin still has in the bag?
Answer
For each friend that Justin gives marbles, you need to calculate the fraction of the original quantity of marbles that Justin keeps in his possession (keep), the fraction of the original quantity of marbles that Justin gives to that friend (give), and the prime factorization of the denominator of the fraction given to that friend (factor).Mike: keep 1 * 1/2 = 1/2, give 1 * 1/2 = 1/2, factor 2 = 2
Paul: keep 1/2 * 2/3 = 1/3, give 1/2 * 1/3 = 1/6, factor 6 = 2 * 3
Dave: keep 1/3 * 3/4 = 1/4, give 1/3 * 1/4 = 1/12, factor 12 = 2 * 2 * 3
Bert: keep 1/4 * 4/5 = 1/5, give 1/4 * 1/5 = 1/20, factor 20 = 2 * 2 * 5
Kyle: keep 1/5 * 5/6 = 1/6, give 1/5 * 1/6 = 1/30, factor 30 = 2 * 3 * 5
Andy: keep 1/6 * 6/7 = 1/7, give 1/6 * 1/7 = 1/42, factor 42 = 2 * 3 * 7
Rick: keep 1/7 * 7/8 = 1/8, give 1/7 * 1/8 = 1/56, factor 56 = 2 * 2 * 2 * 7
Fred: keep 1/8 * 8/9 = 1/9, give 1/8 * 1/9 = 1/72, factor 72 = 2 * 2 * 2 * 3 * 3
Will: keep 1/9 * 9/10 = 1/10, give 1/9 * 1/10 = 1/90, factor 90 = 2 * 3 * 3 * 5
The lowest common multiple of the denominators of the "give" fractions is:
2 * 2 * 2 * 3 * 3 * 5 * 7 = 2520
The quantity of marbles that Justin still possesses is:
2520 * 1/10 = 252
Justin starts with 2520 marbles.
Mike gets 1/2 of 2520, or 1260. Justin still has 1/2 of 2520, or 1260.
Paul gets 1/3 of 1260, or 420. Justin still has 2/3 of 1260, or 840.
Dave gets 1/4 of 840, or 210. Justin still has 3/4 of 840, or 630.
Bert gets 1/5 of 630, or 126. Justin still has 4/5 of 630, or 504.
Kyle gets 1/6 of 504, or 84. Justin still has 5/6 of 504, or 420.
Andy gets 1/7 of 420, or 60. Justin still has 6/7 of 420, or 360.
Rick gets 1/8 of 360, or 45. Justin still has 7/8 of 360, or 315.
Fred gets 1/9 of 315, or 35. Justin still has 8/9 of 315, or 280.
Will gets 1/10 of 280, or 28. Justin still has 9/10 of 280, or 252.
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Comments
This was great. I opened Excel and did a semi-brute force, looking for the path with no decimals. I got the right answer.
Nice! - On the last step Justin gives still has (9/10)*(8/9)*(7/*...*(1/2) of his original stash. This reduces to (1/10) through cancellation.
His original stash is (5*2)*(3*3)*(2*2)*7. This can be found by factoring all the denominators until you get to the largest prime (7), and eliminating any repeats. (10 = 5*2, and 8 = 2*2*2). Since we've already accounted for one of the 2's in 8, we only need to add two more 2's to the product. Continuing to factor beyond the largest prime is redundant (I think - i'm probably wrong, but it seems to apply in this case). Anyway, he starts with 2520, so 1/10th of that is 252!!
His original stash is (5*2)*(3*3)*(2*2)*7. This can be found by factoring all the denominators until you get to the largest prime (7), and eliminating any repeats. (10 = 5*2, and 8 = 2*2*2). Since we've already accounted for one of the 2's in 8, we only need to add two more 2's to the product. Continuing to factor beyond the largest prime is redundant (I think - i'm probably wrong, but it seems to apply in this case). Anyway, he starts with 2520, so 1/10th of that is 252!!
wow -- don't put an 8 next to a close paren in comments.
pzakielarz is wrong about stopping at 7. In seeking the lowest common denominator and, in this case, the smallest possible number of marbles involved, every denominator must be considered. Stepping through the numbers 2 through 10, the list of factors is built this way:
2, 3, 2, 5, nc, 7, 2, 3, nc.
No change (nc) occurs (appears) only when the denominator is the product of primes listed earlier and in sufficient number to not need repetition.
2, 3, 2, 5, nc, 7, 2, 3, nc.
No change (nc) occurs (appears) only when the denominator is the product of primes listed earlier and in sufficient number to not need repetition.
I agree that my earlier post is incorrect (I wasn't sure) -- It just happened to work out this time that you can stop at 7.
Imagine Justin had 11 friends, whom he gave (1/12,1/11...etc). I clearly couldn't stop at the largest prime (11).
Thanks for helping me clarify this!
Imagine Justin had 11 friends, whom he gave (1/12,1/11...etc). I clearly couldn't stop at the largest prime (11).
Thanks for helping me clarify this!
If you stop at 7, you must have slipped and taken both factors of 6. If you work carefully, you would note that a 2 and a 3 are already in the list of factors and you have to examine #8 to realize 2 appears as factor three times. Likewise, #9 has 3 as factor twice.
Nice puzzle!
I did it similarly, but working backward. To give 1/10 the number of marbles at that point must have the factors 2 & 5. To give 1/9 it must have factors 3 & 3 plus the 2 & 5 needed for the next step. To give 1/8 it must have 2, 2, 2 so the factors are 2, 2, 2, 3, 3, 5 at this point. For 1/7 the factors are 2, 2, 2, 3, 3, 5, 7. At this point all of the numbers from 6 to 2 can be made from the existing factors so no new factors need to be added.
(2 x 2 x 2 x 3 x 3 x 5 x 7) x
(1/2 x 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9 x 9/10)
= 252
I did it similarly, but working backward. To give 1/10 the number of marbles at that point must have the factors 2 & 5. To give 1/9 it must have factors 3 & 3 plus the 2 & 5 needed for the next step. To give 1/8 it must have 2, 2, 2 so the factors are 2, 2, 2, 3, 3, 5 at this point. For 1/7 the factors are 2, 2, 2, 3, 3, 5, 7. At this point all of the numbers from 6 to 2 can be made from the existing factors so no new factors need to be added.
(2 x 2 x 2 x 3 x 3 x 5 x 7) x
(1/2 x 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9 x 9/10)
= 252
I see pzakielarz did it the same way by working backward. However, you can't stop at the largest prime.
Where 1/n is the number of marbles given to the last person you would need to check all the numbers from n to floor[(n/2)+1] (floor[x] = x rounded down to the next integer) in order to guarantee you find all prime factors. This can be seen by setting n = 2p - 1 where p is a prime number.
In this case it means you need to check until at least 10/2 + 1 = 6.
Where 1/n is the number of marbles given to the last person you would need to check all the numbers from n to floor[(n/2)+1] (floor[x] = x rounded down to the next integer) in order to guarantee you find all prime factors. This can be seen by setting n = 2p - 1 where p is a prime number.
In this case it means you need to check until at least 10/2 + 1 = 6.
This one took me a while, but I got it Very fun.
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